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MrRissso [65]
3 years ago
12

Prove that √2 +√5 is irrational

Mathematics
1 answer:
Sindrei [870]3 years ago
4 0

We have to prove that \sqrt{2}+\sqrt{5} is irrational. We can prove this statement by contradiction.

Let us assume that \sqrt{2}+\sqrt{5} is a rational number. Therefore, we can express:

a=\sqrt{2}+\sqrt{5}

Let us represent this equation as:

a-\sqrt{2}=\sqrt{5}

Upon squaring both the sides:

(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}

Since a has been assumed to be rational, therefore, \frac{a^{2}-3}{2a} must as well be rational.

But we know that \sqrt{2} is irrational, therefore, from equation \sqrt{2}=\frac{a^{2}-3}{2a} the expression \frac{a^{2}-3}{2a} must be irrational, which contradicts with our claim.

Therefore, by contradiction,  \sqrt{2}+\sqrt{5} is irrational.

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