Mike_bb's latest activity

This seems logical to me. But after rereading, your second post confused me: How is it possible "to multiply number by no a's"? I...

Equation 1 has no solution: ##2^x > 0## for all ##x \in \mathbb{R}##. The only "value" satisfying ##2^x = 0## is ##x = -\infty##, which...

Hello! I decided to solve following equations: 1) Let ##2^x=0##: ##2^x=2^{2x}## and then ##x=0##. ##2^0=0## 2) Let ##2^x=1##...

Because the ones that haven't been proved are either unknown or too difficult. You could say the same about learning to program. Or...

Also, so that we learn how to proceed in order to establish that something is correct. This is essential if the game is to acquire new...

So that we understand the proofs, and can satisfy ourselves that they are indeed correct proofs. We don't take other people's word for...

Hello! If theorems always work and they are true statements then why should we need to re-prove theorems that have been proved already...

We can also work backwards: $$a^3 \times a^{-1} = a^2$$$$a^2 \times a^{-1} = a^1 = a$$$$a \times a^{-1} = 1 (= a^0)$$$$1 \times a^{-1} =...

Exactly. So you want "multiply by no ##a##s" to be "multiply by 1".

It should stay the same.

If you have a number and you choose not to multiply it by anything, should it turn into a zero? Or should it stay the same?

I guess the rationale is this: Start with ##0##, add ##a## you get ##a##. Add ##a## again you get ##2a##. In this case you start with...

Thx. But I'm interested in the explanation that mentions empty product.

Why not define ##a^0 = 0##? And take it from there. For example: The usual index rule for all integers ##m, n## is: $$x^mx^n = x^{m +...

Could you explain how does it work? How is it possible to combine an empty product with original factors (see below)? Thanks. "The...