Improve Your Irrational Numbers Skills Today

What type of number is this $\frac{\sqrt{2}}{2}$?

What's the difference between a rational number and an irrational number?

If $r\ne <!--\; \ne \; -->0$ is rational and $i$ is irrational, then $ri$ is irrational?

Does $f([0,1])$ must always contain an irrational number?

Is there any irrational number in decimal with a rational binary expansion?

Prove that if a and b are real numbers such that the product of ab is an irrational number, then either a or b must be an irrational number.

Prove that $\sqrt{7}+\sqrt{3}$ is irrational

Show that arbitrarly close to any rational number there is a real (non-rational) number. In other words, show that to each real $\mathrm{\epsilon <!--\; \epsilon \; -->}>0$ and each rational $r\in <!--\; \in \; -->\mathbb{Q}$ there exists $x\in <!--\; \in \; -->\mathbb{R}\setminus <!--\; \setminus \; -->\mathbb{Q}$ with $|x-<!--\; -\; -->r|<<!--\; <\; -->\mathrm{\epsilon <!--\; \epsilon \; -->}$

Prove that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y<z<x$.

Let $a$ and $b$ be irrational numbers. when do we have $a\cdot <!--\; \cdot \; -->b$ = rational number?
for example $\sqrt{2}\cdot <!--\; \cdot \; -->\sqrt{2}=2$.

Is φ i.e., ${\displaystyle \frac{1+\sqrt{5}}{2}}$ the most irrational number?

Given an irrational number $0<i<1,$. Find two rational numbers a and b such that $0<a<i<b<1$.

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document I suggested the following problem to my friend: prove that there exist irrational numbers a and b such that ab is rational.
Now, his inital solution was like this: let's take a rational number r and an irrational number i. Let's assume $a=r^i$ b=\frac{1}{i}$$
So we have
$a^b=(r^i{)}^{\frac{1}{i}}=r$
which is rational per initial supposition. b is obviously irrational if i is. My friend says that it is also obvious that if r is rational and i is irrational, then $r^i$ is irrational. I quickly objected saying that $r=1$ is an easy counterexample. To which my friend said, OK, for any positive rational number r, other than 1 and for any irrational number i $r^i$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

How can I prove that
nonzero integer linear combination of two rational independent irrational numbers is still a irrational number?That is to say, given two irrational numbers a and b, if a/b is a irrational number too, then for any m,n is nonzero integer, we have that the number ma+nb is a irrational number, why?

Show that there is a rational number and an irrational number between any two real numbers.

The set $\{n\sqrt{2}mod1:<!--\; :\; -->n\in <!--\; \in \; -->\mathbb{N}\}$ is dense in $(0,1)$, it follows from the irrationality of $\sqrt{2}$ (a well-known fact, that fractional parts of multiples of irrational number form a dense in $(0,1)$).

Is $\{n\sqrt{2}mod\sqrt{3}:<!--\; :\; -->n\in <!--\; \in \; -->\mathbb{N}\}$ dense in $(0,\sqrt{3})$ ?

Is $\{n\sqrt{2}mod\mathrm{\pi <!--\; \pi \; -->}:<!--\; :\; -->n\in <!--\; \in \; -->\mathbb{N}\}$ dense in $(0,\mathrm{\pi <!--\; \pi \; -->})$ ?

Probably the answer is yes. Can it be derived from the above, or this is a completely different result requiring much more advanced theory?

Justify that $\tan <!--\; \; -->1^{\circ <!--\; \circ \; -->}$ is an irrational number.

How do you calculate the decimal expansion of an irrational number?

Is there a "positive" definition for irrational numbers?

$n^x$, $x$ is irrational number what is the result?