MATH SOLVE

2 months ago

Q:
# prove that β2 +β5 is irrational

Accepted Solution

A:

We have to prove that [tex]\sqrt{2}+\sqrt{5}[/tex] is irrational. We can prove this statement by contradiction.Let us assume that [tex]\sqrt{2}+\sqrt{5}[/tex] is a rational number. Therefore, we can express:[tex]a=\sqrt{2}+\sqrt{5}[/tex]Let us represent this equation as:[tex]a-\sqrt{2}=\sqrt{5}[/tex]Upon squaring both the sides:[tex](a-\sqrt{2})^{2}=(\sqrt{5})^{2}\\a^{2}+2-2\sqrt{2}a=5\\a^{2}-2\sqrt{2}a=3\\\sqrt{2}=\frac{a^{2}-3}{2a}[/tex]Since a has been assumed to be rational, therefore, [tex]\frac{a^{2}-3}{2a}[/tex] must as well be rational.But we know that [tex]\sqrt{2}[/tex] is irrational, therefore, from equation [tex]\sqrt{2}=\frac{a^{2}-3}{2a}[/tex] the expression [tex]\frac{a^{2}-3}{2a}[/tex] must be irrational, which contradicts with our claim.Therefore, by contradiction, Β [tex]\sqrt{2}+\sqrt{5}[/tex] is irrational.