1 Explain the steps in the green boxes to show that is irrational.A SIMPLE PROOF THAT IS IRRATIONALIVAN NIVENLet =ab, the quotient of positive integers. We define the polynomialsf(x)=xn(a-bx)nn!F(x)=f(x)-f(2)(x)f(4)(x)-cdots (-1)nf(2n)(x)the positive integer n being specified later. Since n!f(x) has integral coefficients and terms in x of degree not less than n,f(x) and its derivatives f(x)(x) have integral values for x=0; also fo x==ab, since f(x)=f(ab-x) By elementary calculus we haveddx{F'(x)sinx-F(x)cosx}=F''(x)sinxF(x)sinx=f(x)sinxand(1)0f(x)sinxdx=[F'(x)sinx-F(x)cosx]0**=F()F(0).Now F()F(0) is an integer since f()() and f()(0) are integers. But for n0so that the integral in(1)is positive, but arbitrarily small for n sufficiently large. Thus (1)is false and sois our assumption that is rational.*integral values means integer values.0,0so that the integral in(1)is positive, but arbitrarily small for n sufficiently large. Thus (1)is false and sois our assumption that is rational.*integral values means integer values