Spivak Calculus 3rd. Edition Chapter 1 Problem 12 (iii) and (iv) Proofs CritiqueHere are my "proofs" for Spivak's Calculus Chapter 1 Problem 12. I am new to this level of rigour and I am attempting to intimate myself with more advanced topics of mathematics to prepare for next year. I apologize in advance, as these so-called "proofs" are not likely to be nearly as rigorous as they should be. Any assistance on how to write the proofs better or any critiques on faulty logic would be greatly appreciated.12) iii) Prove that | x | | y | = | x y | My proof is: | x | | y | = | x | ⋅ | y − 1 | , therefore by (i), | x | ⋅ | y − 1 | = | x ⋅ ( y − 1 ) | = | x y | 12) iv) Prove that | x − y | ≤ | x | + | y | My proof is: | x | + | y | ≥ | x + y | , replacing y with ( − y ) yields | x | + | − y | ≥ | x + ( − y ) | = | x | + | y | ≥ | x − y | , which is what we wished to prove.

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Spivak Calculus 3rd. Edition Chapter 1 Problem 12 (iii) and (iv) Proofs CritiqueHere are my &quot;proofs&quot; for Spivak&#039;s Calculus Chapter 1 Problem 12. I am new to this level of rigour and I am attempting to intimate myself with more advanced topics of mathematics to prepare for next year. I apologize in advance, as these so-called &quot;proofs&quot; are not likely to be nearly as rigorous as they should be. Any assistance on how to write the proofs better or any critiques on faulty logic would be greatly appreciated.12) iii) Prove that                     |            x              |                            |            y              |              =      |          x      y        |  My proof is:                    |            x              |                            |            y              |              =      |    x      |    ⋅      |        y          −      1            |  , therefore by (i),       |    x      |    ⋅      |        y          −      1            |    =      |    x  ⋅  (      y          −      1        )      |    =      |          x      y        |  12) iv) Prove that       |    x  −  y      |    ≤      |    x      |    +      |    y      |  My proof is:      |    x      |    +      |    y      |    ≥      |    x  +  y      |  , replacing   y with   (  −  y  ) yields       |    x      |    +      |    −  y      |    ≥      |    x  +  (  −  y  )      |    =      |    x      |    +      |    y      |    ≥      |    x  −  y      |  , which is what we wished to prove.

relientaaho2 · Accepted Answer

You can always just do cases. Assume   x  ,  y  &amp;gt;  0 and without loss that   x  &amp;gt;  0  ,  y  &amp;lt;  0. This will do it for you. In the case when   x  =  0 the result follows immediately.

Spivak Calculus 3rd. Edition Chapter 1 Problem 12 (iii) and (iv) Proofs Critique Here are my "proof

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