Why does x=y > f(x)=f(y)?
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Why does x=y > f(x)=f(y)?
One of the first rules of basic algebra is that if you have an equation, you can transform both sides with the same function to produce another valid equation. However, from a foundations perspective, which axiom(s) actually imply that this is true?
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 Soupspoon
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Re: Why does x=y > f(x)=f(y)?
Not sure about official axioms, but from the head of the Wiki article, "In mathematics, a function[1] is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output."
Whether or not two nonequal x and y can still f(x)=f(y), equal x and y must, through f(), arrive at the same result, at least within the domain of mathematical functions (as opposed to computational ones, which are slippier beasts less strict about this). That result may be "undefined", or even composite (±2, 3±1, all cube roots of 1+i, etc), but would be identical or an identical set.
If it is a composite result, then further winnowing of f(x) might produce differences to further winnowing of f(y), but only with unequal winnowing functions, so I don't think that invalidates this. BICBW.
Whether or not two nonequal x and y can still f(x)=f(y), equal x and y must, through f(), arrive at the same result, at least within the domain of mathematical functions (as opposed to computational ones, which are slippier beasts less strict about this). That result may be "undefined", or even composite (±2, 3±1, all cube roots of 1+i, etc), but would be identical or an identical set.
If it is a composite result, then further winnowing of f(x) might produce differences to further winnowing of f(y), but only with unequal winnowing functions, so I don't think that invalidates this. BICBW.
Re: Why does x=y > f(x)=f(y)?
Huh, according to wikipedia that's a substitution axiom (schema) for functions as part of firstorder logic. Well, there it's described for functions with any number of parameters, but one of those is exactly x=y > f(x)=f(y).
I would've expected an axiom that simply said "if x=y you can replace x by y wherever you want" and build x=y, f(x)=f(x) > f(x)=f(y) from there, but it seems mathematicians prefer the former.
[edit]ninjaing within a minute is just unfair
I would've expected an axiom that simply said "if x=y you can replace x by y wherever you want" and build x=y, f(x)=f(x) > f(x)=f(y) from there, but it seems mathematicians prefer the former.
[edit]ninjaing within a minute is just unfair
 Soupspoon
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Re: Why does x=y > f(x)=f(y)?
Flumble wrote:[edit]ninjaing within a minute is just unfair
You clearly did more work than me, who only went on what I thought instinctively and obviously missed the next link. I think you deserve all the credit. (I spent too much time wondering whether to attempt to write out the cube roots of 1+i in full... As if that had anything to do with the price of fish^{1}.)
^{1} Or maybe "poisson distribution"?
 Eebster the Great
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Re: Why does x=y > f(x)=f(y)?
It depends on what rules of inference or transformation you are using. There are many equivalent formal systems, some of which include an axiom (or axiom schema) of substitution.
The axioms used in algebra class typically are special cases or simplifications of more general axioms used by mathematicians. For example, you may know about the "field axioms" like the associative property for addition (a+b)+c = a+(b+c), which can be simply taken as an axiom defining addition or which can be proved for a specific construction of addition as in Peano arithmetic.
The axioms used in algebra class typically are special cases or simplifications of more general axioms used by mathematicians. For example, you may know about the "field axioms" like the associative property for addition (a+b)+c = a+(b+c), which can be simply taken as an axiom defining addition or which can be proved for a specific construction of addition as in Peano arithmetic.
Re: Why does x=y > f(x)=f(y)?
Yeah it basically comes down to how we define a function.
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 doogly
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Re: Why does x=y > f(x)=f(y)?
And how you define =. If you say x=y but f(x) != f(y), then there must be something notquitex about y. Which generally is not a notion supported by =. Try ~?
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Re: Why does x=y > f(x)=f(y)?
doogly wrote:And how you define =. If you say x=y but f(x) != f(y), then there must be something notquitex about y. Which generally is not a notion supported by =. Try ~?
A little of both, I suppose, but this feels more like a property of functions than a property of equality to me. The first example that comes to mind is a function on rational numbers such that f(a/b) = a. It's perfectly reasonable to define equality of rational numbers so that 1/2 = 2/4, but f(1/2) = 1 != 2 = f(2/4), so f is not a function.
 Xenomortis
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Re: Why does x=y > f(x)=f(y)?
I don't think that makes sense.
Your function there is really f: N^{2} > N, not Q > N.
