# What is the flaw in this fake proof of $1 = 0$ proof?

Someone told me this statement $$1^0=1\\ 1^1=1 \\ \implies 1^0=1^1 \implies 0=1$$ Then told that $g(x) := a^x ,a>0$ is strictly increasing and $1>0$ so we we can conclude $1=0$.

Now my answer: I told him that $f(x)=1^x$ is not one to one function so we can not say, $\text{if } 1^a=1^b \implies a=b$.

My question : Is my reason ok ? Can we make better description ?

• But $a^x$ is not even strictly increasing for $a\in [0,1]$. It is strictly decreasing for $a\in (0,1)$, the values zero and one are special and produce constant maps. (Setting aside $0^0$). So yes, you are right, the statement is just wrong. – Mathematician 42 Sep 7 '17 at 6:30
• Then told that a^x, a>0 is strictly increasing Don't trust all you're told without verifying. For example $a = 1/2 \gt 0$ but $a^x$ is decreasing on $\mathbb{R}$. And $a=1$ gives a constant function, of course. – dxiv Sep 7 '17 at 6:32
• @dxiv :I said every thing he told me . But thank you for clarifying . – Khosrotash Sep 7 '17 at 6:34
• @Khosrotash You did well to dispute the conclusion for $a=1$. The point of my comment was that the entire "premise" a^x, a>0 is strictly increasing is false. – dxiv Sep 7 '17 at 6:36

His statement $a^x; a> 0$ is strictly increasing is ludicrously false. If $0 < a < 1$ then $a^x$ is strictly decreasing. If $a = 1$ then $a^x = 1^x$ is constant. And it is only $a > 1$ that $a^x$ is strictly increasing.

To beat the horse senseless,

If $0 < a < 1$ then $a^0 > a^1$ and so we can conclude because $a^x$ is strictly monotonic that $0\ne 1$. And as $a^x$ is decreasing we can conclude $0 < 1$. Although our reason is utterly circular. $0< 1 \implies a^0 > a^1 \implies 0 < 1$.

If $a = 1$ then $1^0 = 1^1$ and as $a^x$ is constant we can conclude that .... $0, 1 \in \mathbb R$. That's all we can conclude.

If $a > 0$ and $a^0 < a^1$ and as $a^x$ is strictly monotonic $0 \ne 1$. And as $a^x$ is increasing, $0 < 1$.

All are perfectly consistant... and trivial... and useless.... but consistent.

Exactly right, your friend used a function that is not a one to one function, and for those functions, we cannot claim $f(a)=f(b)\implies a=b$. In order to do that, your friend would first have to demonstrate that the function he is using is one to one, but what he in fact did was show that it is not one to one, since he discovered two values $a,b$ such that $a\neq b$ and $f(a)=f(b)$.

So, what your friend thought he did:

Take this one-to-one function, and voila, I have shown that $1=0$

What he really did:

Take this function, and assume that it is one-to-one. Then it follows that $1=0$.

and this is just one step away from a proof by contradiction that the function is actually not one-to-one.

The actual case is:

$a^x$ is not strictly increasing for $a>0$. It is strictly increasing for $a>1$, and is constant for $a=1$ (and actually strictly decreasing for $0<a<1$!)

You are right. A litte bit more accurate:

Let $f_a(x)=a^x$ for $a>0$.

If $a>1$, then $f_a$ is strictly increasing,

If $a<1$, then $f_a$ is strictly decreasing,

If $a=1$, then $f_a$ is constant !

Generally speaking, $$f(a)=f(b)\implies a=b$$ is true for a strictly increasing function.

But $1^x$ is not one ! (The statement "$a^x,a>0$ is strictly increasing" is wrong.)

• It's true also for a strictly decreasing function as well, but yeah – 5xum Sep 7 '17 at 8:05

$1^x = 1$ for all $x$ and is therefore not strictly increasing as claimed.

That's literally all there is to this.