# Denseness of irrationals in R

• fishturtle1
In summary, the proof first shows that the set of real numbers that are not rational is contained within the set of irrational numbers. Then, it uses the fact that irrationals are dense in the real numbers to show that for any two real numbers a and b where a is less than b, there exists an irrational number x between them. However, the proof fails when trying to show that the mean of a and b is irrational because of a mistake in defining one of the variables.

## 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 that's 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?

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:
fishturtle1 said:

## 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 that's 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 said:
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!
Math_QED said:
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...

fishturtle1 said:
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?

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