# How do I prove an either/or inequality?

1. Feb 12, 2017

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Given: $a,b,c∈ℤ$,
Prove: If $2a+3b≥12m+1$, then $a≥3m+1$ or $b≥2m+1$."

2. Relevant equations
$P:a≥3m+1$
$Q:b≥2m+1$
$R:2a+3b≥12m+1$

3. The attempt at a solution
Goal: $~(P∨Q)≅(~P)∧(~Q)⇒~R$

Assume that $a<3m+1$ and $b<2m+1$. Then $2a+3b<2(3m+1)+3(2m+1)=12m+5$. But this doesn't necessarily imply that $2a+3b>12m+1$. Can someone help me connect the dots?

2. Feb 12, 2017

### FactChecker

The problem doesn't ask you to prove that. Your proof by contradiction is complete as it is.
CORRECTION: I should have said proof by contrapositive.
CORRECTION 2: I completely missed the point that the OP proves <12m+5 but it needs <12m+1

Last edited: Feb 13, 2017
3. Feb 12, 2017

### Q.B.

Hi,

Your statement must be wrong since $a=5,b=1$ and $m=1$ clearly violate it.

4. Feb 12, 2017

### FactChecker

Note. Tildas are difficult to do in LaTeX. I don't know how to do them. Here is a readable version of your Goal.

5. Feb 12, 2017

### Eclair_de_XII

I was trying to do proof by contra-positive, but I ended up with an inequality in my original post that says nothing about my original statement. I figured that it was only conditionally true. So I ended up with a statement just saying that the statement is false when at least one of the inequalities is false; then I gave a counter-example.

6. Feb 13, 2017

### Ray Vickson

Since $a, b, m$ are all integers, you can re-write the three inequalities in the original question by first removing the "+1" on all three right-hand-sides and replacing "≥" by ">". That works because both sides are integers. It is worth doing---try it and see.

7. Feb 13, 2017

### FactChecker

If a < 3m+1 and you are restricted to integers, then you know that a ≤ A = 3m. Likewise b < 2m+1 implies b ≤ B = 2m. Compare 2a+3b with A+B and 12m+1.
CORRECTION: Should have said "Compare 2a+3b with 2A+3B and 12m+1."

Last edited: Feb 14, 2017
8. Feb 13, 2017

### Eclair_de_XII

Well, that simplified my argument by a whole lot more. Thanks, too, @FactChecker.