MIT18_014F10_pset1sols

# MIT18_014F10_pset1sols - 18.014 Problem Set 1 Solutions...

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Unformatted text preview: 18.014 Problem Set 1 Solutions Total: 24 points Problem 1: If ab = 0, then a = 0 or b = 0. Solution (4 points) Suppose ab = and b = 0. By axiom 6, there exists a real number y such that by = 1. Hence, we have a = 1 a = a 1 = a ( by ) = ( ab ) y = 0 y = 0 using axiom 4, axiom 1, axiom 2, and Thm. I.6. We conclude that a and b cannot both be non-zero; thus, a = 0 or b = 0. Problem 2: If a &lt; c and b &lt; d , then a + b &lt; c + d . Solution (4 points) By Theorem I.18, a + b &lt; c + b and b + c &lt; d + c . By the commutative axiom for addition, we know that c + b = b + c,d + c = c + d . Therefore, a + b &lt; c + b,c + b &lt; c + d . By Theorem I.17, a + b &lt; c + d . Problem 3: For all real numbers x and y , | x | | y | | x y | . Solution (4 points) By part (i) of this exercise, | x || y | | x y | . Now notice that ( | x || y | ) = | y || x | . By definition of the absolute value, either || x || y || = | x || y | or || x || y || = | y || x | . In the first case, by part (i) of this problem, we see that || x | | y || | x y | . In the second case, we can interchange the x and y from part (i) to get || x = | | y || | y | | x | | y x | = | x y | , where the last equality comes from part (c) of this problem. Thus, || x | | y || | x y | . Problem 4: Let P be the set of positive integers. If n,m P , then nm P . Solution (4 points) 1 Fix n P . We show by induction on m that nm P for all m P ....
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## This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.

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MIT18_014F10_pset1sols - 18.014 Problem Set 1 Solutions...

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