# Have I just proven $0=1$?

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:

$$r+r = r \times r = r^r$$ 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

• Why not just define $x=0$ and divide both sides by $x$ to get $1=0$? – Michael Oct 15 '18 at 2:02

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.

• I knew the explanation had to be something really basic that I was missing. Thank you – Martino Ciaramidaro Oct 14 '18 at 23:28

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.