Math 2510: Real Analysis I – Homework Solutions §4
Fall 2008
§4 Techniques of Proof: II
4.4 Prove: There exists a rational number x such that x2 + 3x/2 = 1. Is this rational number unique?
Take x = −2. Then x is a rational number and x2 + 3x/2 = (−2)2 + 3(−2)/2 = 4 − 3 = 1.
This rational number is not unique. Take x = 1/2. Then −2 6= 1/2, x is a rational number
and x2 + 3x/2 = (1/2)2 + 3(1/2)/2 = 1/4 + 3/3 = 1.
4.10 Prove: If x is a real number, then |x − 2| ≤ 3 implies that −1 ≤ x ≤ 5.
Suppose that |x − 2| ≤ 3. Then −3 ≤ x − 2 ≤ 3. Adding 2 to each term we get −1 ≤ x ≤ 5
as desired.
4.13 Suppose x and y are real numbers. Prove the following.
(b) If x is rational and y is rational, then xy is rational.
Suppose that x and y are rational. Then x = p/q where p and q are integers and q 6= 0.
Also, y = m/n where m and n are integers and n 6= 0. Then xy = (pm)/(qn). Now pm
is an integer, qn is an integer, and qn 6= 0 because q 6= 0 and n 6= 0. Therefore xy is a
rational number.
4.14 Suppose x and y are real numbers. Prove or give a counterexample.
(c) If x is irrational and y is irrational, then xy is irrational.
√
√
This is false: Take x = 2 and y = 2. Then x and y are irrational but xy = 2 is
rational.
4.19 Prove or give a counterexample: The sum of any five consecutive integers is divisible by five.
Proof: Suppose n is an integer. Then the five consecutive integers beginning with n are n,
n + 1, n + 2, n + 3, and n + 4. The sum is n + n + 1 + n + 2 + n + 3 + n + 4 = 5n + 10 = 5(n + 2),
which is a multiple of 5. Therefore, the sum of any five consecutive integers is divisible by
five.