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## Here's the Solution to this Question

(i) If n is an even integer, then it can be presented as n=2k, where k is an integer. Then -n=-2k is also even.

(ii) If n is an even integer, then it can be presented as n=2k, where k is an integer. Then 3n+5=6k+5=2(3k+2)+1. It is clear, that the latter is odd.

(iii) If n is odd, then n=2k+1, where k is an integer. $n^2+3n=4k^2+4k+1+6k+3=2(2k^2+5k+2)$ . The latter is even.

(iv)  If m is an even integer and n is an odd integer, then m=2k, n=2s+1, where k, s are integers. Then $m^2-2n=4k^2-4s-2$ The latter is even.

(v) If m and n are even integers, then their product mn is also even. The sum mn+r is odd, since r is odd.

(vi) Two odd integers m and n can be presented as $m=2k+1$ and $n=2s+1$ , where k and s are integers. Then $m+n=2(k+s+1)$ is even.

(vii) Suppose that $\sqrt{x}>\sqrt{y}$ . Then we can multiply both sides by $\sqrt{x}$ and use the inequality again. Then we get: $x>\sqrt{y}\sqrt{x}>y$. It contradicts to inequality $x\leq y$ . Thus, $\sqrt{x}\leq \sqrt{y}$

(viii) If a divides b, we can write it as b=ka for some integer k. Similarly, we get that c=sb for some integer s. Then, c=ska. It means that a divides c.