Solutions-PSA - Problem Set Partial Solution PDF

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MAT102F - Introduction to Mathematical Proofs - UTM - FALL 2017 Solutions to Selected Problems from Problem Set A 1.5.22 To prove that ||x| − |y|| ≤ |x − y| we use triangle inequality twice. Namely, |x| = |(x − y) + y| ≤ |x − y| + |y|,

|x − y| ≥ |x| − |y|,

|y| = |(y − x) + x| ≤ |x − y| + |x|,

|x − y| ≥ |y| − |x|.

Therefore, −|x − y| ≤ |x| − |y| ≤ |x − y|. Equivalently, ||x| − |y|| ≤ |x − y|. Q1. Consider D = (−a)2 − 4(a − 1) = a2 − 4a + 4 = (a − 2)2 . (a) By Quadratic Formula Theorem, the equation has exactly one root if and only if D = 0. This means that (a − 2)2 = 0, a = 2; (b) By Quadratic Formula Theorem, the equation has exactly two real roots if and only if D > 0. Therefore, a 6= 2. Q2. First we note that the equation has two distinct roots if and only if D = (−4a)2 −4·5a = 4a(4a−5) > 0. Let x1 , x2 be two distinct roots. The sum of their squares x21 + x22 must be equal to 6. By Vieta’s formulas (see 1.5.1) we have x1 + x2 = 4a, x1 · x2 = 5a. Therefore, x12 + x22 = (x1 + x2 )2 − 2x1 x2 = (4a)2 − 10a = 16a2 − 10a. Solving the equation 16a2 − 10a = 6 we get two values for a: a = 1 and 3 3 a = − . When a = 1, D < 0. Hence, this value is not good for us. When a = − , D > 0. 8 8 3 Answer: a = − . 8 Q3. Re-write this inequality in the equivalent form: p |2b|2 + (b + 1) ≥ |2b|2 (b + 1). 2 This holds true by AGM with u = |2b|2 and v = (b + 1) (both u, v ≥ 0.)...