produce(x: number) { return x * 2; },
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Let’s assume that’s the case; that would mean that s1 = s2 regardless of our choice of K2. However, in the irrational case, we’ve established that s1 is necessarily a positive number. We can make the 1 / K2 fraction arbitrarily small by increasing K2, so no matter how small s1 is, the right-hand fraction can be made smaller to flip the inequality. This produces a contradiction. It must be that for a given irrational r and some chosen K2 K1, the equation will produce a new, distinct a2 / b2 such that s2
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