In Q, you don't differentiate 1/2 and 2/4.
Your function there is really f: N^{2} > N, not Q > N.
In Q, you don't differentiate 1/2 and 2/4.
 Eebster the Great
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Re: Why does x=y > f(x)=f(y)?
1/2 and 2/4 are, by the usual definitions, either elements of the same rational number (where rational numbers are defined as equivalence classes of ordered pairs of integers, with the second integer nonzero, and a/b means (a,b)) or representations of the same rational number (using an axiomatic definition). Technically, they are not themselves rational numbers.
You could define a function over elements of rational numbers, which is a subset of Z^{2}, in which case 1/2 != 2/4, so it is perfectly possible that f(1/2) != f(2/4) for some function f.
You could define a function over elements of rational numbers, which is a subset of Z^{2}, in which case 1/2 != 2/4, so it is perfectly possible that f(1/2) != f(2/4) for some function f.

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Re: Why does x=y > f(x)=f(y)?
Sure, I wasn't being very precise about my definitions of rational numbers, but my point was that such an f is not a function under any standard definitions of the terms. (In fact, I remember this being one of the textbook examples in my undergrad "introduction to proofs" class.) Admittedly, I did not express myself very well; I'll try again.
Doogly said that if x = y but f(x) != f(y), then there must be something notquitex about y, which is generally not a notion supported by =. I think we all agree that such an f cannot exist under the normal definitions of "equals" and "function". As a parallel, if x = y but f(x) != f(y), then there must be something about f that cares whether you called the input x or y, which is generally not a notion supported by "function". Again, I think we all agree that such an f cannot exist under the normal definitions of "equals" and "function".
The domain of a function is always (as far as wikipedia and I know) a set, so if x and y are both in the domain and x = y, then x and y are just two different names for the same element. If we pretend there is a function such that x = y and f(x) != f(y), then either x is not really equal to y, so we have a contradiction, or f cares how we name elements, which is also a contradiction.
Again, I think we all agree that my example was not a function (and not very well explained) and that no such function exists with standard terminology. I was just trying to point out that it makes as much sense to blame the definition of "function" as it does to blame the definition of "equals".
Doogly said that if x = y but f(x) != f(y), then there must be something notquitex about y, which is generally not a notion supported by =. I think we all agree that such an f cannot exist under the normal definitions of "equals" and "function". As a parallel, if x = y but f(x) != f(y), then there must be something about f that cares whether you called the input x or y, which is generally not a notion supported by "function". Again, I think we all agree that such an f cannot exist under the normal definitions of "equals" and "function".
The domain of a function is always (as far as wikipedia and I know) a set, so if x and y are both in the domain and x = y, then x and y are just two different names for the same element. If we pretend there is a function such that x = y and f(x) != f(y), then either x is not really equal to y, so we have a contradiction, or f cares how we name elements, which is also a contradiction.
Again, I think we all agree that my example was not a function (and not very well explained) and that no such function exists with standard terminology. I was just trying to point out that it makes as much sense to blame the definition of "function" as it does to blame the definition of "equals".
 Soupspoon
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Re: Why does x=y > f(x)=f(y)?
In computing terms, inequality could happen in a number of circumstances.
...you can doubtless think of others (f(whatever)=time(now())?) and most mean that even f(x)!=f(x) is entirely possible.
But none of those are valid pure mathematical functions. Most are stateful (randomiser included) and so are f(value, state), when unpacked.
Code: Select all
sub f(value) = return rand(value);
// Same input value brings forth (almost always) different 0<=return<value, or similar constraints
Code: Select all
sub f(value) = return { while value {out=pop list}; out};
//...or similar stateful returns like traversing a linkedlist circuit or static variable memory between calls
Code: Select all
sub f(valueRef) = return valueRef.address
// Assuming the compiler hasn't optomised x and y as same address whilst equal...
...you can doubtless think of others (f(whatever)=time(now())?) and most mean that even f(x)!=f(x) is entirely possible.
But none of those are valid pure mathematical functions. Most are stateful (randomiser included) and so are f(value, state), when unpacked.
 Eebster the Great
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Re: Why does x=y > f(x)=f(y)?
Equality is treated pretty loosely in most languages too, especially with 0 and 0, for instance. A function IsNegative(x) that returns true for 0 and false for 0 is easy to define, but 0==0 returns true.
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