Mike_bb's latest activity

I don't blindly trust authorities. Limits are used when we want to check whether function continuous at the point or not but limits...

This approach doesn't explain why ##a^0=1##.

I wrote about Leibniz's infinitesimals. Such infinitesimals are still widely used in Physics.

That's a great point!

Infinitesimals like ##dx=1/N## (Leibniz's) work in Physics but most of mathematicians refuse them because such infinitesimals don't have...

This may not be germane to the subject of this thread, but epsilon-delta proofs have been the bane of many calculus students, myself...

We can use empty product to define ##a^0=1##.

Trying to understand why ##a^0 = 1## by using the positive-integer interpretation ##a \times a \times a \times \cdots## is the wrong...

For ##a^0=1## we have symmetry around ##a^0##: ##\frac{1}{a^{-1}} = a## ##\frac{a^1}{a^{-1}} = a \cdot a## For ##a^0\neq1## we don't...

Ah, that's what you're talking about. :smile:

Well, the first one is ##a\cdot a\cdot a\cdot a\cdot a##. So what's the third one? Or alternatively, note that the third one is...

I don't understand again. ##a\cdot a\cdot a\cdot a^2## ##a\cdot a\cdot a\cdot a^1## ##a\cdot a\cdot a\cdot a^0## What is the next step?

When you wrote post#2 you probably meant that "to multiply number by no a's" = ##a*a*a*a^0##. I can't understand how did you infer that...

I can't understand how does it work if ##x*x*x*x^0##? x³ = 1 × x × x × x x² = 1 × x × x x¹ = 1 × x x⁰ = 1 (we multiply by x zero...

What's the difference? Multiplication is commutative, you can write 1 wherever you want. My thought: you are overcomplicating things.