# Where's the problem with a false “proof”: $\;1^0 = 1^2 \overset{?}\implies 0 = 2$

What's wrong with this:

$$\large 1^0=1^2$$

Since bases are same, therefore

$$\large 0=2$$

My thinking:

Since the function $\,f(x)=1^x\,$ is not one to one, therefore whenever $\,f(x)=f(y),\,$ $x\,$ need not be equal to $\,y$.

Question:

Is my reasoning sound?

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Apparently it is. Otherwise, we'd have $0=2$. – Hagen von Eitzen Jul 22 '14 at 12:52
A proof is not true or false, it's valid or invalid. – lhf Jul 22 '14 at 13:03
Consider: $f(x)=x\cdot 0$. Then $f(0)=f(2)$... – Thomas Andrews Jul 22 '14 at 13:04
some of the false proofs I have seen is that they often ignore basic arithmetic, order of operations and this will not be the case if $(1^0)=(1^2)$ the way a novice like me will look at the question – Gobabis Jul 22 '14 at 13:12
Almost every one of these proofs of $1=2$ works exactly the same way: they take some non-injective function $f$, observe that $f(a) = f(b)$, and then erroneously conclude that $a=b$. Division by zero is a special case of this pattern, where one observes that $0\cdot a = 0 \cdot b$ and then erroneously concludes that $a=b$. – MJD Jul 22 '14 at 15:09

Yes. That's fine reasoning.

Indeed $f(x) = 1^x = 1 \;\forall x \in \mathbb R$.

Certainly, as you note, $f$ fails to be injective, so it is NOT the case that $\forall x, y \in \mathbb R, \; f(x) = f(y) \implies x = y$.

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I'm not so sure. What if the argument was that the conjunction of "$$\large 1^1=1^2$$ and $$\large 0^1=0^2$$ is true, therefore 1=2? – Doug Spoonwood Jul 22 '14 at 15:23
@DougSpoonwood: Then the argument would fail for the same reason, since neither $x \mapsto 1^x$ nor $x \mapsto 0^x$ is injective. – Clive Newstead Jul 22 '14 at 15:45
One could alternatively say such cases are all counterexamples to the claim of it being injective. – Semiclassical Jul 22 '14 at 15:47

I like to think of these things in terms of divide-by-zero errors. In that sense, if you take the $\log$ of both sides you get $0\log 1=2\log 1$, and the argument that the bases are the same implying equality is essentially cancelling out the $\log 1$ from each side. This would be valid for $\log b$ for any $b\neq 1$, but $\log 1=0$.

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Your reasoning is fine. Your reasoning also explains why $\log_b$ is not defined for $b=1$.

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I'm not sure your reasoning is sound if you define x$^y$ as a binary function f(x,y). The bases being the same also doesn't imply anything about cancellation of the basis. One could have picked -1, or 0 as the base also, or even the imaginary unit "i" and we'd have the same sort of thing going on. For a binary function "A" we only have to have one instance of A(b, d)=A(b,c) where c does not equal d, and "b" is the base of the exponent here for something like this to happen.

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The best possible explanation could be that f(x)=1^x is not bijective hence making the inverse non existent and so x = y does not mean any thing.

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I disagree. OP already observed that $f$ is not injective, which is sufficient. Your observation that it is not bijective consists of two things: 1. It is not injective. 2. It is not surjective. But the non-surjectivity is irrelevant here, so OP's explanation is better than yours. – MJD Jul 22 '14 at 15:11