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# Suppose k is an approximation to sq. root of 2.?

Prove that q= (2+k)/(1+k) is a better approximation to sq. root of 2.

### 1 Answer

- IndicaLv 78 years agoFavorite Answer
Assume |k−√2| < ϵ. Need to prove that | (k+2)/(k+1)−√2| < tϵ where 0<t<1

Starting from |k−√2| < ϵ assemble an expression for (k+2)/(k+1)−√2

|k−√2| < ϵ → √2−ϵ < k < √2+ϵ → 1+√2−ϵ < k+1 < 1+√2+ϵ

→ 1/(1+√2+ϵ) < 1/(k+1) < 1/(1+√2−ϵ)

→ 1/(1+√2+ϵ) + 1−√2 < (k+2)/(k+1) − √2 < 1/(1+√2−ϵ) + 1−√2

→ (1+1−2+ϵ(1−√2)) / (1+√2+ϵ) < (k+2)/(k+1)−√2 < (1+1−2−ϵ(1−√2)) / (1+√2−ϵ)

→ −ϵ(√2−1) / (1+√2+ϵ) < (k+2)/(k+1)−√2 < ϵ(√2−1) / (1+√2−ϵ)

Assume that initially that ϵ<1 (0.5<k<2.4 will do this) so then

→ −ϵ(√2−1) / (1+√2) < (k+2)/(k+1)−√2 < ϵ(√2−1) / (1+√2−1)

→ −ϵ(√2−1) / √2 < (k+2)/(k+1)−√2 < ϵ(√2−1) / √2

→ | (k+2)/(k+1)−√2| < ((√2−1)/√2)ϵ which is better because (√2−1)/√2 < 1