.Prove that √5+√2 is irrational.
Answers
Answer:
Let us assume that √2+√5 is a rational number.
A rational number can be written in the form of p/q where p,q are integers and q≠0
√2+√5 = p/q
On squaring both sides we get,
(√2+√5)² = (p/q)²
√2²+√5²+2(√5)(√2) = p²/q²
2+5+2√10 = p²/q²
7+2√10 = p²/q²
2√10 = p²/q² – 7
√10 = (p²-7q²)/2q
p,q are integers then (p²-7q²)/2q is a rational number.
Then √10 is also a rational number.
But this contradicts the fact that √10 is an irrational number.
Our assumption is incorrect
√2+√5 is an irrational number.
Hence proved.
Answer:
Let us assume that √2+√5 is a rational number.
A rational number can be written in the form of p/q where p,q are integers and q≠0
√2+√5 = p/q
On squaring both sides we get,
(√2+√5)² = (p/q)²
√2²+√5²+2(√5)(√2) = p²/q²
2+5+2√10 = p²/q²
7+2√10 = p²/q²
2√10 = p²/q² – 7
√10 = (p²-7q²)/2q
p,q are integers then (p²-7q²)/2q is a rational number.
Then √10 is also a rational number.
But this contradicts the fact that √10 is an irrational number.
Our assumption is incorrect
√2+√5 is an irrational number.
Hence proved.