<p>“as an integer times a non-integer is a non-integer”</p>
<p>(1/2) is not an integer. 2*(1/2) (which is an integer times a non-integer) is 1, which is an integer–this statement from your proof is incorrect.</p>
<p>I want to say this is true if you replace “non-integer” with “irrational”–that is, an integer times an irrational is an irrational (the sketch of a proof I’m imagining would go something like, suppose x is irrational and y is a nonzero integer, but x<em>y is an integer; then you could write x as the ratio of two integers (x</em>y)/y, which is rational by definition, contradiction on x’s irrationality). But…I don’t really see a clean proof that cosC is irrational instead of non-integer. I suspect (though I could very easily be wrong here, as I don’t know math) there isn’t a particularly easy way, since there are infinitely many values of C such that cosC is a non-integer, rational number between 0 and 1.</p>