Exercise 6A

Date: 2023-02-06

first we prove that any number equals zero (UNRELATED AND WRONG BUT FUNNY)

Assume that a=b, where a and bRab=a2aba2=a2a2a(ba)=0a=0all real numbers are equal to 0QED

Question 1a

Assume that m is even and n is even.Then m=2k and n=2l where k and l are integers.m+n=2k+2l=2(k+l)Hence m+n is even, since k+l is an integer.QED

Question 1b

Assume that m is even and n is even.Then m=2k and n=2l where k and l are integers.mn=2k×2l=4kl=2(2kl)Hence mn is even, since 2kl is an integer.QED

Question 2

Assume that m is odd and n is odd.Then m=2k+1 and n=2l+1, where k and l are integers.m+n=2k+2l+2=2(k+l+1)Hence m+n is even, since k+l+1 is an integer.QED

Question 3

Assume that m is even and n is odd.Then m=2k and n=2l+1, where k and l are integers.mn=4kl+2=2(kl+1)Hence mn is even, since kl+1 is an integer.QED

Question 4a

Assume that m is divisible by 3 and n is divisible by 7.Then m=3k and n=7l, where k and l are integers.mn=21klHence mn is divisible by 21.QED

Question 4b

Assume that m is divisible by 3 and n is divisible by 7.Then m=3k and n=7l, where k and l are integers.m2n=63k2lHence m2n is divisible by 63.QED

Question 5

Assume that m and n are perfect squares.Then m=k2 and n=l2, where k and l are integers.mn=k2l2=(kl)2Hence mn is a perfect square, since kl is an integer.QED

Question 6

Assume that m and n are integers.Using the binomial expansion, we can turn (m+n)2(mn)2 into m2+2mn+n2m2+2mnn2=4mnHence, (m+n)2(mn)2 is divisible by 4, since mn is an integer.QED

Question 7

Assume n is an even integer.Then n=2m, where m is an integer.Subtituting n for 2m in the expression, we get (2m)212m+5=2(2m26m)+5Hence, n26n+5 is odd, since 2m26m is an integer.QED

Question 8

Assume n is an odd integer.Then n=2m+1, where m is an integer.Subtituting n for 2m+1 in the expression, we get(2m+1)2+16m+8+3=4m2+20m+12=2(2m2+10m+6)Hence, n2+8m+3 is even, since 2m2+10m+6 is an integer.QED

Question 9

First assume that n is an even number.Then n=2m, where m is an integer.Subtituting n for 2m in the expression, we get5(2m)2+6m+7=20m2+6m+7=2(10m2+3m+3)+1Hence, 5n2+3n+7 is odd if n is even, since 10m2+3m+3 is an integer.Now assume that n is an odd number.Then n=2m+1, where m is an integer.Substituting n for 2m+1 in the expression, we get=5(2m+1)2+3(2m+1)+7=5((2m)2+4m+1)+6m+3+7=20m2+26m+15=2(10m2+13m+7)+1Hence, 5n2+3n+7 is odd if n is odd, since 10m2+13m+7 is an integer.Since in both cases, the expression is odd, the expression is always odd.QED

Question 10

Assume that x and y are positive real numbers.


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