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MATH 2001 Homework Han-Bom Moon Homework 3 Solution Section 1.5. • Do not abbreviate your answer. Write everything in full sentences. • Write your answer neatly. If I couldn’t understand it, you’ll get 0 point. • You may discuss with your classmates. But do not copy directly. 1. Prove that the sum of any two rational numbers is rational. a c Let x, y be two rational numbers. Then x = and y = for some a, b, c, d ∈ Z b d and b, d 6= 0. Then a c ad + bc x+y = + = b d bd and bd 6= 0. Because ad + bc, bd are integers, x + y is also a rational number. 2. Let a, b, and c be three integers such that a|b and a|c. Prove that for any integers x and y, a|(xb + yc). Let a, b, c ∈ Z. Suppose that a|b and a|c. Because a|b, b = ak1 for some k1 ∈ Z. Similarly, from a|c, c = ak2 for some k2 ∈ Z. Then for every x, y ∈ Z, xb + yc = xak1 + yak2 = a(xk1 + yk2 ). Therefore a|(xb + yc). 3. Let a, b be two positive real numbers. Show that √ 2ab ab ≥ . a+b (The left side is called the geometric mean and the right side is called the harmonic mean of a and b.) Let a, b be two positive real numbers. Then (a − b)2 ≥ 0. a2 − 2ab + b2 ≥ 0 (a + b)2 = a2 + 2ab + b2 ≥ 4ab Because a + b is a positive number, p √ √ a + b = (a + b)2 ≥ 4ab = 2 ab. √ √ √ ab(a + b) ≥ 2 ab · ab = 2ab √ 2ab ab ≥ a+b 1 MATH 2001 Homework Han-Bom Moon √ 2ab if and only if a = b. a+b Let a, b be two positive real numbers. Suppose that a = b. Then 4. Prove that for two positive real numbers a and b, √ √ ab = Conversely, suppose that and √ a2 = a = ab = 2a2 2ab = . 2a a+b 2ab . Then a+b √ (a + b) ab = 2ab, ab = √ ab (a + b) = 2 √ = 2 ab. ab √ (a + b)2 = (2 ab)2 = 4ab a2 + 2ab + b2 = 4ab (a − b)2 = a2 − 2ab + b2 = 0 Therefore a − b = 0, so a = b. 5. By using contraposition, show that if n2 is odd, then n is odd. Let n ∈ Z. Suppose that n is even. Then n = 2k for some k ∈ Z. n2 = (2k)2 = 4k 2 = 2 · 2k 2 . Therefore n2 is even. 6. Let x and y be two real numbers. Prove that if x is rational and y is irrational, then x + y is irrational. Let x be a rational number and y be an irrational number. Suppose, on the cona trary, that x + y is a rational number. Then x + y = for some a, b ∈ Z. Because b c x is rational, x = for some c, d ∈ Z. Thus d y= a c ad − bc a −x= − = . b b d bd So y is rational. It contradicts to the fact that y is irrational. Therefore x + y is an irrational number. 7. Let’s assume the following statement: Any integer n can be written exactly one of following three forms: 3k, 3k + 1, and 3k + 2 for some integer k. (We will discuss its generalization later.) 2 MATH 2001 Homework Han-Bom Moon (a) By using above statement, show that for a integer n, if n2 is a multiple of 3, then n is a multiple of 3. Suppose, on the contrary, that n is not a multiple of 3. Then n = 3k + 1 or n = 3k + 2 for some k ∈ Z. If n = 3k + 1, then n2 = (3k + 1)2 = 9k 2 + 6k + 1 = 3(3k 2 + 2k) + 1. If n = 3k 2 + 2, then n2 = (3k + 2)2 = 9k 2 + 12k + 4 = 3(3k 2 + 4k + 1) + 1. Therefore in any case, n2 is not a multiple of 3. (b) Show that the following statement is not true: If n2 is a multiple of 4, then n is a multiple of 4. Consider n = 2. Then n2 = 4 = 4 · 1, so n2 is a multiple of 4. However, n is not a multiple of 4. 8. Let a, b be two integers. Consider the following statement: If a|b and b|a, then a = b. Is it true? Prove it if it is true, and give a counterexample if it is false. Let a = 3 and b = −3. Then a = 3 = (−1) · (−3) = (−1) · b, so b|a. Similarly, because b = −3 = (−1) · 3 = (−1) · a, a|b. But 3 6= −3. Therefore the statement is false. 3