So it's been a WHILE
I started this post in September, and I was so close to finishing it, but I obviously forgot to actually post it! I've had actually NO time to do anything since maybe September, and I'm still really tired right now because I had a competition that was a couple of hours away, and I had to do stuff, and yeah.
I'm probably going to past at least 2 more times over Thanksgiving break, so don't worry about consistency. On second thought, worry about consistency, because you know it's really bad.
Let's get into the post.
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Prove that the square root of two is irrational.
What we're going to do here is assume that the opposite is true: Sqrt(2) is rational. By that logic, Sqrt(2) can be written as the fraction (a/b), where (a) and (b) are integers. Let's also assume that the fraction is simplified to its lowest terms. If it is, then both of them can't be even-- it could be simplified more.
If (a/b)=2, then (a^2/b^2)=2. Multiplying both sides by (b^2), you get (a^2)=2(b^2). Therefore (a) is even.
Because both (a) and (b) can't be even, then proving that (b) is even would prove that Sqrt(2) is irrational.
Since a is even, we can write (a) as (2*k), where k is half of a. Therefore:
(2k)^2=2(b^2)
4(k^2)=2(b^2)
2(k^2)=(b^2)
So b is even too which proves that Sqrt(2) is irrational.
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Prove that there are an infinite amount of prime numbers.
For this one, we're going to do the same thing- assume the opposite. There 'are' a finite amount of prime numbers.
If that's true, then we can multiply all of the prime numbers together to get some number c.
However, c+1 must be prime, because every prime number would have a remainder of 1 when dividing it. So that kind of proves it wrong, and it's a relatively easy proof.
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Stay coolio,
John
