Hi guys, having problems solving this problem. Question is prove by contradiction that is irrational. Any help would be highly appreciated.
Ok here is my attempt.
Suppose that is rational. Therefore = for some integers m, n with n 0.
We may suppose without loss of generality (wlog) that m, n have no common factor (i.e. are coprime). Now = and so is even. Put . Then = and so =
which implies that n is even. This contradicts the assumption that m, n are coprime.
Hence must be irrational.
Is this correct?
Hello, jvignacio!
You've mixed in 2's, 4's, and even numbers.
Assume that is rational.Prove by contradiction that is irrational.
Then: . for positive integers and , where and are coprime. .[1]
Square the equation: .
Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: . .[2]
The equation becomes: .
Since the right side is divisible by 3, the left side is also divisible by 3.
. . Hence: . .[3]
But [2] and [3] say that . . . which contradicts [1].
. .
That's because, if is divisible by 3, then so is . In order for 3 to divide , one could try breaking 3 into two factors: so that and , but ( means " divides "). However, 3 is prime, and this is impossible.dont understand how you got to the a = 3c part?
Since , we have , and we can divide both sides of by 3.