Mike_bb's latest activity

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...

In arithmetic, ##x^k## is defined recursively. The recursive step is ##x^k=x^{k-1}x##. What is the base case then, ##x^0##? For the...

$$a^x:=\exp(x\ln a),\quad \exp(x):=1+\sum_{k=1}^\infty\frac{x^k}{k!},\quad a>0$$

##a^0## can't be ##0## because if ##a^0=a\cdot 0 =0## for every ##a##, then ##0^0=0##, that is not true in mathematics ... In addiction...

Space-time curvature.

This can only represent "gravitational fields" on the chosen background. It can't, for example, represent a black hole spacetime...

This has several excellent answers and I cannot improve on them. I just have this persistent nagging voice wanting to express this...