Math, asked by roy2005tia, 3 months ago

prove that
 \sqrt{6}  +  \sqrt{8}
is an irrational number. ​

Answers

Answered by Anonymous
9

Assume √6 + √8 = a where a ∈ Q.

And √6 + √8 can be written in form of p/q where p, q are integers and q ≠ 0.

∴ a² = (√6 + √8)²

⇒ a² = 14 + 8√3

⇒ (a² - 14)/8 = √3

That means, √3 is a rational number. But this contradicts the fact that (√3) is an irrational number.

∴ (√6 + √8) is an irrational number.

More:-

Any real number which cannot be expressed in form of p/q where (p, q) ∈ Z and q ≠ 0 are called irrational numbers.

Expressions in form of p where p is any prime number are irrationals.

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