First of all, let's make the following observations:
15 is an integer, so in particular it is rational. The sum of two rational is rational. This means that [tex]15+17\sqrt{3}[/tex] is irrational if and only if [tex]17\sqrt{3}[/tex] is irrational.
Similarly, since 17 is rational and the multiplication of two rationals is rational, we have that [tex]17\sqrt{3}[/tex] is irrational if and only if [tex]\sqrt{3}[/tex] is irrational.
The proof that [tex]\sqrt{3}[/tex] is irrational is the following: suppose by contradiction that we could write
[tex]\sqrt{3}=\dfrac{a}{b}[/tex]
where a and b are two integers with no common factors.
Squaring both sides, we have
[tex]\dfrac{a^2}{b^2}=3 \iff a^2 = 3b^2[/tex]
So, [tex]a^2[/tex] is a multiple of 3. This happens if and only if a itself is a multiple of 3. So, we can write [tex]a=3k[/tex] and the expression becomes
[tex](3k)^2 = 3b^2 \iff 9k^2 = 3b^2 \iff 3k^2 = b^2[/tex]
So, [tex]b^2[/tex] is a multiple of 3 as well, and this happens if and only if b is itself a multiple of 3.
So, we started with the assumption that we could write [tex]\sqrt{3}[/tex] as a reduced fraction, but this assumption led to a contradiction.
We deduce that [tex]\sqrt{3}[/tex] is irrational, and so are [tex]17\sqrt{3}[/tex] and [tex]15+17\sqrt{3}[/tex]