Can Irrational Numbers Be Found Between Any Two Real Numbers?

[fishturtle1]

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 $\lbrace r + \sqrt{2} : r \epsilon \mathbb{Q}\rbrace \subset \mathbb{I}$)

Homework Equations

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?

 

[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.

 

[tnich]

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 $\lbrace r + \sqrt{2} : r \epsilon \mathbb{Q}\rbrace \subset \mathbb{I}$)

Homework Equations

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.

 

[tnich]

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 $a$.

 

[fishturtle1]

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]

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?

 

[fishturtle1]

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 $a < x < b$.
$\square$.

 

[tnich]

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