Square root of 2 and the Rationals - Math 100 Square Roots Martin H Weissman 1 The real square root of two Since we have not thoroughly discussed it we

# Square root of 2 and the Rationals - Math 100 Square Roots...

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Math 100Square RootsMartin H. Weissman1.The real square root of twoSince we have not thoroughly discussed it, we begin by mentioning theintermediate value theoremforpolynomials:Suppose thatP(x) is a polynomial with real coefficients, with one variablex. Suppose thata, bR, and letu=P(a), andv=P(b). Suppose thata < b, andu < v. Ifu < w < v,then there existscR, such thata < c < b, andP(c) =w.Rather than proving such a theorem, we take it as anaxiom: a fact about numbers which we assume. Thisaxiom guarantees that every positive real number has a real square root:Theorem 1.Ifyis a real number, andy >0, then there exists a real numberxsuch thatx2=y.Proof.Suppose thatyis a positive real number. LetPbe the polynomialP(x) =x2. Then, 0< y, andP(0)< P(y+ 1), sinceP(0) = 0 andP(y+ 1) = (y+ 1)(y+ 1) and the product of positive numbers ispositive.We claim that 0< y < P(y+ 1). Note that 0< ysinceyis positive. Moreover, we can compareP(y+ 1)andyas follows:P(y+ 1) =y2+ 2y+ 1, and soP(y+ 1)-y=y2+y+ 1. Since 1,y, andy2are all positive,it follows thatP(y+ 1)-yis also positive. HenceP(y+ 1)-y >0, and soP(y+ 1)> y. We have proventhat 0< y < P(y+ 1).Now, we may apply the intermediate value theorem: sinceP(0)< y < P(y+ 1), and 0< y+ 1, we canchoosexRsuch that 0< x < y+ 1 andP(x) =y. Thereforex2=y./Note that the above proof used many of the basic axioms for real numbers. Moreover, it was not a completely“straightforward” proof – at one point, the expressiony+ 1 was used, even thoughy