Chapter 2

# Chapter 2 - B.1 Sguare roots and the existence of irrational numbers I Deﬁnition If b2 = a then we say that b is a sguare m of a A negative

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Unformatted text preview: B.1 Sguare roots, and the existence of irrational numbers. I Deﬁnition. If b2 = a, then we say that b is a. sguare m of a. A negative number has no square root (see Theorem I .20), and the number 0 has ' only one square root,'namely 0. We shall show that a positive real number has exactly two square roots, one positive and one ne ative. 7 Theorem. Let a > 0. Then there is a number b > 0 such that b2 = a. Proof. Step I. Let x and y be positive numbers. Then 1: < y if and only ifx2 < If x < y, we multiply both sides, ﬁrst by x and then by y, to obtain the inequalities x«x<y-x and y-x<y-y. 2 2 Thus x < y2 . Conversely if x2 < y , then it cannot be true that x = y (for that would 2 imply x y2 < x2). Hence we must have x < y. yz), or that y < x (for that would imply, by what we just proved, that Step 2. We construct b as follows: Consider the set S={x|x>0 and x2 < a}. The set S is nonempty; indeed if x is a number such that 0 < x g 1 and x < a, then x2<axga-1=a, so that x is in S. Furthermore, S is bounded above; indeed, I + a is an upper bound on S: B.2 For if x is in S, then x2 < a; since a < 1 + 23. + a2 = (1+a)2, it follows from Step 1 that x < 1 + a. Let b denote the supremum of S; we show that b2 = a. We verify this fact by showing that neither inequality b2 < a or b2 > a can hold. Step 3. Assume ﬁrst that b2 < a. We shall show that there is a positive number h such that (b+h)2 < a. It then follows that b + h belongs to S (by deﬁnition of S), contradicting the fact that b is an upper bound for S. To ﬁnd h, we proceed as follows: The inequality (b+h)2 < a is equivalent to the inequality h(2b+h) < a—b2. Now a —- b2 is positive; it seems reasonable that if we take h to be sufﬁciently small, this inequality will hold. Speciﬁcally, we ﬁrst Specify that h g 1; then we have ' h(2b+h) g h(2b+1). It is then easy to see how small h should be; if we choose h < (a—bz) / (2b+1), then h(2b+1) < a — b2 and we are ﬁnished. Step 4. Now assume that b2 > a. We shall show that there is a number h such that 0 < h < b and (b—h)2 > a. It follows that b —— h is an upper bound for S: For B.3 if x is in s, then a > x2, so that (b—h)2 > x2, whence by Step 1, b — h > x. This contradicts the fact that b is the leg upper bound for S. To ﬁnd h, we proceed as follows: The inequality (b—h)2 > a is equivalent to the inequality h(2b—h) < b2 — 3.. Now b2 — a is positive; it seems reasonable that if h is sufﬁciently small, this; inequality will hold. Our ﬁrst requirement is that’O < h < b. Then we note that h(2b——h) = 2hb ~— 11}2 < 2hb. It is now easy to see how small h should be; if we choose h < (b2—a)/2b, then 2hb < b2 — a and we are ﬁnished. 0 Corollary. I f a > 0, then a has exactly two square roots. We denote the positive square root of a by JE. Proof. Let b > O and b2 = a. Then (—b)2 = 3.. Thus a has at least two square roots, b and ——b. Conversely, if c is any square root of a, then c2 = a, whence (b+c)(b-c)-= b2 ~ c2 = o. It follows that c = —-b or c = b. a We now demonstrate the existence of irrational numbers. Theorem. Leta be a positive integer; let b 2 J5. Then either b is a. positive integer or b is irrational. Proof. suppose that b = J5 and b is a rational number that is not an integer. We derive a contradiction. Let us write b = m/n, where m and n are positive integers and n is as small as possible. (I.e., we choose 11 to be the smallest positive integer such that nb is an integer, and we set m = nb.) Choose q to be the unique integer such that B.4 q < m/n < q+1. Then qn<m<qn+n, or (*) 0 < m —— qn < n. We compute as follows: ' (m/n)2 = b2 = a, m2 = n23. m(m——qn) = n(na——qm). Then using (*), we can write b _ E _ najm " n — m—qn' This equation expresses b as a ratio of positive integers; and by (*) the denominator is less than n. Thus we reach a contradiction. ' 0 Corollary. ﬂ is irrational. Proof. Let b = ﬂ. Then b cannot be an integer, for the square of 1 equals 1 while the square of any integer greater than 1 is at least 4. It follows that b is irrational. n The same proof shows that the number Jii is irrational whenever n is a positive integer less than 100 that is not one of the integers 1, 4, 9, 16, 25, 36, 49, 64, or 81. ...
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## This note was uploaded on 04/26/2009 for the course MATH CALC taught by Professor Brubaker during the Spring '09 term at MIT.

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Chapter 2 - B.1 Sguare roots and the existence of irrational numbers I Deﬁnition If b2 = a then we say that b is a sguare m of a A negative

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