# Proving an inequality by contraposition

Let $a, b, m$ be integers. Prove that if $2a+3b \geq 8$, then $a \geq 3m+3$ or $b \geq −2m+2$.

This my textbook solution:

We will prove the contrapositive implication, which is: If $a < 3m + 3$ and $b < −2m + 2$, then $2a + 3b < 8$. So suppose that $a < 3m + 3$ and $b < −2m + 2$.

(1)Then $a \leq 3m + 2$ and $b \leq 6 −2m + 1$, because all numbers are integers.

(2)Then $2a \leq 6m + 4$ and $3b \leq −6m + 3$ (multiply by $2$ and $3$, respectively).

(3)Then $2a + 3b \leq (6m + 4) + (−6m + 3) = 7$. Then $2a + 3b < 8$, as required.

I pretty much understood everything except for the step when we said that "because all numbers are integers": What do we mean by that and why did we subtracted $1$ and change the inequality? Thank you

"All numbers are integers" just means that every variable within the scope of the problem is by construction an integer (the first line of the problem is "let $a$, $b$, and $m$ be integers.")
As to the apparent subtraction by one, think about it like this: which number is less than or equal to the integer that is less than three? It's two. You'll notice we go from $a<3m+3$ to $a\leq3m+2$. The "subtraction" is really just a result of going from $<$ to the $\leq$.