What allows us to use imaginary numbers?Do complex numbers really exist?Why don't we define “imaginary” numbers for every “impossibility”?Where is my error in trying to find Pythagorean triples with matching areas?Manhattan distance problem with infinite zig zagsUse comma to split numbersImaginary NumbersWhat proof is there that all numbers behave as the numbers we commonly use?Imaginary number$pi$ in imaginary numbers?What exactly are the numbers we use everyday?Does Collatz Rule $3x+5$ have a bias for certain loops, or are my results faulty?Imaginary numbers - Is there a shortcut?Why are imaginary numbers defined the way they are?
Are there really no countries that protect Freedom of Speech as the United States does?
Escape Velocity - Won't the orbital path just become larger with higher initial velocity?
Unconventional examples of mathematical modelling
Can someone with Extra Attack do a Commander Strike BEFORE he throws a net?
If a person claims to know anything could it be disproven by saying 'prove that we are not in a simulation'?
The more + the + comparative degree
When was "Fredo" an insult to Italian-Americans?
What is the opposite of "hunger level"?
Solving a maximum minimum problem
Airline power sockets shut down when I plug my computer in. How can I avoid that?
Locked room poison mystery!
How can I shoot a bow using Strength instead of Dexterity?
Did Michelle Obama have a staff of 23; and Melania have a staff of 4?
How do I call a 6-digit Australian phone number with a US-based mobile phone?
Help, I cannot decide when to start the story
Would the USA be eligible to join the European Union?
Graphs for which a calculus student can reasonably compute the arclength
How can I communicate my issues with a potential date's pushy behavior?
Number in overlapping range
When did Bilbo and Frodo learn that Gandalf was a Maia?
How can I find files in directories listed in a file?
What would cause a nuclear power plant to break down after 2000 years, but not sooner?
Sum Square Difference, which way is more Pythonic?
How can I find an old paper when the usual methods fail?
What allows us to use imaginary numbers?
Do complex numbers really exist?Why don't we define “imaginary” numbers for every “impossibility”?Where is my error in trying to find Pythagorean triples with matching areas?Manhattan distance problem with infinite zig zagsUse comma to split numbersImaginary NumbersWhat proof is there that all numbers behave as the numbers we commonly use?Imaginary number$pi$ in imaginary numbers?What exactly are the numbers we use everyday?Does Collatz Rule $3x+5$ have a bias for certain loops, or are my results faulty?Imaginary numbers - Is there a shortcut?Why are imaginary numbers defined the way they are?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
number-theory
$endgroup$
|
show 12 more comments
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
number-theory
$endgroup$
1
$begingroup$
Examining the history is enlightening. See here for an interesting snippet.
$endgroup$
– Bill Dubuque
8 hours ago
3
$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$beginpmatrixa & -b \ b & aendpmatrix?$$
$endgroup$
– Bungo
8 hours ago
7
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
7 hours ago
6
$begingroup$
Interestingly, one can introduce $sqrt-1$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
7 hours ago
3
$begingroup$
Also, in the beginning there were only rational numbers. The discovery that $sqrt2$ was irrational apparently cost the discoverer his life.
$endgroup$
– Lee Mosher
7 hours ago
|
show 12 more comments
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
number-theory
$endgroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
In the beginning, when there were just reals, these operations were defined for them. Then, i was created, literally a number whose value is undefined, like e.g. one divided by zero is undefined.
Does anyone know about how mathematical operations' ranges and domains were expanded to include imaginaries?
EDIT: An interesting comment notes a first use of complex numbers where,
those values would cancel in the end.
But can I refute that with, "from an inconsistency, anything is provable"?
A corollary question: Could I define a new number z which is 1/0 and simply begin using it? Seems ludicrous.
number-theory
number-theory
edited 7 hours ago
Dogweather
asked 8 hours ago
DogweatherDogweather
1414 bronze badges
1414 bronze badges
1
$begingroup$
Examining the history is enlightening. See here for an interesting snippet.
$endgroup$
– Bill Dubuque
8 hours ago
3
$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$beginpmatrixa & -b \ b & aendpmatrix?$$
$endgroup$
– Bungo
8 hours ago
7
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
7 hours ago
6
$begingroup$
Interestingly, one can introduce $sqrt-1$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
7 hours ago
3
$begingroup$
Also, in the beginning there were only rational numbers. The discovery that $sqrt2$ was irrational apparently cost the discoverer his life.
$endgroup$
– Lee Mosher
7 hours ago
|
show 12 more comments
1
$begingroup$
Examining the history is enlightening. See here for an interesting snippet.
