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Unformatted text preview: Irrational Square Roots: J2 and J; when n is a Positive Integer and Not a Perfect Square Theorem 3.7.1: J2 is irrational. Proof: [Proof by Contradiction] Suppose, by way of contradiction, that J2 is rational. (***) Since J2 lS ratlonal and pos1t1ve, there exrst p051t1ve integers m and n w1th n ¢ 0 such that J2 = — ,
n and we can assume that — lS written 1n lowest terms, so that m and n have no common prime factor.
n [ The author mistakenly says that m and n “have no common factor”, but 1 is always a common factor. ] Sincex/E =11: , 2
E H N /-’\
\% II N] 5., by substitution . Since 2=—-, 2n2 = m. [ The contradiction that we will establish is that 2 | m and 2 l n ,
which contradicts the fact that m and n have no common prime factor. ]
Since m2 = 2n2 , 2 1 ml, by deﬁnition of “divides”.
Since 2 l m2 , In2 > I, and therefore, m > 1. [We need to know m > I to apply Thm (NIB) 3.] Since 2 is prime and 2 I m2 , 2 I m , by Theorem (NIB) 3. 2 k, by deﬁnition of “divides”. Recall that 2 n2 = m2 . There exists an integer k such that m
.22112 = (2k)2 = 2(2k2). n 2 = (2 k2 ) , and so 2 I n 2 , by deﬁnition of “divides”. 2 > 1, and therefore, 11 > 1. Since 2 I n2 , n Since2 is prime and 2 I n2 , 2 I n, by Theorem (NIB)3.
2 I m and 2 | n , which contradicts the fact that m and n have no common prime factors. Therefore, J2 is irrational. [Considering the initial supposition (***) above] QED To Prove: For all positive integers n , if n is not a perfect square, then J; is irrational.
[ This is the statement to prove in Problem #22 of Section 4.7, ]
Proof: [by Contraposition] Let n be any positive integer. Suppose that Jr; is rational. [ We need to show that n is a perfect square. ] Since J; is rational and positive, there exist positive integers a and b with b 9e 0 such that x/iz— = g , a . . . .
and we can assume that 3 is wrltten 1n lowest terms, so that a and b have no common prlme factor. Sincex/n=% , n II
E M II a 2 a2 a2
[—j = — . Since n — bzn = a2. [We next prove that b = 1 using aproof-by-contradiction .]
Suppose, by way of contradiction, that b 9e 1. (***)
.‘.Sinceb > O and baé 1, b >1.
.'.by Theorem 3.3.2 , there exists some prime number p such that p I b .
Since bzn = b(bn) , bI bzn by deﬁnition of “divides”. p I b2n , by transitivity of divisibility. Recall that bzn = a2. p I a2 , by substitution. Since p is prime and p I a2 , p I a, by Theorem (NIB)3. p | a and p I b , which contradicts the fact that a and b have no common prime factor. b = 1. [Considering the initial supposition (***) above] a a .
a2, and, therefore, 11 IS a perfect square. I
I .'.For all positive integers n, if J; is rational, then n is aperfect square. .'.For all positive integers n, if n is not a perfect square, then «5 is irrational, by contraposition. QED ...
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