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A long time ago I noticed that $2+2 = 2 \times 2 = 2^2$, which is pretty cool because it’s the 3 basic arithmetical operations. It then recently occurred to me to try to prove that $2$ is the only real number for which this is true. Here is what I came up with:

I start with this double equality:

$r+r = r \times r = r^r$ or rewritten as $r+r = r^2 = r^r$.

Just examining the first part of the double equality:

$$r+r=r^2,$$ I divide by $r$ and get:

$$1+1 = r \implies r=2.$$

Looking at the second part of the double equality:

$$r^2=r^r$$

I divide by $r^2$ and get:

$$1=r^{r-2}$$

Next, I take the logarithm of both sides:

$$\ln(1) = \ln\left(r^{r-2}\right) \implies 0 = (r-2)\ln(r).$$

The only numbers that make this true are $r=2$ and $r=1$, since substituting in any other real number would mean that two non-zero numbers multiplied together would make $0$, which is clearly false. Furthermore $r=1$ does not satisfy the first part of the double equality so it has to be $2$. QED.

I was pretty proud of myself for solving this (and yes I'm sure to most of you this is no big deal but I'm not a math person). However a few hours later a serious problem occurred to me. Going back to this step:

$$0=(r-2)\ln(r).$$

What if I divide both sides by $(r-2)\ln(r)$, then I get:

$$\dfrac{0}{(r-2)\ln(r)} = \dfrac{(r-2)\ln(r)}{(r-2)\ln(r)} \implies 0 = 1.$$

I can't explain this away as division by zero since it's in the numerator.
Can someone tell me what I'm doing wrong?

Thank you

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    $\begingroup$ Why not just define $x=0$ and divide both sides by $x$ to get $1=0$? $\endgroup$ – Michael Oct 15 '18 at 2:02
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If $0=(r-2)\ln(r)$, then you can't divide by $(r-2)\ln(r)$, since it is equal to $0$ so you would be dividing by $0$.

More generally, any time you divide by an expression, that step is only valid under the assumption that the expression is not equal to $0$. If the expression involves a variable, this may be true for some values of the variable.

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  • $\begingroup$ I knew the explanation had to be something really basic that I was missing. Thank you $\endgroup$ – Martino Ciaramidaro Oct 14 '18 at 23:28
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You have not proven that $0=1$

you have divided $0$ by $0$ and the result was $1$

As you know $0/0$ is not a real number because you can assign whatever value that you like to it.

For example you may argue that $0/0=5$ and cross multiply to get the correct result $5\times 0=0$

The same goes for any other number.

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