How do I prove an either/or inequality?

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Homework Help Overview

The discussion revolves around a proof involving inequalities with integer variables. The original statement requires proving that if \(2a + 3b \geq 12m + 1\), then either \(a \geq 3m + 1\) or \(b \geq 2m + 1\). Participants are exploring the implications of the inequalities and the validity of their approaches.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss proof techniques, including proof by contradiction and contrapositive. There are attempts to analyze the implications of assuming the negation of the original statement. Some participants question the validity of specific examples and the assumptions made regarding integer values.

Discussion Status

There is ongoing exploration of the proof structure and the implications of the inequalities. Some participants have provided insights that may clarify the original poster's reasoning, while others have raised counterexamples that challenge the assumptions. The discussion remains open with various interpretations being examined.

Contextual Notes

Participants note the challenge of working with integer constraints and the potential simplifications that arise from manipulating the inequalities. There is mention of specific values that violate the original statement, prompting further examination of the conditions under which the inequalities hold.

Eclair_de_XII
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Homework Statement


"Given: ##a,b,c∈ℤ##,
Prove: If ##2a+3b≥12m+1##, then ##a≥3m+1## or ##b≥2m+1##."

Homework Equations


##P:a≥3m+1##
##Q:b≥2m+1##
##R:2a+3b≥12m+1##

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?
 
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Eclair_de_XII said:
But this doesn't necessarily imply that ##2a+3b>12m+1##.
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:
Hi,

Your statement must be wrong since ##a=5,b=1## and ##m=1## clearly violate it.
 
Eclair_de_XII said:

The Attempt at a Solution


Goal: ## not(P∨Q)≅( not P)∧( not Q)⇒ not R##
Note. Tildas are difficult to do in LaTeX. I don't know how to do them. Here is a readable version of your Goal.
 
FactChecker said:
Your proof by contradiction is complete as it is.

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.
 
Eclair_de_XII said:

Homework Statement


"Given: ##a,b,c∈ℤ##,
Prove: If ##2a+3b≥12m+1##, then ##a≥3m+1## or ##b≥2m+1##."

Homework Equations


##P:a≥3m+1##
##Q:b≥2m+1##
##R:2a+3b≥12m+1##

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?

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.
 
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:
Ray Vickson said:
Since a,b,ma, 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 ">".

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

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