# Denseness of irrationals in R

## Homework Statement

Let ##\mathbb{I}## be the set of real numbers that are not rational; elements of ##\mathbb{Z}## are called irrational numbers. Prove if ##a < b##, then there exists ##x \epsilon \mathbb{I}## such that ##a < x < b##. (Hint: First show ##\lbrace r + \sqrt{2} : r \epsilon \mathbb{Q}\rbrace \subset \mathbb{I}##)

## The Attempt at a Solution

Well I can show the hint is true...
Proof: Let ##r = \frac mn \epsilon \mathbb{Q}##. Suppose, by way of contradiction, that ##r + \sqrt{2} = \frac pq \epsilon \mathbb{Q}##. Then ##\sqrt{2} = \frac pq - \frac mn = \frac{pn - mq}{qn} = \frac{p'}{q'}## where ##p' = pn - mq## and ##q' = qn##. Thus, ##\sqrt{2}## is rational, a contradiction. We can conclude ##r + \sqrt{2}## is irrational, thus ##\lbrace r + \sqrt{2} : r \epsilon \mathbb{Q} \rbrace \subset \mathbb{I}##. ##\square##

So I'm thinking given any a, b such that ##a < b##, we need to add/subtract something starting at ##\sqrt{2}## so that we end up somewhere between ##a## and ##b##..

Proof: We will show the irrationals are dense in ##\mathbb{R}##. Suppose ##a, b \epsilon \mathbb{R}## such that ##a < b##. Then there exists ##l_1, l_2 \epsilon \mathbb{R}## such that ##a = \sqrt{2}\cdot l_1## and ##b = \sqrt{2}\cdot l_2##. So ##a < \frac{a+b}{2} = \frac{\sqrt{2}(l_1+l_2)}{2} < b##. So ##a < \frac{l_1+l_2}{\sqrt{2}} < b##.

Now we show ##\frac{l_1+l_2}{\sqrt{2}}## is irrational. We proceed by contradiction. Suppose ##\frac{l_1+l_2}{\sqrt{2}}## is rational. Then ##\frac{l_1+l_2}{\sqrt{2}} = \frac pq## where ##p,q## are relatively prime integers. Then ##l_1+l_2 = \frac{p\sqrt{2}}{q}##. So ##\sqrt{2} = \frac{q(l_1+l_2)}{p}##. Let ##h = q(l_1+l_2) \epsilon \mathbb{Z}##. So ##\sqrt{2} = \frac hp##. Thus ##\sqrt{2}## is rational, a contradiction.

We can conclude there exists an irrational number ##x## such that ##a < x < b##.
##\square##

But this can't be right because the number we showed to be irrational was the mean of ##a## and ##b## so thats saying ##3 < 5## implies 4 is irrational...

I didn't really think about fractions being in simplified form because isn't the set of rationals a bunch of equivalence classes? like 4/10 isn't in reduced form but its still rational.... But I'm thinking maybe this is why my proof fails for some reason?

My question is where did my proof go wrong?

## Answers and Replies

member 587159
Well I can show the hint is true...

Proof: Let ##r = \frac mn \epsilon \mathbb{Q}##. Suppose, by way of contradiction, that ##r + \sqrt{2} = \frac pq \epsilon \mathbb{Q}##. Then ##\sqrt{2} = \frac pq - \frac mn = \frac{pn - mq}{qn} = \frac{p'}{q'}## where ##p' = pn - mq## and ##q' = qn##. Thus, ##\sqrt{2}## is rational, a contradiction. We can conclude ##r + \sqrt{2}## is irrational, thus ##\lbrace r + \sqrt{2} : r \epsilon \mathbb{Q} \rbrace \subset \mathbb{I}##. ##\square##

Easier:

Suppose ##r + \sqrt{2}## is rational. Then, since the rationals are closed under addition, it follows that ##\sqrt{2} = r + \sqrt{2} + (-r)## is rational. An obvious contradiction.

Let ##h = q(l_1+l_2) \epsilon \mathbb{Z}##.

This line is wrong.

Did you learn about the sequential characterisation of closures? If you did, here's a proof that uses this, but it is not difficult to find more elementary proofs.

You have to prove that ##\operatorname{cl}(\mathbb{I}) := \overline{\mathbb{I}} = \mathbb{R}##.

So, let ##x \in \mathbb{R}##. We will show that we can write ##x## as a limit of irrationals, using your hint.

If ##x## was irrational, just take a constant sequence. If ##x## is rational, no worries. Define the sequence ##(x_n)_n## by ##x_n:= x + \frac{\sqrt{2}}{n}##. Then, ##(x_n)_n## is a sequence that lives in the irrationals with limit ##x##, and you are done.

