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# Proof: Invertibility implies a unique solution to f(x)=y

Proof: Invertibility implies a unique solution to f(x)=y for all y in co-domain of f. Created by Sal Khan.

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• The function to square a number cannot be considered to have Invertiblity, if the domain is the set of real numbers, because its inverse, the square root of a number, maps to 2 values, one positive and the other negative. Is it necessary then to restrict the domain to only positive numbers to make it have Invertiblity?
• If y=2x+1 is the original function, why is (y-1)/2=x considered the inverse? From where I sit (y-1)/2=x is the same function, just rewritten. Shouldn't the inverse function be y=(x-1)/2?
• Whether you call it y(x) or x(y) or indeed p(z) doesn't matter. It's still the same function, just using different variable names. So y=(x-1)/2 and x=(y-1)/2 are the same thing.
• At , Sal states that Y is the codomain. Shouldn't Y be the range? Couldn't you have a matrix that doesn't span the entire codomain, but it is still invertible, and has unique mappings from domain->range and from range->domain?
• For any function f: X -> Y, the set Y is called the co-domain. The subset of elements in Y that are actually associated with an x in X is called the range of f. Since in this video, f is invertible, every element in Y has an associated x, so the range is actually equal to the co-domain. So yes, Y is the co-domain as well as the range of f and you can call it by either name.
I didn't fully understand your second question, but if you're asking that can any element in the co-domain can be left out for an invertible function, then the answer is no by definition.
• At , I think Sal meant to say f(x) = y instead of f(x) = x. Please correct me if I am wrong.
• Can any one explain F: X → Y is invertible ↔ ∀y∈Y :∃ unique solution to the f(x)=y in a simple layman language...language which everyone can easily relate to?!
• In laymen terms: Distinct objects in the domain map to distinct objects in the range, and every object in the range is the image of an object in the domain. The first is simply injectivity or 1-1ness, the second is surjectivity or onto-ness.
• At , why is it that if f(x)=f(a), then x has to be equal to a? If f(x)=abs(x), then both +x and -x could give the same f(x). Did I miss any point here?
• Bowen,

The theorem that you quoted in your question only holds for an invertible function, and abs(x) is not invertible. In order to be invertible, every x in the domain must map to a single y in the range, AND every y in the range must be mapped to by only a SINGLE x in the domain. Basically, the function has to be "one-to-one" and "onto." The absolute value function isn't one-to-one, because although every x in the domain maps to only one y in the range, not every y in the range is mapped to by only one x.

But IF the function is invertible, then f(x)=f(a) will imply x=a.

Does that help make it make more sense?
• Just to be clear, does invertibility imply that there is a unique y that satisfies f^-1(y)=x for every x in X as well as vice versa?

Alternatively, would this condition being false imply that f is not a function?
• Yes, part of the conditions to be invertible is that the function be ono-to-one, that means that for every element in the domain there is one (and only one) corresponding element in the co-domain, and vice-versa.

But just the failure to fulfil this condition is not enough to disqualify a function. For example, the function `f(x) = x²` is a valid function, but it's not one-to-one, since there are multiple values in the domain that map to the same value in the co-domain, so it's not invertible.
• Sal says if f(x) = y then no 2 x can have same y.
Ok fine.
But what if 2 y's have same unique x. Then you can not construct invertible function.
• If the same x maps to two different y's, then f isn't a function.
• If f(x)=f(a) =/=> x=a

cos(0) = cos(2pi) ... 0 =/= 2pi

At Sal alludes to this being true