Question

prove that
$\sqrt{2} \\ \sqrt{3} \\ \sqrt{5}$
are irrational numbers

Answer 1

Hints:

Suppose there exist coprime a,b∈Za,b∈Z s.t.

2–√+3–√=ab⟹6–√=a22b2−52=a2−5b22b2
2+3=ab⟹6=a22b2−52=a2−5b22b2
If you already know 6–√6 is irrational then you're already done, otherwise prove it as with 2–√2 , say:

6–√=pq,(p,q)=1⟹6q2=p2⟹2∣p
6=pq,(p,q)=1⟹6q2=p2⟹2∣p
and thus we can write

6–√=2p′q⟹2∣qalso , and this is a contradiction

Answer 2

proof: let us assume that
$\sqrt{2}$
is rational
so, we can find integers r and s (s not equal 0) such that
$\sqrt{2} = \frac{r}{s}$
suppose r and s have a common factor other than 1, then we devide by the common factor to get
$\sqrt{2} = \frac{a}{b}$
where a and b are coprime
so
$b \sqrt{2} = a$
squaring both side and rearranging, we get
$2 {b}^{2} = {a}^{2}$
therefore, 2 divides
${a}^{2}$
now by theorem 1.3, it follows that 2 divides a.
so, we can write a=2c for some integer c.
*but this contradicts the fact that a and b have no common factors other than 1.
this contradiction has arisen because of our incorrect assumption that
$\sqrt{2}$
is a rational. so, we conclude that
$\sqrt{2}$
is irrational