Question

Explain your answer whenever possible:

4. Prove the following theorem: n is even if and only if n² is even.

5. Prove: if m and n are even integers, then mn is a multiple of 4.

         6. Prove: |xy| = |x||y|, where x and y are real numbers. (recall that |a| is the absolute value of a, equals (a) if a>0 and equals (–a) if a<0 )

Answer 1

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frove- h is is even if and only if 2. heed to to prove this, we prove both ways j is even then n² is even. n=aK n²= 4K2 n²= 2(26²) thence, n² is eveni ;) if n² is even then h is We will prove prove by contra- diction Suppose, n² is but n is odd. Now, n-ak ti (2K+1) 2 n² = 4k² +4kti your writing partner

n²= 4(x²+k) + Here, n² n² is odd. Hence + our assumption was wrong. n is even so, n² is even then Hence, we have proved both ways 5.) Prove if are even then mn is multiple of u maak n= Qe QK.ae ucke) which is multiple of 4 Hence proved beyl= |aellyl, where where a real numbers. 7 6.) Prove We will prove by cases.

Case- Both positive | ги) = 1 ч а) (@| >ə then a = х 4 Proved Cause I -> Case THI Both negative Те») - (-4)| - - - - - х. ) One negative « Eч) || |-| | - | say Proved ч х)