Last edited by a moderator:
tnich
Homework Helper

## Homework Statement

Let ##\mathbb{I}## be the set of real numbers that are not rational; elements of ##\mathbb{Z}## are called irrational numbers. Prove if ##a < b##, then there exists ##x \epsilon \mathbb{I}## such that ##a < x < b##. (Hint: First show ##\lbrace r + \sqrt{2} : r \epsilon \mathbb{Q}\rbrace \subset \mathbb{I}##)

## The Attempt at a Solution

Well I can show the hint is true...
Proof: Let ##r = \frac mn \epsilon \mathbb{Q}##. Suppose, by way of contradiction, that ##r + \sqrt{2} = \frac pq \epsilon \mathbb{Q}##. Then ##\sqrt{2} = \frac pq - \frac mn = \frac{pn - mq}{qn} = \frac{p'}{q'}## where ##p' = pn - mq## and ##q' = qn##. Thus, ##\sqrt{2}## is rational, a contradiction. We can conclude ##r + \sqrt{2}## is irrational, thus ##\lbrace r + \sqrt{2} : r \epsilon \mathbb{Q} \rbrace \subset \mathbb{I}##. ##\square##

So I'm thinking given any a, b such that ##a < b##, we need to add/subtract something starting at ##\sqrt{2}## so that we end up somewhere between ##a## and ##b##..

Proof: We will show the irrationals are dense in ##\mathbb{R}##. Suppose ##a, b \epsilon \mathbb{R}## such that ##a < b##. Then there exists ##l_1, l_2 \epsilon \mathbb{R}## such that ##a = \sqrt{2}\cdot l_1## and ##b = \sqrt{2}\cdot l_2##. So ##a < \frac{a+b}{2} = \frac{\sqrt{2}(l_1+l_2)}{2} < b##. So ##a < \frac{l_1+l_2}{\sqrt{2}} < b##.

Now we show ##\frac{l_1+l_2}{\sqrt{2}}## is irrational. We proceed by contradiction. Suppose ##\frac{l_1+l_2}{\sqrt{2}}## is rational. Then ##\frac{l_1+l_2}{\sqrt{2}} = \frac pq## where ##p,q## are relatively prime integers. Then ##l_1+l_2 = \frac{p\sqrt{2}}{q}##. So ##\sqrt{2} = \frac{q(l_1+l_2)}{p}##. Let ##h = q(l_1+l_2) \epsilon \mathbb{Z}##. So ##\sqrt{2} = \frac hp##. Thus ##\sqrt{2}## is rational, a contradiction.

We can conclude there exists an irrational number ##x## such that ##a < x < b##.
##\square##

But this can't be right because the number we showed to be irrational was the mean of ##a## and ##b## so thats saying ##3 < 5## implies 4 is irrational...

I didn't really think about fractions being in simplified form because isn't the set of rationals a bunch of equivalence classes? like 4/10 isn't in reduced form but its still rational.... But I'm thinking maybe this is why my proof fails for some reason?

My question is where did my proof go wrong?
You defined ##l_1## and ##l_2## as real numbers and ##q## as an integer and then said ##q(l_1+l_2) \epsilon \mathbb{Z}##. So now you have forced ##l_1+l_2## to be EDIT: rational, which means ##a+b = \sqrt 2(l_1+l_2)## has to be irrational.

Last edited:
tnich
Homework Helper
You defined ##l_1## and ##l_2## as real numbers and ##q## as an integer and then said ##q(l_1+l_2) \epsilon \mathbb{Z}##. So now you have forced ##l_1+l_2## to be EDIT: rational, which means ##a+b = \sqrt 2(l_1+l_2)## has to be irrational.
I suggest that you consider two cases: 1) ##a## is rational, and 2) ##a## is irrational. Then choose an irrational number or rational number, respectively, less than ##b-a## and add it to ##a##.

Thanks for both replies!
Easier:

Suppose r+√2r+2r + \sqrt{2} is rational. Then, since the rationals are closed under addition, it follows that √2=r+√2+(−r)2=r+2+(−r)\sqrt{2} = r + \sqrt{2} + (-r) is rational. An obvious contradiction.
thats awesome

just for the sake of it... Let ##a## be irrational and ##r## be rational. Suppose, by way of contradiction, that ##a+r## is rational. Since rationals are closed under addition, we can see ##(a+r)+-r = a+(r+-r) = a +0 = a##. So ##a## is rational, a contradiction. We can conclude the sum of a rational and irrational is irrational.