$endgroup$
– Bill Dubuque
8 hours ago
3
$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$beginpmatrixa & -b \ b & aendpmatrix?$$
$endgroup$
– Bungo
8 hours ago
7
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
7 hours ago
6
$begingroup$
Interestingly, one can introduce $sqrt-1$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
7 hours ago
3
$begingroup$
Also, in the beginning there were only rational numbers. The discovery that $sqrt2$ was irrational apparently cost the discoverer his life.
$endgroup$
– Lee Mosher
7 hours ago
1
1
$begingroup$
Examining the history is enlightening. See here for an interesting snippet.
$endgroup$
– Bill Dubuque
8 hours ago
$begingroup$
Examining the history is enlightening. See here for an interesting snippet.
$endgroup$
– Bill Dubuque
8 hours ago
3
3
$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$beginpmatrixa & -b \ b & aendpmatrix?$$
$endgroup$
– Bungo
8 hours ago
$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$beginpmatrixa & -b \ b & aendpmatrix?$$
$endgroup$
– Bungo
8 hours ago
7
7
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
7 hours ago
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
7 hours ago
6
6
$begingroup$
Interestingly, one can introduce $sqrt-1$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
7 hours ago
$begingroup$
Interestingly, one can introduce $sqrt-1$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
7 hours ago
3
3
$begingroup$
Also, in the beginning there were only rational numbers. The discovery that $sqrt2$ was irrational apparently cost the discoverer his life.
$endgroup$
– Lee Mosher
7 hours ago
$begingroup$
Also, in the beginning there were only rational numbers. The discovery that $sqrt2$ was irrational apparently cost the discoverer his life.
$endgroup$
– Lee Mosher
7 hours ago
|
show 12 more comments
4 Answers
4
active
oldest
votes
$begingroup$
The necessary and sufficient axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
$endgroup$
1
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
5 hours ago
add a comment |
$begingroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
$endgroup$
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$beginalign
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) endalignqquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
$endgroup$
add a comment |
$begingroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3324375%2fwhat-allows-us-to-use-imaginary-numbers%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The necessary and sufficient axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
$endgroup$
1
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
5 hours ago
add a comment |
$begingroup$
The necessary and sufficient axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
$endgroup$
1
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
5 hours ago
add a comment |
$begingroup$
The necessary and sufficient axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
$endgroup$
The necessary and sufficient axioms to define the complex numbers are
$$(a,b)+(a',b')=(a+a',b+b')$$
$$(a,b)cdot(a',b')=(aa'-bb',ab'+a'b).$$
(Subtraction and division can be defined as the inverses of addition and multiplication, as usual.)
In particular,
$$(a,b)+(0,0)=(a,b)$$ so that $(0,0)$ is the zero and
$$(a,b)cdot(1,0)=(a,b)$$ so that $(1,0)$ is the unity.
As you can check, $(a,b)$ can also be represented as the expression $a+ib$, where $i$ is a reserved symbol, with the usual computation rules on polynomials (with $i$ seen as the variable).
Using this notation,
$$(0,1)cdot(0,1)=(-1,0)$$
translates to the famous
$$i^2=-1.$$
As you can check, the "pair" representation and the "$i$" representation are completely interchangeable. $i$ has a simple geometric interpretation: in a 2D plane, multiplication by $i$ corresponds to a rotation around the origin by a quarter turn.
Note that there are absolutely no undefined operations here.
edited 6 hours ago
answered 6 hours ago
Yves DaoustYves Daoust
144k10 gold badges88 silver badges244 bronze badges
144k10 gold badges88 silver badges244 bronze badges
1
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
5 hours ago
add a comment |
1
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
5 hours ago
1
1
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
5 hours ago
$begingroup$
It's not clear what you mean by "necessary and sufficient", but it certainly is not necessary to use this construction of $,Bbb C,,$ e.g. another construction is $,Bbb R[x]/(x^2+1),,$ or via matrix rep's, or term algebra, etc.
$endgroup$
– Bill Dubuque
5 hours ago
add a comment |
$begingroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
$endgroup$
add a comment |
$begingroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
$endgroup$
add a comment |
$begingroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
$endgroup$
We can do anything we want!
Specifically, we can define anything we want (as long as our definitions don't contradict each other). So if we want to allow ourselves to use imaginary numbers, all we have to do is write something like the following:
Define a complex number as an ordered pair of the form $(a, b)$, where $a$ and $b$ are real numbers.
