Try this for example:
Suppose a, b, m are integers such that a < 3m + 1
and b < 2m + 1. Then a ≤ 3m and b ≤ 2m. Therefore 2a ≤ 6m and 3b ≤ 6m. Adding these
inequalities, 2a + 3b ≤ 12m. It follows that 2a + 3b < 12m + 1.
We assumed a < 3m + 1 and b < 2m + 1, and deduced 2a + 3b < 12m + 1. This proves the
contrapositive: If 2a + 3b ≥ 12m + 1, then a ≥ 3m + 1 or b ≥ 2m + 1.
Remarks. This is not true for a, b, m real numbers instead of integers. For example, take
m = 0, a = 1/2, b = 1/3. Then 2a + 3b = 2 and 12m + 1 = 1, so 2a + 3b ≥ 12m + 1, yet
a < 3m + 1 and b < 2m + 1. So the hypothesis that a, b, m are integers can not be ignored.
Also note that the negation of “a ≥ 3m + 1 or b ≥ 2m + 1” is “a < 3m + 1 and b < 2m + 1”.
The original conclusion (a ≥ 3m+1 or b ≥ 2m+1) is true if either condition is true, or both;
it is false if both conditions are false, i.e., a < 3m + 1 and b < 2m + 1.
Here is a different way to understand it. Assume that it is NOT true that a ≥ 3m + 1 or
b ≥ 2m + 1. In other words we are assuming NOT(a ≥ 3m + 1 or b ≥ 2m + 1). Then neither
one of them can be true, so both a < 3m + 1 and b < 2m + 1.
I hope this helps you. This example will help your algebra.