I don't think I learned about sequential characteristics of closure yet but metric space is in a few chapters so yea.. I'd like to come back to this.

Using post #4's suggestion
Proof: We will show ##\mathbb{I}## is dense in ##\mathbb{R}##. Let ##a < b## and consider 2 cases:

case 1: ##a \epsilon \mathbb{I}##. We know there are infinitely many rationals in ##(a,b)##. Let ##q_1, q_2## be rationals in ##(a,b)## such that ##q_1 < q_2##. Then ##\frac{q_2 - q_1}{2} < b - a##. We know ##\frac{q_2 - q_1}{2}## is rational, so ##a + \frac{q_2 - q_1}{2}## is irrational and in ##(a,b)##.

case 2: ##a \epsilon \mathbb{Q}##. It seems ##(\sqrt{2} + (b-a)\cdot l) \epsilon (a,b)## for some ##l \epsilon \mathbb{Z}##. Still working on proving this...

tnich
Homework Helper
Thanks for both replies!

thats awesome

just for the sake of it... Let ##a## be irrational and ##r## be rational. Suppose, by way of contradiction, that ##a+r## is rational. Since rationals are closed under addition, we can see ##(a+r)+-r = a+(r+-r) = a +0 = a##. So ##a## is rational, a contradiction. We can conclude the sum of a rational and irrational is irrational.

I don't think I learned about sequential characteristics of closure yet but metric space is in a few chapters so yea.. I'd like to come back to this.

Using post #4's suggestion
Proof: We will show ##\mathbb{I}## is dense in ##\mathbb{R}##. Let ##a < b## and consider 2 cases:

case 1: ##a \epsilon \mathbb{I}##. We know there are infinitely many rationals in ##(a,b)##. Let ##q_1, q_2## be rationals in ##(a,b)## such that ##q_1 < q_2##. Then ##\frac{q_2 - q_1}{2} < b - a##. We know ##\frac{q_2 - q_1}{2}## is rational, so ##a + \frac{q_2 - q_1}{2}## is irrational and in ##(a,b)##.

case 2: ##a \epsilon \mathbb{Q}##. It seems ##(\sqrt{2} + (b-a)\cdot l) \epsilon (a,b)## for some ##l \epsilon \mathbb{Z}##. Still working on proving this...
Let's simplify case 1. What is a rational number ##q < b-a##? Try ##q = \frac 1 n## where ##n \in \mathbb{Z^+}##. How can you find a value of ##n## that satisfies the inequality?

Let's simplify case 1. What is a rational number ##q < b-a##? Try ##q = \frac 1 n## where ##n \in \mathbb{Z^+}##. How can you find a value of ##n## that satisfies the inequality?
So ##\frac 1n < b - a## is equivalent to ##n > \frac{1}{b-a}##. To find an integer ##n## satisfying this we can take ##n = \lceil \frac{1}{b-a} \rceil##.

So for case 1: Let ##a## be irrational. Observe that ##\frac 1n < b - a## is equivalent to ##n > \frac{1}{b-a}##. An integer solution for this is ##n := \lceil \frac{1}{b-a} \rceil##. Since a is irrational and ##\frac 1n## is rational, we can conclude ##a + \frac 1n## is irrational. Also ##a < a + \frac 1n < b##.

case 2: Let ##a## be rational. From our case 1, we can see ##n := \lceil \frac{1}{b-a} \rceil + \sqrt{2}## also satisfies ##\frac 1n < b - a##. Also, ##n## is the sum of a rational number and irrational number, thus ##\frac 1n## is irrational. By the same logic, ##a + \frac 1n## is irrational. Also ##a < a + \frac 1n < b##.

We can conclude for all ##a,b \epsilon \mathbb{R}## such that ##a < b##, there exists an irrational number ##x## such that ##a < x < b##.
##\square##.

Last edited:
tnich
Homework Helper
So ##\frac 1n < b - a## is equivalent to ##n > \frac{1}{b-a}##. To find an integer ##n## satisfying this we can take ##n = \lceil \frac{1}{b-a} \rceil##.

So for case 1: Let ##a## be irrational. Observe that ##\frac 1n < b - a## is equivalent to ##n > \frac{1}{b-a}##. An integer solution for this is ##n := \lceil \frac{1}{b-a} \rceil##. Since a is irrational and ##\frac 1n## is rational, we can conclude ##a + \frac 1n## is irrational. Also ##a < a + \frac 1n < b##.
Right, now can you modify that argument so prove case 2?

fishturtle1
Right, now can you modify that argument so prove case 2?
I edited case 2 into the previous post, thank you for your help