# Prove in any interval there exists an irrational z

• cmajor47
In summary, the conversation discusses how to prove that there exists an irrational number in any given interval. The suggested approach is to add a rational number to a known irrational number in order to get another irrational number. Further clarification is provided on how to choose the rational number to ensure it falls within the given interval. The final proof presented involves choosing a rational number in the interval (r-a, s-a), where r and s are the endpoints of the given interval and a is a known irrational number.
cmajor47

## Homework Statement

Prove that in any interval there exists an irrational z.

## The Attempt at a Solution

My professor wrote this for me when trying to explain how to prove this:

$a \notin Q$, $\epsilon$ rational

$[r, s]\in a$

$l([r, s])<\frac{\epsilon}{2}$

$s-a<\frac{\epsilon}{2}$

$s-\frac{\epsilon}{2}<a$

$s-\frac{\epsilon}{2}+\epsilon<a+\epsilon$

$s<s+\epsilon<a+\epsilon$

I don't see why you use $[r, s]\in a$ or how my professor went from $s-\frac{\epsilon}{2}+\epsilon<a+\epsilon$ to $s<s+\epsilon<a+\epsilon$

I also don't know where to go from here to show what needs to be proved.
I'd really appreciate any help.

As people were saying in the previous thread, the problem appears to be that much of what you have written down is nonsense ($$[r, s] \in a$$, which is backward, and the dubious inequality deduction you mention).

What might be going on here is an attempt to make the following argument:
Given that there is a rational number in every interval, and the existence of one single irrational number $$a$$, prove that there is an irrational number in every interval.

($$a$$ might be $$\sqrt{2}$$, using the standard proof.)

Here is how this argument would go. Let $$(r,s)$$ be the interval in which we wish to find an irrational number. We try to do this by adding a rational number $$q$$ to $$a$$, because the sum of a rational and an irrational is irrational. So we need to find $$q$$ so that $$r < q + a < s$$. Clearly it suffices to have $$r - a < q < s - a$$, i.e., choose $$q$$ to be any rational number in $$(r - a, s - a)$$. We can do this by hypothesis, so the proof is done.

Does this make sense?

I think so. Is this proof correct:
Let $a\in$ Q, $\epsilon$ rational, where $\epsilon\in[r-a, s-a]$.
Then $r-a<\epsilon<s-a$, so $r<\epsilon+a<s$
So $\epsilon+a\in[r, s]$ and $\epsilon+a$ is irrational.
Therefore in any interval, there exists an irrational number.

But, since $a \in \mathbb{Q}$ and $\epsilon \in \mathbb{Q}$ clearly $a + \epsilon \in \mathbb{Q}$. Thus $a + \epsilon$ cannot be irrational.

Oops, meant to write $a\notin$ Q

## What does it mean to prove that there exists an irrational z in any interval?

To prove that there exists an irrational z in any interval means to show that there is at least one irrational number that lies within a given interval on the real number line. This is typically done by providing a specific example of an irrational number within the given interval.

## Why is it important to prove the existence of irrational numbers in any interval?

Proving the existence of irrational numbers in any interval is important because it helps us better understand the structure of the real number line. It also allows us to make more accurate and precise mathematical calculations and models.

## What techniques are commonly used to prove the existence of irrational numbers in any interval?

There are several techniques that can be used to prove the existence of irrational numbers in any interval. These include proof by contradiction, using the decimal expansion of irrational numbers, and utilizing the properties of rational and irrational numbers.

## Can you give an example of a proof for the existence of an irrational number in any interval?

Yes, one example of a proof for the existence of an irrational number in any interval is to show that the square root of 2 is irrational. This can be done by assuming that the square root of 2 is rational and then showing that this leads to a contradiction, therefore proving it must be irrational.

## How does the existence of irrational numbers in any interval relate to the completeness of the real number system?

The existence of irrational numbers in any interval is closely related to the completeness of the real number system. This is because the completeness property states that between any two real numbers, there exists another real number. This can include irrational numbers, which are necessary to fill in the gaps between rational numbers and make the real number line complete.

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