Recent content by Jamesandthegi

Yes, but that doesn't prove that it's measurable. To prove its so, you have to write it as A U B, A borel and B null. Then the measure part I get.

Ok, I figured out A, but I'm not sure on B. Any help?

I am trying to learn about Lebesgue measure. One of the questions I couldn't solve is this; Show that the following sets are Lebesgue measurable and determine their measure A = {x in [0,1) : the nth digit in the decimal expansion is equal to 7} B = {x in [0,1) : all but finitely many...

i do not get it. what is b? what i meants was say a^2 = 1 then a^2*a^2*...*a^2 = 1 but that' can not happen because we knows a^54 = -1 i don't see what the b is and we are not given a^54 is not a quad res just that a is

i don't know how you can automatically say that. we kniw a^54 ~=1 by euler's criterion. so say a^2 = 1, then if you multiple out a^2 27 times you would get a^54 is 1 is rthat right?

Maybe you misunderstand? It is trivial that the order of a must divide phi(109). Great so then we have to consider the cases {a, a^2, a^3, a^4, a^6, a^9, a^12, a^18, a^36, a^54} only, I kniw that

Well of course we know that it a divisor of 108, this is trivial fact no? It's not trivial to show that a^6 ~= 1 or a^4 ~= 1 or a^36 ~=1

That seems like it has nothing to do with problem. where does 105 come in. No group,s no lagrange's it should be proven using number theory

It factors as 1 + 2^2*3^3 i know that

What is this gorup? This is a number theory question not a group question? Can you show it with the number theory ? What is this group?

Please prove that if x is quadratic nonResidue modulo 109 and x also cubic nonresidue modulo 109 than x is guaranteed to be primitive root modulo 109 thanks you very much