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problem set solutions

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Department
Mathematics
Course
MAT102H5
Professor
Shay Fuchs
Semester
Fall

Description
MAT102S - Introduction to Mathematical Proofs - UTM - Spring 2011 Solutions to Selected Problems from Problem Set B ▯ ▯ ▯ To prove that jxj ▯ jyj ▯ jx ▯ yj we use the triangle inequality ja + bj ▯ jaj + jbj twice as follows: If we set a = x ▯ y and b = y we get jx ▯ y + yj ▯ jx ▯ yj + jyj which becomes jxj ▯ jyj ▯ jx ▯ yj . If we set a = y ▯ x and b = x we get jy ▯ x + xj ▯ jy ▯ xj + jxj which becomes jyj ▯ jxj ▯ jx ▯ yj . ▯ ▯ ▯ ▯ Now since jxj ▯ jyj must be equal to either jxj▯jyj or jyj▯jxj we conclude that jxj ▯ jyj ▯ jx▯yj . 1.20. If r;s are two real distinct solutions of ax + bx + c = 0, then from the quadratic formula we have p p ▯b + b ▯ 4ac ▯b ▯ b ▯ 4ac r = and s = 2a 2a (or vice versa). A direct computation shows that 2 2 r + s = ▯2b = ▯ b and rs = b ▯ (b ▯ 4ac) = a : 2a a 4a2 c 1.31. (a) For any two numbers we have 0 ▯ (t ▯ s) which is equivalent to 2ts ▯ t + s . 2 2 2 2 2 Using this result for t = xy and s = zw we get that 2xyzw ▯ x y + z w , or
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