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Mathematics 37 Online
mhchen:

Prove that the sqrt(n) + sqrt(n+1) is irrational.

mhchen:

In other words, prove \[\sqrt{n} + \sqrt{n+1} \notin Q\]

mhchen:

We shall prove this by contradiction. Assume it IS a rational number, then we can rewrite it as \[\sqrt{n}+\sqrt{n+1} = \frac{p}{q} , p,q \in \mathbb{Z}\] where p,q has no common divisors \[n + 2 \sqrt{n} \sqrt{n+1} + (n+1) = (\frac{p}{q})^{2}\] \[ \sqrt{n(n+1)} = \frac{\frac{p^{2}}{q^{2}} - 1-2n}{2}\] Since rational numbers are closed under multiplication, division, addition, and subtraction, the right side is still rational, so we can rewrite it again as: \[\sqrt{n(n+1)} = \frac{p}{q},p,q \in \mathbb{Z}\] Case I: n is ODD Then n(n+1) = odd * even = odd Case II: n is EVEN Then n(n+1) = even * odd = odd I'm not sure where to go from here...

mhchen:

Wait I'm an idiot, even * odd = even. LOL

mhchen:

Since they're both even, we can substitute n(n+1) = 2k where k is any integer Thus \[(\sqrt{2k})^{2} = \frac{p^{2}}{q^{2}}\] \[2kq^{2} = p^{2}\] Thus p^2 is even, and since even^2 = even*even = even, and odd^2 =odd * odd = odd, then p is also even. Thus we can re-write p as 2x where x is any integer Then \[2kq^{2} = (2x)^{2}\] \[kq^{2}=2x^{2}\] \[q^{2} = 2(\frac{x^{2}}{k^{2}})\] Thus q^2 is even, and by the previous logic, q is also even. But if p is even and q is even, then they have a common divisor of 2, which contradicts our assumption that p and q don't have a common divisor! Thus our assumption is false.

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