How do I prove an either/or inequality?

  • Thread starter Eclair_de_XII
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In summary, the goal is to prove that if 2a+3b≥12m+1, then either a≥3m+1 or b≥2m+1. This can be re-written as ~(P∨Q)≅(~P)∧(~Q)⇒~R, where P:a≥3m+1, Q:b≥2m+1, and R:2a+3b≥12m+1. By assuming a<3m+1 and b<2m+1, it can be shown that 2a+3b<12m+5, but this does not necessarily imply that 2a+3b
  • #1
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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  • #2
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:
  • #3
Hi,

Your statement must be wrong since ##a=5,b=1## and ##m=1## clearly violate it.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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:
  • #8
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.
 

1. How do I prove an either/or inequality using mathematical induction?

To prove an either/or inequality using mathematical induction, you will need to start by showing that the inequality holds true for the base case. Then, you will need to assume that the inequality holds true for an arbitrary value of n and use this assumption to show that it also holds true for n+1. This will complete the inductive step and prove the either/or inequality for all values of n.

2. What is the difference between an either/or inequality and a regular inequality?

An either/or inequality is a type of inequality that presents two possible options or outcomes, whereas a regular inequality presents a single statement that must be true. In an either/or inequality, one of the options must be true, but not necessarily both. This means that the inequality can be proven by showing that at least one of the options is true.

3. Can I use contradiction to prove an either/or inequality?

Yes, you can use contradiction to prove an either/or inequality. To do this, you would assume that both options in the inequality are false and then show that this leads to a contradiction. This would prove that at least one of the options must be true, thus proving the either/or inequality.

4. Are there any common strategies for proving either/or inequalities?

Yes, there are a few common strategies that can be used to prove either/or inequalities. These include direct proof, contradiction, and mathematical induction. Each strategy may be more effective for certain types of inequalities, so it is important to choose the best approach for the specific inequality you are trying to prove.

5. How can I apply either/or inequalities in real life situations?

Either/or inequalities are commonly used in decision-making processes, such as choosing between two options based on certain criteria. They can also be used in probability and statistics to determine the likelihood of certain events occurring. For example, an either/or inequality can be used to determine the probability of getting heads or tails when flipping a coin.

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