$\sqrt{2}\sqrt{2}$ 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?
if $a$ and $b$ are irrational and $a \neq b$, then is $ab$ necessarily irrational?
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proof-verification
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2$\sqrt{2}\sqrt{3}$ – 2017-01-16
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0Do 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. – 2017-01-16
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1Take $a=\pi$ and $b=1/\pi$. – 2017-01-16
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0@barakmanos: the OP is asking for $ab$ irrational. – 2017-01-16
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0@YvesDaoust: Oh... From the title, it seems that OP is asking if $ab$ is necessarily irrational... – 2017-01-16
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0@barakmanos: I agree, the question is a little puzzling. I was trapped mysef. – 2017-01-16
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0Yes I messed this up, I meant to ask must ab always be irrational or is there a case where ab is rational. – 2017-01-16
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0sqrt(2) * sqrt(8) ???? – 2017-01-16
4 Answers
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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 $ab = 2-\sqrt{2} \notin \mathbb{Q}$.
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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}
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0I like ur Glasses. + – 2017-01-16
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0@BabakS. Thanks. – 2017-01-16
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0Implication 2 is wrong (think $a=0$). Implications 3, 4, 7, 8 are not implications. – 2017-01-16
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0What? Implications are implications, not examples. If you do not mean implications, do not use $\implies$. – 2017-01-16
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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.
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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.