Prove that the set of sequence with rational terms is uncountable.
Does this mean that rational numbers are countable? do i solve this by contradiction?
There are seventy thousand ways to solve this. Try proving that the set of all $\displaystyle f:\mathbb{N}\to\{0,1\}$ is uncountable (Cantor's Diagonalization method).
Or, note that this is equipotent to the power set of the reals.
Or, note that $\displaystyle \aleph_0^{\aleph_0}=\mathfrak{c}$
etc.
And no. It means that the set of all functions from the naturals to the rationals is uncountable.
A really easy way to solve this may be to notice that the reals are an uncountable set. If you have done some basic analysis then you know that every real number has a sequence of rationals that converges to it. This means there is a surjection from the set of sequence of rational terms into the reals, thus, its cardinality is at least that of the reals.