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

• Jamin2112
In summary, the homework statement is that there is a rational number between a and b. The attempt at a solution is to find a useful k such that: a < k sqrt(2) < 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.

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.

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:
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.

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.

## 1. What does it mean for a number to be irrational?

An irrational number is a number that cannot be expressed as a ratio of two integers. This means that it cannot be written as a simple fraction and its decimal representation is non-terminating and non-repeating.

## 2. How do you prove that there is an irrational number between two given numbers?

To prove that there is an irrational number between two given numbers, we can use the fact that between any two distinct real numbers, there exists an infinite number of rational and irrational numbers. We can also use the theorem that states that the square root of any prime number is irrational.

## 3. What is the significance of proving that there is an irrational number between two numbers?

Proving that there is an irrational number between two numbers is significant because it shows that there is a greater complexity in the number system than just rational numbers. It also helps to understand the density of real numbers and their relationship with each other.

## 4. Can you give an example of two numbers where there is an irrational number between them?

Yes, for example, between 2 and 3, there is an irrational number such as √2, which is approximately 1.414. This number cannot be expressed as a ratio of two integers and is therefore irrational.

## 5. Is it possible to have more than one irrational number between two given numbers?

Yes, it is possible to have an infinite number of irrational numbers between two given numbers. This is because the irrational numbers are infinitely dense in the real number system, meaning that between any two irrational numbers, there are infinitely many other irrational numbers.

• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
3K
• Calculus and Beyond Homework Help
Replies
15
Views
5K
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Precalculus Mathematics Homework Help
Replies
30
Views
2K
• Calculus and Beyond Homework Help
Replies
7
Views
2K
• Precalculus Mathematics Homework Help
Replies
13
Views
1K