Define $i$ as the complex number $(0, 1)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) + (c, d)$ as $(a + c, b + d)$.
If $(a, b)$ and $(c, d)$ are complex numbers, define $(a, b) cdot (c, d)$ as $(ac - bd, ad + bc)$.
And define subtraction and division in similar ways.
Is that it? Are we done? No, there's still more that we want to do. There are a lot of useful theorems about real numbers that also apply to the complex numbers, but we don't know that they apply to the complex numbers until we prove them. For example, one very useful theorem about real numbers is:
Theorem: If $a$ and $b$ are real numbers, then $a + b = b + a$.
The analogous theorem about the complex numbers is:
Theorem (not yet proven): If $a$ and $b$ are complex numbers, then $a + b = b + a$.
This theorem is, in fact, true, but we didn't know that it was true until somebody proved it.
Once we've proved all the theorems that we want to prove, then we can say that we're "done."
(Do we have to prove these theorems? No, we don't have to if we don't want to. But without these theorems, complex numbers aren't very useful.)
As for your corollary question:
Could I define a new number $z$ which is $1/0$ and simply begin using it? Seems ludicrous.
Yes, you absolutely can! All you have to do is write:
Assume that there is a value $z$. Define $1/0$ as $z$.
And that's perfectly valid; this definition doesn't contradict any other definitions. This is completely legal, acceptable and proper.
Is that it? Are we done? Probably not; there's more we'd like to do. For example, what do you suppose $z cdot 0$ is? There are a couple of theorems here we might like to use, but we can't. Let's take a look at them:
Theorem: If $x$ is a real number, then $x cdot 0 = 0$.
Theorem: If $x$ and $y$ are real numbers, and $y ne 0$, then $(x / y) cdot y = x$.
Do you see why we can't use these theorems?
Does the first theorem tell us that $z cdot 0 = 0$? No, because we don't know that $z$ is a real number. So the first theorem doesn't apply.
How about the second theorem? We know that $z = 1/0$. Does the second theorem tell us that $(1 / 0) cdot 0 = 1$ (and therefore $z cdot 0 = 1$)? No, because the second theorem is only applicable when the denominator is not $0$, and here, the denominator is $0$. So the second theorem doesn't apply, either.
If we want, we can add more definitions and maybe make some of these theorems "work" for $z = 1/0$, just like we have a lot of theorems that "work" for the complex numbers. But when we do this, we encounter a lot of problems. Rather than dealing with these problems, most mathematical writers simply refuse to define $1/0$. (That's what the sentence "$1/0$ is undefined" means: the expression $1/0$ is an undefined expression, because we have refused to define it.)
answered 6 hours ago
Tanner SwettTanner Swett
5,24820 silver badges40 bronze badges
5,24820 silver badges40 bronze badges
add a comment |
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$beginalign
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) endalignqquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
$endgroup$
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$beginalign
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) endalignqquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
$endgroup$
add a comment |
$begingroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$beginalign
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) endalignqquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
$endgroup$
What axiom or definition says that mathematical operations like +, -, /, and * operate on imaginary numbers?
It is set theory that allows us to give a rigorous foundation for complex numbers. In particular, as explained here the axiom of pairing plays a crucial role, It allows us to construct the product set $,Bbb R^2,$ and then reduce complex arithmetic to arithmetic on pairs of reals - as Hamilton did when he gave the first rigorous construction of $,Bbb C,,$ representing $,a + b,i $ by the pair $,(a,b),$ with operations
$$beginalign
(a!+!bi) + (c!+!di) &= , a!+!c!+! (b!+!d)i\[.2em]
rightsquigarrow, (a, b) + (c, d) &= (a!+!c, b+d)\[.4em]
(a!+!bi)times (c!+!di) &= ac!-!bd!+!(ad!+!bc)i\[.2em]
rightsquigarrow, (a, b) times (c, d), &= (ac!-!bd, ad!+!bc) endalignqquadqquad$$
This reduces the consistency of $,Bbb C,$ to the consistency of $,Bbb R,$ i.e. any contradiction derived in $,Bbb C,$ would yield a contradiction on such pairs of reals, so a contradiction in $,Bbb R.$
Further, a major accomplishment of the set-theoretical construction of $,Bbb C,$ (and
algebraic structures) is that it eliminates imprecise syntax and semantics in informal approaches. The imprecise term $, a + b, i, $
is replaced it by its rigorous set-theoretic representation
$,(a,b),$ - which eliminates many ambiguities, e.g. doubts about the meaning of symbols $,i,$ and $,+,$ and $,=,$ in complex arithmetic. Such questions were rampant in the early development of complex numbers, and without set theory or any other rigorous foundation it was difficult to provide convincing precise answers. For example below is how Cauchy tried to explain them.
