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Unformatted text preview: MATH 290 – WRITING ASSIGNMENT 1 – SOLUTIONS 1. Prove that for all real numbers a and b ,  a + b  ≤  a  +  b  . (Hint: This will be an example of a proof by cases.) Solution: Let a and b be real numbers. We must show that  a + b  ≤  a  +  b  . In order to do this we will first prove the following claim. Claim. For all real numbers x , x  ≤ x ≤  x  . To prove this claim, let x be a real number. There are two cases. Assume first that x ≥ 0. Since x ≥ 0,  x  = x so that x ≤  x  . Since x =  x  ≥ 0 we have that x ≤ 0, hence that x ≤ ≤  x  and finally that x  ≤ x . Therefore in this case, x  ≤ x ≤  x  . Assume next that x < 0. Since x < 0,  x  = x , so that x  ≤ x . Since  x  ≥ 0 and x < 0, x < ≤  x  , so that x ≤  x  . Therefore in this case, x  ≤ x ≤  x  , and the claim is proved. To finish the proof there are two cases to consider. Assume that a + b ≥ 0....
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This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.
 Fall '08
 Alligood,K
 Advanced Math, Real Numbers

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