Prove that if a>b, there is an irrational number between a and b

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Homework Help Overview

The discussion revolves around proving that if \( a > b \), there exists an irrational number between \( a \) and \( b \). Participants explore various mathematical concepts and reasoning related to this proof, particularly focusing on the properties of rational and irrational numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants discuss the midpoint \( (a+b)/2 \) and question whether it can be replaced with an irrational number. Others suggest considering the implications of rational sums and products with irrational numbers.

Discussion Status

The discussion includes various attempts to find a suitable approach to the proof. Some participants provide hints and partial insights, while others express confusion or question the relevance of certain statements. There is a recognition of the need to handle specific cases, such as when \( b + a = 0 \) or when \( b + a \) is irrational.

Contextual Notes

Participants note the importance of finding a value \( k \) such that \( a < k \sqrt{2} < b \) and discuss the density of rational numbers in the real numbers. There is also mention of the Archimedean Property in relation to the proof.

Jamin2112
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Homework Statement



I can't figure this part out

Homework Equations



In the previous part of this problem, I proved that there is a rational number between a and b.

The Attempt at a Solution



Maybe 1 < √2 < 2 ---> a < √2 a < 2a ---> a < √2 a < 2b ... then somehow morph that into a < q < b, where q is an irrational number.

Ideas, please??
 
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In general if you have two points a,b and a<b where is (b+a)/2 ?

Can we put an irrational number in the place of 2 ?
 
╔(σ_σ)╝ said:
In general if you have two points a,b and a<b where is (b+a)/2 ?

Can we put an irrational number in the place of 2 ?

Can we put an irrational number in the place of 2 ? ... I don't think so.

Of course, (a+b)/2 is the midpoint of a and b
 
Well if a +b is rational is (b+a)/ sqrt(2) rational ?

Now if b+a where irrational this may not work but try to think about something that works.

Hint

1/n can be made as small as we want .
 
╔(σ_σ)╝ said:
Well if a +b is rational is (b+a)/ sqrt(2) rational ?

Now if b+a where irrational this may not work but try to think about something that works.

Hint

1/n can be made as small as we want .

I'm know that a rational number multiplied by an irrational number equals an irrational number.
 
Okay, that means I have just provided you 1/3 of your proof. :-) :-)

Handle the case when b+a=0 and when
b+a is irrational.

by the way b+a/sqrt(2) is sitting somewhere between a and b :-)
 
╔(σ_σ)╝ said:
Okay, that means I have just provided you 1/3 of your proof. :-) :-)

Handle the case when b+a=0 and when
b+a is irrational.

by the way b+a/sqrt(2) is sitting somewhere between a and b :-)

Consider a and b approach each other (e.g. both near 1). Then neither b + (a/sqrt(w)) nor (a+b)/sqrt(2) are between a and b.

Consider instead, finding a useful value of k, where:

a < k sqrt(2) < b
 
If a approaches b and b approaches a you do not have an interval.
 
╔(σ_σ)╝ said:
If a approaches b and b approaches a you do not have an interval.

Think. Try 1 and 1.01. Your statements are incorrect and not really relevant. Maybe what you meant was:

(a+b*sqrt(2))/(1+sqrt(2)) is between a and b, but that is not very useful.
 
  • #10
PAllen said:
Think. Try 1 and 1.01. Your statements are incorrect and not really relevant. Maybe what you meant was:

(a+b*sqrt(2))/(1+sqrt(2)) is between a and b, but that is not very useful.

What is should have said is this ... since the rationals are dense in R.
a<x<y<b

take

q = x +(y-x)/sqrt(2)

That is correct.

I see I succeeded in confusing myself in my post by using a,b when I shouldn't have.
 
Last edited:
  • #11
Look, all the OP needs to think about is finding a useful k such that:

a < k sqrt(2) < b

Nothing more complex is needed to solve the problem. I don't want to say more, else I do the whole problem.

Ok, since I see the prior post basically give the answer, I'll give mine. There is a rational between

a/sqrt(2) and b/sqrt(2). Call one such a rational k. Then k*sqrt(2) is an irrational between a and b.
 
  • #12
Okay ... I get it.

a < b implies a/√2 < b/√2. Because a/√2 and b/√2 are real numbers, the Archimedean Property tells us that is a rational number r such that a/√2 < r < b/√2. Therefore a < √2 r < b. Because the product of an irrational number and a rational number equals an irrational number, √2 r is an irrational number and the proof is complete.
 

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