In analysis, we call a symbolic expression any combination of symbols or algebraic signs which means nothing by itself but which one attributes a value different from the one it should naturally be [...] Similarly, we call symbolic equations those that, taken literally and interpreted according to conventions generally established, are inaccurate or have no meaning, but from which can be deduced accurate results, by changing and altering, according to fixed rules, the equations or symbols within [...] Among the symbolic expressions and equations whose theory is of considerable importance in analysis, one distinguishes especially those that have been called imaginary. -- Cauchy, Cours d'analyse,1821, S.7.1
It's no surprise that Cauchy's peers were not persuaded by such handwaving, e.g. Hankel replied
If one were to give a critique of this reasoning, we can not
actually see where to start. There must be something "which
means nothing," or "which is assigned a different value than
it should naturally be" something that has "no sense" or is
"incorrect", coupled with another similar kind, producing
something real. There must be "algebraic signs" - are these
signs for quantities or what? as a sign must designate something
- combined with each other in a way that has "a meaning." I do
not think I'm exaggerating in calling this an unintelligible
play on words, ill-becoming of mathematics, which is proud
and rightly proud of the clarity and evidence of its concepts. $quad$-- Hankel
Hamilton's elimination of such "meaningless" symbols - in favor of pairs of reals - served as a major step forward in placing complex numbers on a foundation more amenable to his contemporaries. Although there was not yet any theory of sets in which to rigorously axiomatize the notion of pairs, they were far easier to accept naively - esp. given the already known closely associated geometric interpretation of complex numbers.
See said answer for further discussion of this and related topics (above is excerpted from there).
edited 5 hours ago
answered 5 hours ago
Bill DubuqueBill Dubuque
221k30 gold badges210 silver badges680 bronze badges
221k30 gold badges210 silver badges680 bronze badges
add a comment |
add a comment |
$begingroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
$endgroup$
add a comment |
$begingroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
$endgroup$
add a comment |
$begingroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
$endgroup$
We can always use the usual rules for doing arithmetic with real numbers on complex numbers too, provided we always substitute $-1$ for $i^2$ whenever we encounter it. It follows that once we allow the imaginary unit, we have a consistent algebra which obeys the usual laws.
Thus, we discover that if we extend all the usual laws of addition and multiplication on binomials of the form $a+ib,$ with $i^2=-1,$ everything goes on smoothly. One way to formally do this is by using ordered pairs, as William Hamilton first did, but the idea I think you need is that if we allow ourselves to calculate with complex numbers as we did with real numbers, only remembering to replace $i^2$ with $-1,$ then we have a consistent algebra. Mathematicians usually call the system so defined a field. There are many other fields apart from the ones formed by the real or complex numbers with addition and multiplication as usually defined, but that's another story. The gist is just that we can define two operations over the complex numbers similar in behavior to the usual addition and multiplication over the real numbers -- and with a certain lightness we may think of these as extensions of the usual addition and multiplication. Thus, the $+$ and $×$ of the complex system is technically different from that of the reals, but of course for symbolic economy (and also because they behave very similarly), we retain the same symbols. In general we use these symbols for the operations in any field, too, whether the members of the field are numbers or people.
answered 6 hours ago
AllawonderAllawonder
3,8898 silver badges18 bronze badges
3,8898 silver badges18 bronze badges
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3324375%2fwhat-allows-us-to-use-imaginary-numbers%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Examining the history is enlightening. See here for an interesting snippet.
$endgroup$
– Bill Dubuque
8 hours ago
3
$begingroup$
Would it bother you less simply to think of $a + bi$ as shorthand for the matrix $$beginpmatrixa & -b \ b & aendpmatrix?$$
$endgroup$
– Bungo
8 hours ago
7
$begingroup$
You may be surprised to learn that the real numbers are not really older: when $i$ was first considered, negative numbers were not generally accepted as numbers either.
$endgroup$
– Robert Israel
7 hours ago
6
$begingroup$
Interestingly, one can introduce $sqrt-1$ so that all operations 'behaves the usual way', while introducing $1/0$ will definitely break some familiar laws. Nevertheless, it can be done..
$endgroup$
– Berci
7 hours ago
3
$begingroup$
Also, in the beginning there were only rational numbers. The discovery that $sqrt2$ was irrational apparently cost the discoverer his life.
$endgroup$
– Lee Mosher
7 hours ago