Question

prova that root5 plus root7 is a irrational number

Answer 1

assume its rational
then
\(\large \sqrt 5+ \sqrt 7=\frac{a}{b} \)
\(\large b\sqrt 5+ b\sqrt 7=a \)
hence both
\(b\large \sqrt 5 \) is rational
and
\(b\large \sqrt 7 \) is rational

Answer 2

using rational root test :
\(x^2 - 5\) has no rational roots \(\implies \sqrt{5}\) is irrational
\(x^2 - 7\) has no rational roots \(\implies \sqrt{7}\) is irrational

since sum of two irrational numbers is irrational, \(\sqrt{5} + \sqrt{7}\) is irrational.

Answer 3

two irrational does not apply the sumation irrational
like
1/pi - 1/pi =0

Answer 4

*sum of two positive irrational numbers

Answer 5

the contradiction u got , was since sqrt 5 irrational and b is integer and
we assumed
\(b \sqrt 5 \) is integer , but
we got
\(b \sqrt 5 \) ( integer * rational \(\neq \) integer )
which is a contradiction