Respuesta :
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, [tex]\sqrt{2}+\sqrt{5}[/tex] is irrational.
Step-by-step explanation:
Let us assume [tex]\bf \sqrt{2}+\sqrt{5} [/tex] as rational.
[tex]\therefore \: \:\bf \sqrt{2}+\sqrt{5} = \dfrac{a}{b}[/tex]
- (where a and b are coprimes)
[tex]\implies\:\:\bf \sqrt{5} = \dfrac{a}{b} - \sqrt{2} [/tex]
[tex]\implies\:\:\bf(\sqrt{5})^{2} = \bigg( \dfrac{a}{b} -\sqrt{2}\bigg)^{2} [/tex]
- squaring on both sides
[tex]\implies\:\:\bf 5 = \dfrac{a^{2} }{b^{2}} - 2 \bigg(\dfrac{a}{b}\bigg) (\sqrt{2}) + (\sqrt{2})^{2}[/tex]
[tex]\implies\:\:\bf 5 = \dfrac{a^{2} }{b^{2}} - 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg) + 2[/tex]
[tex]\implies\:\:\bf 5-2 = \dfrac{a^{2}}{b^{2}}- 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)[/tex]
[tex]\implies\:\:\bf 3 = \dfrac{a^{2}}{b^{2}}- 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)[/tex]
[tex]\implies\:\:\bf 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)= \dfrac{a^{2}}{b^{2}} -3[/tex]
[tex]\implies\:\:\bf 2 \sqrt{2} \bigg( \dfrac{a}{b} \bigg)=\dfrac{a^{2}-3b^{2}}{b^{2}}[/tex]
[tex]\implies\:\:\bf 2 \sqrt{2} = \bigg( \dfrac{a^{2}-3b^{2}}{b^{2}} \bigg) \bigg( \dfrac{b}{a}\bigg) [/tex]
[tex]\implies\:\:\bf \sqrt{2} =\dfrac{a^{2}-3b^{2} }{2ab}[/tex]
- [tex]\bf Irrational \cancel{=} Rational [/tex]
[tex]\textsf Hence, our assumption is wrong [/tex]
[tex] \therefore \:\:\sf {{\bf{\sqrt{2}+\sqrt{5}}} is irrational }[/tex]
[tex]\mathfrak\purple{{\purple{\bf{H}}}ence\: {\purple{\bf{P}}}roved}[/tex]
