Show that $\gcd(2^m-1, 2^n-1) = 2^ {\gcd(m,n)} -1$ [duplicate]
Possible Duplicate: Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ $\gcd(b^x - 1, b^y - 1, b^ z- 1,…) = b^{\gcd(x, y, z,…)} -1$ I'm trying to figure this out: ...
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So if I have a matrix and I put it into RREF and keep track of the row operations, I can then write it as a product of elementary matrices. ...
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I am asked to prove the density of irrationals in $\mathbb{R}$. I understand how to do this by proving the density of $\mathbb{Q}$ first, na...
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Can someone just explain to me the basic process of what is going on here? I understand everything until we start adding 1's then after ...
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