I know that, in general, numbers in the base $b$ system are of the form: $$\sum_{k=0}^n a_kb^k $$ For example, The number $1729$ written under base ten is written $6C1$ in basis sixteen: $$6C1_{16} = (6\cdot 16^2 ) + (\underbrace{12}_C \cdot16^1)+(1\cdot16^0 ) = 1536+192+1 = 1729_{10}$$
Now, how can I convert a irrational numbers , for example $\sqrt{7}$, to another base $b \in \mathbb{N}$, (for example $2$)?
Thanks
We can use decimals like we normally would. In your example, we do the following.
We search for the largest power of $2$ that fits in $\sqrt7$; that's $2$ itself. Now we look for the largest power that fits in the leftover $\sqrt7-2$; that's $2^{-1}$. Now the largest power that fits in $\sqrt7-2-2^{-1}$ is $2^{-3}$. We won't ever be done computing the digits, since $\sqrt7$ is irrational (we can't do this in base 10 either). In the end, we get
$$\sqrt7=2+2^{-1}+2^{-3}+2^{-6}+2^{-8}+2^{-10}+\cdots$$
and that converts to
$$\sqrt7=(10.1010010101\cdots)_2$$