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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<yx and yx<yy. 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={xx>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<axga1=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)(bc)= 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.
 Spring '09
 BRUBAKER

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