I want to prove that there are infinitely many irrational numbers! How can I do that?
I don't know where or how to start so any hint is appreciated.
Thanks! :)
I want to prove that there are infinitely many irrational numbers! How can I do that?
I don't know where or how to start so any hint is appreciated.
Thanks! :)
On
Several ideas :
Based on the fact that the set of natural numbers is infinite:
Take your favourite irrational number, say $\;\pi\;$, and now look at the set $\;\{n\pi\;;\;n\in\Bbb N\}\;$ . Prove this set is infinite and all its elements are irrational
Based on the fact that $\;|\Bbb R|=2^{\aleph_0}>\aleph_0=|\Bbb Q|=|\Bbb N|\;$ : By difference of cardinalities , if $\;Irr=$ the set of irrational numbers, then
$$|Irr|=|\Bbb R\setminus\Bbb Q|=2^{\aleph_0}-\aleph_0=2^{\aleph_0}\;\;\text{(yes, that difference is ugly...)}$$
On
The radical of any order $\geqslant2$ out of any prime is irrational, for the same reasons that $\sqrt2$ also is. And since the number of primes is infinite... $($Unless what you wanted to prove is that the number of irrationals is uncountable, in which case this approach is not enough...$)$
HINT: Show that there is an injection from an infinite set, e.g. $\Bbb N$, into the set of irrational numbers. That is, send every natural number $n$ to some irrational number. Also, recall that if we take a non-zero rational number, its sum and product with any irrational number is irrational.