Prove that sqrt(1 + sqrt(3)) is irrational!
assume it is rational. then we can write for , .
we square both sides to obtain:
now the right side of the equation is a rational number (by our assumption). thus we must have that is rational. therefore, we can prove the claim by showing that this is false.
now proving is irrational is also done by contradiction, and usually done the same way is proven to be irrational (see here)
but there is a shorter way.
consider the equation
by the rational roots theorem, if rational roots exist, they will be one of . but none of these solve the equation. however, solves the equation and is therefore not a rational number.
so seeing that is not rational, we have our contradiction
If you're assuming the usual thing, where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.
This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
Definition: Given a positive integer and a prime we define ( )as the number of times that divides , i.e. but . (Note, this definition is a good definition because eventually powers of will exceeded and therefore will not divide it any further thus such an has to exist).
Lemma 1: Given two positive integers with prime we have that .
Proof: If then but . If then but . This means that but if then (or divides (or respectively) which is a contradiction.
Lemma 2: Given a positive integer the integer is a power of if and only if is divisible by for all primes .
Proof: If then . Conversely, if is divisible by then by unique factorization we know that and thereofore each exponent is prime and divisible by hence is a power of .
Theorem: If is not a power of then is irrational.
Proof: Proof by contrapositive. Suppose that for positive integers . Then, thus . Now for any prime we have thus . This shows us that is divisible by for any prime . Thus, by Lemma 2 we have that is a power of .