Alternating pattern of root 2 convergents

Question

Is there an easy way to see why successive convergents to root 2 are alternately bigger and smaller than it? (I've looked at the continued fractions and the recursive formula and plotted them as gradients.)

Answer 1

I've managed a 1-page proof by contradiction, leading to the result that either a posited intermediate convergent coincides with an existing one or the denominator is too big.

Answer 2

If p/q is a convergent of root 2 [gcd(p,q) = 1], then (x,y) = (p,q) is a solution of the Brahmagupta-Pell's equation x^2 - 2y^2 = plusminus 1. For example, (x,y) = (3,2) is a solution of x^2 - 2y^2 = 1. This indicates that 3/2 will be larger than root 2. Similarly (x,y) = (7,5) is a solution of x^2 - 2y^2 = -1, indicating that 7/5 is lesser than root 2.

If p/q is a convergent of root 2, then p+2q/p+q is the next convergent.

We can prove that if (p,q) is a solution of x^2 - 2y^2 = 1 then (p+2q, p+q) is a solution of x^2 - 2y^2 = -1, and that if (p,q) is a solution of x^2 - 2y^2 = -1 then (p+2q, p+q) is a solution of x^2 - 2y^2 = 1. The proof of this is mere substitution.

This proves the pattern of the convergents of root 2 bouncing about root 2.