$\sqrt{2}\sqrt{2}$ is rational so it is not the case $\forall$ m,n $\in$ $\Bbb{R} - \Bbb{Q}$ , mn $\in$ $\Bbb{R} - \Bbb{Q}$. What about if m $\neq$ n? Is there a case where m $\neq$ n and $mn$ is rational?
Asked
Active
Viewed 477 times
6
-
2$\sqrt{2}\sqrt{3}$ – Nosrati Jan 16 '17 at 08:04
-
Do you really ask what you're think you're asking? You asking for an example where the product of two irrational numbers is irrational. $\sqrt2$ and $\sqrt3$ is such an example, both are irrational, differing and their product $\sqrt6$ is irrational as well. – skyking Jan 16 '17 at 08:14
-
1Take $a=\pi$ and $b=1/\pi$. – barak manos Jan 16 '17 at 08:19
-
@barakmanos: the OP is asking for $ab$ irrational. – Jan 16 '17 at 08:20
-
@YvesDaoust: Oh... From the title, it seems that OP is asking if $ab$ is necessarily irrational... – barak manos Jan 16 '17 at 08:21
-
@barakmanos: I agree, the question is a little puzzling. I was trapped mysef. – Jan 16 '17 at 08:22
-
Yes I messed this up, I meant to ask must ab always be irrational or is there a case where ab is rational. – Anthony O Jan 16 '17 at 08:35
-
sqrt(2) * sqrt(8) ???? – Saketh Malyala Jan 16 '17 at 08:50
4 Answers
7
Counter example for first case, and an example for second case: $a = \sqrt{2}, b = \dfrac{1}{\sqrt{2}}$. $a \neq b$, yet $ab = 1 \in \mathbb{Q}$. For the other case, take $a = \sqrt{2}-1, b = \sqrt{2}$, $a \neq b$, and $ab = 2-\sqrt{2} \notin \mathbb{Q}$.
DeepSea
- 77,651
1
It's trivial that \begin{eqnarray} && a\in\mathbb{Q}\,,\,b\in\mathbb{Q}\,\Rightarrow ab\in\mathbb{Q}\\ && a\in\mathbb{Q}\,,\,b\in\mathbb{Q}^c\,\Rightarrow ab\in\mathbb{Q}^c\,\,\,,\,a\neq0\\ \end{eqnarray} but with other cases: \begin{eqnarray} && \sqrt{2}\in\mathbb{Q}^c\,,\,\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, 2\in\mathbb{Q}\\ && \sqrt{3}\in\mathbb{Q}^c\,,\,\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, \sqrt{6}\in\mathbb{Q}^c\\ \end{eqnarray} More, that's trivial that \begin{eqnarray} && a\in\mathbb{Q}\,,\,b\in\mathbb{Q}\,\Rightarrow a+b\in\mathbb{Q}\\ && a\in\mathbb{Q}\,,\,b\in\mathbb{Q}^c\,\Rightarrow a+b\in\mathbb{Q}^c\\ \end{eqnarray} but with other cases: \begin{eqnarray} && \sqrt{2}\in\mathbb{Q}^c\,,\,-\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, 0\in\mathbb{Q}\\ && \sqrt{3}\in\mathbb{Q}^c\,,\,\sqrt{2}\in\mathbb{Q}^c\,\,\,\,\text{but}\,\,\, \sqrt{3}+\sqrt{2}\in\mathbb{Q}^c\\ \end{eqnarray}
Nosrati
- 29,995
-
-
-
Implication 2 is wrong (think $a=0$). Implications 3, 4, 7, 8 are not implications. – Did Jan 16 '17 at 09:08
-
What? Implications are implications, not examples. If you do not mean implications, do not use $\implies$. – Did Jan 16 '17 at 09:34
0
You are asking if $\forall m,n\in\mathbb R, m\ne n\implies mn\in \mathbb Q$.
That would mean that $\forall m\in\mathbb R-\mathbb Q, m\ne\sqrt2\implies m\sqrt2\in \mathbb Q$.
There are many more counterexamples than examples, as $\mathbb R-\mathbb Q$ is uncountable, while $\mathbb Q$ is.
0
Be $q$ a rational number other than $0$. Be $a$ an irrational number with $a^2\ne q$. Then $b=\frac{q}{a}$ is an irrational number that is not equal to $a$, but $ab=q$ is rational by construction.
So the product of two different irrational numbers is not always irrational. Indeed, the number of counterexamples is uncountable (as for any given $q$ almost all irrational numbers can be chosen as $a$).
Interestingly this means there are as many pairs of irrational numbers whose product is rational as there are products of irrational numbers whose product is irrational, despite there being more irrational numbers than rational numbers.
celtschk
- 43,384