Good day to all. I apologize for the lengthy post, but I am seeking an opinion on some proofs we have been asked to construct as well as guidance for one that has puzzled me. I have omitted the theorems and definitions used in order to avoid making the post heavier than it already is. I wanted something fairly solid before seeking any advice. Any opinion/suggestion will be greatly appreciated. Here they are:

Question 1: if m\n, then m <= n (m, n are positive integers)

Proof:

if m\n then n = k.m for some positive integer k

Then m = (n/k)

if k = 1 then m = n

if k > 1 then (n/k) < n i.e. m < n

therefore m <= n (Q.E.D)

Question2: Prove: Let a and b be positive integers, gcd(a; b) = a if and only if a \ b.

Use direct proof for =>and contradiction method for <= .

Proof:step1: (Direct proof)

if gcd(a;b) then a belongs in the prime factorization of b

then b = p1^s1 . p2^s2 . ... .pn^sn . a

Let k = p1^s1 . p2^s2 . ... .pn^sn

then b = k . a and a\b Q.E.D.

step2: (By contradiction)

We assume that a\b and that gcd(a;b)<>a

Therefore there exists t (positive integer) such that gcd(a;b) = t.

But t > a, i.e. t does not divide a.

This contradicts our assumption that there exists an larger integer t such that gcd(a;b) = t since gcd(a;b)=t => t\a and t\b.

Q.E.D

Finally, this last question is the one that has puzzled me somewhat. I am searching for some guidance and not the answer (I know this is not complicated but I cannot seem to see the process).

Question3:Use forward-backward method to prove: If a < 1 and b < 1, then ab + 1 > a + b.