Lately I've read Susan Bassein's An Infinite Series Approach to Calculus . She presents an interesting proof, in my opinion, that the positive square root of 2 cannot get expressed exactly in terms of rational numbers. At least once I could understand her language.

Here, we'll define a 2nd power root of n as any number x such that x²=n. The positive square root, denoted sqr(n) hereafter, consequently comes as a positive number which satisfies this equation. Due to the way exponentiation works there only exists one positive number here. With that definition in mind, we'll proceed to show that if b, and c are integers, then sqr(2) ≠ b/c.

First since sqr(2) ≠ b/c, we have that 2 &ne (b/c)². So, 2c² &ne b². We can also derive sqr(2) ≠ b/c from 2c² &ne b², so if we show 2c² &ne b², then the conjecture follows. We'll now develop some seemingly trivial lemmas.

Lemma 1: the product of two odd numbers d, e is another odd number.
Since d=2y+1, and e=2z+1, we have de=(2y+1)(2z+1)=4yz+2y+2z+1=2(yz+y+z)+1. Since yz+y+z equals another integer, 2(yz+y+z)+1 is odd.

Lemma 2: the product of two even numbers, f, g, is another even number.
Since f=2w, g=2v, fg=(2w)(2v)=2(2wv). Since 2wv equals another integer, fg is even.

If one does something even more seemingly "trivial" here, something else can become apparent.

Lemma 3: the square of an even number h is also even.
Since h=2z, h²=(2z)²=2²z²

Now suppose that b in 2c² &ne b² is odd. So, b² is odd also. Since 2c² is even this implies this inequality as correct.

Suppose that b is even, and c is odd. Then, the inequality holds since 2c² is even also and can only get divided by 2 once, and b²=(2n)²=2²n which implies it can get divided by 2 twice.

Now suppose that b in 2c² &ne b² is even and c is even also. So, both b² and c² are even also. All even numbers can get written as a product 2^xo, where o denotes an odd number. Consequently, b and c can get divided by 2 x times, or we could say that b and c have x even factors. So, now treating y as any integer, y²=[(2^x)o]²=(2^x)²*o². So, we have 2^2x*o²=2^x*2^x*o². Thus, b and c have x+x=2x even factors. Since this holds, 2c² has an odd number of even factors. So, the inequality holds.

Therefore, for all integers b, c, sqr(2) ≠ b/c.

As far as I can tell, there's no hidden reductio ad absurbum lurking here, but it might come as trickier to spot than I believe.