True? So... 2^(1/n) is irrational for n≥ 3. Proof (Schultz, 2003): assume 2^(1/n)= p/q; then p^n= q^n + q^n is this contradicting Fermat's Last Theorem?
This is what some people would call using a big hammer on a small nail. Yes, given Fermat's Last Theorem, this constitutes a proof that 2^(1/n) is irrational for n at least 3.
haha That's actually quite an astute observation I wish I had made first.
As Layra says, it's a little excessive. You can also just produce a contradiction by assuming p/q is the irreducible form of 2^(1/n) and then contradict that fact.
2 = p^n/q^n p^n = 2*q^n --> implies p^n and p are even 2 = (2*t)^n/q^n 2 = 2^n*t^n/q^n 2*q^n = 2^n*t^n q^n = 2^(n-1)*t^n --> implies that q^n AND q are even
if p and q are even, then p/q reduces, contradicting our assumption. Therefore, 2^(1/n) is irrational.
This is what some people would call using a big hammer on a small nail. Yes, given Fermat's Last Theorem, this constitutes a proof that 2^(1/n) is irrational for n at least 3.
ReplyDeletehaha That's actually quite an astute observation I wish I had made first.
ReplyDeleteAs Layra says, it's a little excessive. You can also just produce a contradiction by assuming p/q is the irreducible form of 2^(1/n) and then contradict that fact.
2 = p^n/q^n
p^n = 2*q^n --> implies p^n and p are even
2 = (2*t)^n/q^n
2 = 2^n*t^n/q^n
2*q^n = 2^n*t^n
q^n = 2^(n-1)*t^n --> implies that q^n AND q are even
if p and q are even, then p/q reduces, contradicting our assumption. Therefore, 2^(1/n) is irrational.
Logan Collins Danke...for the answer, I'll keep the big hammer ;)
ReplyDeleteIt's irrational for n=2 too.
ReplyDelete