TITLE: Proving an inequality by contraposition
QUESTION [0 upvotes]: 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

REPLY [1 votes]: "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 $\leq$.
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