Solutions-PSC

# Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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MAT102S - Introduction to Mathematical Proofs - UTM - Spring 2010 Solutions to Selected Problems from Problem Set C 1. 1.17. The domain of y = | x | is R (every number has an absolute value). The image is [0 , ) since every y 0 is the absolute value of itself: y = | y | . 1.35. x/y + y/x 2 if and only if x and y have the same sign: If x or y is 0, then the expression is undeﬁned. If they have opposite signs, then the left side is negative. If they have the same sign, then multiplying by xy yields x 2 + y 2 2 xy , equivalent to x 2 - 2 xy + y 2 0, equivalent to ( x - y ) 2 0. The last inequality holds whenever x and y have the same sign, so this necessary condition is also suﬃcient. 1.49. (a) TRUE. If f and g are bounded, then there exist M 1 and M 2 , such that | f ( x ) | ≤ M 1 and | g ( x ) | ≤ M 2 for all x R . Using the triangle inequality, we conclude that | f ( x ) + g ( x ) | ≤ | f ( x ) | + | g ( x ) | ≤ M 1 + M 2 for all x R , and hence f + g is a bounded function. (d) FALSE. Let f ( x ) = x 2 + 1 and g ( x ) = 1 x 2 +1 . Then f · g = 1 is a bounded function, but f is not. (e) TRUE. We know that f + g and fg are bounded. Let M 1 and M 2 be bounds for f + g and fg respectively. Then for all
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Unformatted text preview: x R we have: f 2 ( x ) + g 2 ( x ) = | [ f ( x ) + g ( x )] 2-2 fg | | f ( x ) + g ( x ) | 2 + 2 | fg | M 2 1 + 2 M 2 . This implies that | f ( x ) | = p f 2 ( x ) p f 2 ( x ) + g 2 ( x ) q M 2 1 + 2 M 2 , and thus f is bounded. Similarly, g is also bounded. 3. No it is not a eld. The number 1, for instance, does not have a negative. 4. We showed in class that a b = 1, and we will use this fact here. The element a 2 must be either 0 , 1 ,a or b . We check each of the options: If a 2 = a a = 0, then from the properties of a eld we get that a = 0, which is impossible. If a 2 = a , we can multiply both sides by a-1 to get a = 1, which is impossible. If a 2 = 1, then a 2 = a b , and we can multiply both sides by a-1 to get b = 1, which is impossible. The only option left is therefore a 2 = b ....
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