If g is non-zero. How do I PROVE if (|g|)^0.5=g^m for some integer m, the only solutions are g=1 or -1. I know this is true just by looking at it but am struggling to write out a proof.

Question

amistre64 · Answer

perhaps assume it is false and show that such a notion is absurd

amistre64 · Answer

or try some algebra?  depends on what class this is in
 (|g|)^0.5=g^m

amistre64 · Answer

$$ (|g|)^{0.5}=g^m$$

$$ g^{0.5}=g^m$$
$$ 1=g^{m-0.5}$$

henryblah · Answer

I'm trying to do a guided proof on why R* is not a cyclic group, this is one of the steps. Just not sure how rigorous I have to be.

amistre64 · Answer

by an exponential property:  g^0 = 1  for all g which means that m-.05 is never 0, so this cannot be true for all the other numbers that are not 1 .. but that is not a proof is it, its just saying that 1 is the only hopeful solution

amistre64 · Answer

ah, group theory

henryblah · Answer

If 1 was the only answer, I would be fine with that answer there, but -1 works also, so it doesn't quite fit.

amistre64 · Answer

the absolute value bars define a distance of 1, and distance is never negative

amistre64 · Answer

bad wording ..

the absolute value bars define a distance. and distance is never negative.  |-1| = 1

henryblah · Answer

Ahh I think I see.

henryblah · Answer

Although that g on the right hand side...

henryblah · Answer

1=g^(m−0.5) implies that -1 can't be a solution does it not? because m-0.5 can't be any even number.

amistre64 · Answer

we can algrebra this into:
$$\huge 1=g^{\frac{2m-1}{2}}$$
for any nonzero real number g we have g^0 = 1 by well known exponent properties.

2m-1 is an odd number, but 0 is an even number, so this excludes any real number .... leaving g=1 or -1 as test subjects

amistre64 · Answer

not "any" real number ... but it narrows the dataset to -1 and 1

henryblah · Answer

Ok, I think i'll say that it leaves -1 and 1 to be checked, thought I somehow feel this isn't the neatest way to do it. Thanks for the help!

amistre64 · Answer

lol, i never really have the "neatest" way to approach things ... thats the trouble when you have to make things up on the fly

amistre64 · Answer

but it is something to work with :)

amistre64 · Answer

$$(-1)^m=1$$
for all even values of m