Question

2. (i) Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even.
(ii) Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd.
(iii) Prove that m2 = n2 if and only if m = n or m = -n.
(iv) Prove or disprove that if m and n are integers such that mn = 1, then either m = 1 or else m = -1 and n = -1.

Answer 1

2.(i)$=>$ We prove first the direct implication. Assume $n$ is even. Then $n=2k$ for some integer $k.$ Then $7n+4=7(2k)+4=14k+4=2(7k+2),$ which is even.

$<=$ For the converse, which is “if $7n+4$ is even, then $n$ is even”, we use a proof by contrapositive.

The contrapositive is: “if $n$ is not even (that is, odd), then $7n+4$ is not even (that is, odd)”. If $n$ is odd, then $n=2k+1,$ for some integer $k.$

Then $7n+4=7(2k+1)+4=14k+7+4=14k+11,$ which is odd.

Hence we have proven that $n$ is even if and only if $7n+4$ is even. 

(ii) $=>$ We will use a direct proof on “If $n$ is odd, then $5n+6$ is odd”. Assume $n$ is odd, so $n=2k+1$ for some integer $k.$

Then $5n+6=5(2k+1)+6=10k+5+6$ $=10k+11=2(5k+5)+1.$

Thus, $5n+6$ is odd.

$<=$ We now must prove the converse, “If $5n+6$ is odd, then $n$ is odd.” For this, we will use proof by contrapositive. So the statement becomes “If $n$ is not odd, then $5n+6$ is not odd.” Assume that $n$ is not odd, so $n=2k$ for some integer $k.$

Then $5n+6=5(2k)+6=10k+6=2(5k+3).$

Thus, $5n+6$ is not odd.

Hence we have proven that $n$ is odd if and only if $5n+6$ is odd. 

(iii) $=>$ If $m=n,$ then $m^2=n^2$ follows using basic principles of Algebra.

Similarly, if $m=-n,$ then $m^2=(-n)^2=n^2$ follows using basic principles of Algebra.

$<=$ Suppose that $m^2=n^2.$ Notice that the square of any real number is non-negative, so we can take the square root of each side of this equation in order to obtain: $\sqrt{m^2}=\sqrt{n^2},$ or $|m|=|n|.$ Thus, using basic principles of algebra, either $m=n$ or $m=-n.$

(iv) $mn=1=>m\not=0, n\not=0.$

Case 1: Suppose $|m|>1.$ Regardless of what $n$ is, $mn=1,$ so $n=\dfrac{1}{m}.$ But we said that both $m$ and $n$ are integers. If $|m|>1,$ then $n$ here cannot be an integer. So, contradiction.

Case 2: Suppose $|n|>1.$ Regardless of what $m$ is, $mn=1,$ so $n=\dfrac{1}{m}.$ But we said that both $m$ and $n$ are integers. If $|n|>1,$ then $m$ here cannot be an integer. So, contradiction.

Case 3: Suppose $m=1$ and $n=-1.$ Clearly then $mn=-1.$ So, contradiction.

Case 4: Suppose $m=-1$ and $n=1.$ Clearly then $mn=-1.$ So, contradiction.

Case 5: Suppose $m=1$ and $n=1.$ Then $mn=1.$

Case 6: Suppose $m=-1$ and $n=-1.$ Then $mn=1.$

What techniques did we use? Cases and contradiction