# How can I disprove the following statements?

1) $a \neq b , a^0 = b^0, a = b$

Usually we do $a^n = b^n \implies a= b,$ can't we do it when $n$ is zero?

2) $i = \sqrt{ -1 \over 1} = \sqrt{1 \over -1 } = {1 \over i} \implies -1 = 1$

What's wrong with the above statements?

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Please, show your statements. How you conclude $a^0=b^0\Rightarrow a=b$? – Sigur Jul 6 '12 at 14:43
just so you know, the downvotes on your question aren't because it's bad, but because it's been asked many times before. Use the search function first next time, and then if you don't understand those explanations, post another question asking for clarification on a specific point. – Robert Mastragostino Jul 6 '12 at 15:00
I don't mind downvotes, i knowingly asked this question that i would get downvotes ... i got answer for second problem ... i haven't got answer for first problem. How can I explain first problem to my younger brother who's in 9th grade. – hasExams Jul 6 '12 at 15:04
@testuser: You should mind the downvotes and improve your question accordingly. That's the way SE works. Only if they're legitimate, of course. – Gigili Jul 6 '12 at 15:13
well the question was stupid and so am i ... still i need to know the answer. – hasExams Jul 6 '12 at 15:16

What you're doing is basically this.

$$(-2)^2=4$$ $$2^2=4$$ $$\therefore -2=2$$

both $i$ and $-i$ are square roots of $-1$. Since squaring isn't a one to one function, this doesn't imply that the numbers inside the square roots are equal. This also applies to your first question. The general pattern is

$$f(a)=f(b)\implies a=b$$

only if $f(a)$ is one to one. $f(a)=a^2$ isn't one to one. $f(a)=a^0$ isn't one to one either, since everything gets turned into $1$ (except $0$, sort of, but that's more complicated). Basically you're assuming that there's a nice "undo" operation when one doesn't exist. So what you should ask yourself first is this. "I have the answer. Can I figure out where I came from with this information?" If you can't, then this idea of just getting rid of the function doesn't work. Let's take $f(a)=a^0$.

$$a^0=b^0$$

So far so good. But what does it mean to isolate $a$ and $b$ here? You have to apply an inverse function to both sides. But such a function doesn't exist (i.e. you can't work backwards), so this doesn't work.

As for explaining this to someone in $9^{th}$ grade, I would use the idea of "going backwards". That is, you can't always go backwards, you have to show that you can. Say I have identical balls in both my left and right hands. I put them into different bags. I then point to a bag and say, "which hand did that ball come from?" You can answer. Next I take the balls out again, one in each hand, and put them into the same bag. Now if I pick a ball, you can't say which hand it came from, because they ended up in the same place. You can only undo something if different things always end up in different places, which doesn't happen in either of these.

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If we have $a>0, b>0 \text { then:}$

$$a^n=b^n \implies a=b \text{ is shown by taking nth roots, thus } (a^n)^{\frac 1 n} = (b^n)^{\frac 1 n}$$

But we all know that we can't do $\frac 1 0$, so the normal proof doesn't go through. Complex numbers bring multiple values for roots (not just square roots) - and some care is needed in defining these in the wider context. You should also note that there are some subtleties involved in defining $x^y$ when $y$ is irrational.

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so we say that we are eqating roots .. and a^0 and b^0 does not have any roots?? – hasExams Jul 6 '12 at 15:23
$\,a^n=b^n\Longrightarrow a=b\,$ can be false big time even in the real numbers, as any even power can prove. Within the complex numbers every single natural power greater than 1 provides galore counterexamples, for instance $\,\left(e^{\pi i/3}\right)^3=(-1)^3 \rlap{\;\;\;/}\Longrightarrow e^{\pi i/3}=-1\,$ – DonAntonio Jul 6 '12 at 15:26
$a^0=1$ if $a>0$ - you can take roots of 1, of course, but you can't recover $a$ by doing so - that is lost information. $0^0$ is undefined, and a proper treatment of the roots of negative numbers requires complex numbers (as in $\sqrt{-1}$). I'll edit my answer to specify $a$ and $b$ are positive. – Mark Bennet Jul 6 '12 at 15:42

if we are in complex number system,then $$\sqrt{a}\sqrt b\neq\sqrt{ab}$$

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Please do learn ASAP how to write mathematics with LaTeX in this site, otherwise it becomes really hard to read really fast. – DonAntonio Jul 6 '12 at 14:47

I don't know why someone hasn't just given a counterexample to the first yet.

Suppose we have $2,3$. Then $2^0 = 3^0 = 1$, but $2 \neq 3$.

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An idea: that $\,a^0=b^0\rlap{\;\;\;/}\Longrightarrow a=b\,$ stems from the fact that $\,x^0:=1\,\,,\,0\neq x\in\Bbb R\,$ is a definition which, of course, is based in some logic:

Since we'd like to be able to write $\,a^b=e^{b\log a}\,\,,\,a,b\in \Bbb R\,$, and to get that exponential functions are continuous (i.e., nice), we should have (here comes the reason of the definition):

$$a^0=\lim_{b\to 0}a^b=\lim_{b\to 0}e^{b\log a}=e^{\lim_{b\to 0}b\log a}=e^0=1$$ where $e^0=1\,$ since the inverse function of $\,e^x\,$ is $\,\log x\,$ ,so $$e^0=y\Longleftrightarrow 0=\log y\Longleftrightarrow y=1$$

The above is, of course, not the unique explanation. There's a more elementary one: if $\,0\neq a\in \Bbb R\,$ , then any (natural) consecutive power of $\,a\,$ is obtained by multiplying the precedent one by $\,a\,$ : $$a^3=a\cdot a^2\,\,,\,a^8=a\cdot a^7\,\,,\,etc.$$

By this logic, to obtain $\,a=a^1\,$, we must multiply the precedent power, namely $\,a^0\,$ , by $\,a\,$ to get $\,a\,$...this leaves no choice but $\,a^0=1\,\,,\,a\neq 0$

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I assume in (1) there is supposed to be a $\Rightarrow$ before the last equation. In that case, it's easily disproved by giving a counterexample: Choose $a=2$, $b=3$. Then $a\neq b$, and $a^0 = b^0=1$. Note that $a^n=b^n\Rightarrow a=b$ also is wrong in general, as seen from $(-1)^2=1^2=1$ but $-1\neq 1$. Basically you'd get from $a^n=b^n$ to $a=b$ by applying $x\mapsto x^{1/n}$ on both sides, but that actually gives $(a^n)^{1/n}=(b^n)^{1/n}$. The power law you then want to apply is $(a^b)^c=a^{bc}$, because then $(a^n)^{1/n}=a^{n/n}=a$ which can only work if $n\neq 0$ (because otherwise $1/n$ is not defined. However, even where defined, that power law has some preconditions. It applies if $a>0$, or if $b$ and $c$ are both integers (and for certain integer values of $b$ and $c$ you additionally need that $a\neq 0$), and for some other special cases. This is why you can't conclude from $(-1)^2=1^2$ that $-1=1$: Neither is $-1$ positive, nor is $1/n$ an integer (and it's also not one of those other special cases I mentioned).

Also in (2), it's just a mis-application of power laws. $(a/b)^n = a^n/b^n$ is only true if either $a$ and $b$ both have the same sign (or, if complex, the same argument), or if $a\neq 0$, $b\neq 0$ and $n$ is integer. In your case, neither is the case ($\sqrt{x} = x^{1/2}$, so your exponent is not integer, and $1$ and $-1$ most definitely do not have the same sign).

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