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## 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 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?