Question

i need a mathematical proof of x^0=1

Answer 1

\[x^0=x^{n-n}=\frac{x^n}{x^n}=1\]

Answer 2

as long as x does not =0

Answer 3

see my use of law of exponents! :)

Answer 4

is that an mathematical proof?

Answer 5

Proof:
Assume \[x \neq 0\]
\[x^0=x^{n-n}\]
since n-n=0
\[x^{n-n}=\frac{x^n}{x^n}\]
by law of exponents
\[\frac{x^n}{x^n}=x^nx^{-n}\]
by law of exponents
\[x^{n}x^{-n}=1\]
since \[x^n, x^{-n}\] are multiplicative inverses af each other

so we have thus proved \[x^0=1\].//

Answer 6

what about 0/0?

Answer 7

notice i said assume x does not equal 0

Answer 8

0/0 is indeterminate form
and so is 0^0

Answer 9

thanks :)

Answer 10

\[\lim_{x->0}x^0=1\]

Answer 11

its like haveing online turrets :)

Answer 12

tourette ....