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

[Jamin2112]

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??

 

[╔(σ_σ)╝]

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 ?

 

[Jamin2112]

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

 

[╔(σ_σ)╝]

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 .

 

[Jamin2112]

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.

 

[╔(σ_σ)╝]

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 :-)

 

[PAllen]

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

 

[╔(σ_σ)╝]

If a approaches b and b approaches a you do not have an interval.

 

[PAllen]

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.

 

[╔(σ_σ)╝]

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.

 

[PAllen]

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.

 

[Jamin2112]

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.