A proof of 1/0 = ∞

Question

I have recently come up with a proof of 1/0 = ∞. Here is the proof:

We will first write 1 and 0 as powers of 10.

1 = 10^0

Since the limit of 10^x as x approaches -∞ is equal to 0, 0 = 10^-∞

Therefore, 1/0 = (10^0)/(10^-∞) = 10^(0-(-∞)) = 10^∞ = ∞

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Is this proof correct?

Answer 1

Well... yes and no. It depends on the context in which you set yourself.

For example, consult the operations with the infinity element of the Riemann Sphere: https://en.wikipedia.org/wiki/Riemann_sphere#Arithmetic_operations

(Note that on the Riemann Sphere, also the real projective line, you don't make a difference between $-\infty$ and $+\infty$.)

In simple $\mathbb{R}$, where infinity is not an accepted element, this doesn't work, because you're exiting your base set. It's like trying to prove that $-2$ is a natural number, since $5 - 7 = -2$. However, this is not how $\mathbb{N}$ works: $-$ is not a closed operator in $\Bbb N$.

Answer 2

Who told you that $\frac{1}{0}$ is $\infty$ !!

This expression is simply undefined as division by zero is undefined.

However following is correct

$\displaystyle\lim_{x \to 0^+}\frac{1}{x}=\infty$

As you already said in question that

$\displaystyle\lim_{x \to -\infty}10^{-x}=0$

then how can you say that $10^{-\infty}=0$. Both the expressions are different.

Answer 3

Since the limit of $10^x$ as $x$ approaches $-∞$ is equal to $0$, $0 = 10^{-∞}$

This is not correct. Infinity is not a number, so you cannot perform operations like exponentiation with it.