Prove that √2 √5 as irrational number?
Let √2+√5 be a rational number.
A rational number can be written in the form of p/q where p,q are integers.
√2+√5 = p/q
Squaring on both sides,

p,q are integers then (p^2-7q^2)/2q is a rational number.
Then √10 is also a rational number.
But this contradicts the fact that √10 is an irrational number.
.�. Our supposition is false.
√2+√5 is an irrational number.
Hence proved.
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Prove that √2 √5 as irrational number?
Introduction:
√2 and √5 are both irrational numbers. The product of two irrational numbers is also an irrational number. In this case, √2 * √5 is irrational. In this response, we will prove that √2 * √5 is irrational.
Proof:
Assume that √2 * √5 is a rational number. This means that we can write it in the form of p/q where p and q are integers and q ≠ 0.
√2 * √5 = p/q
Squaring both sides, we get:
2 * 5 = p^2/q^2
10q^2 = p^2
This means that p^2 is a multiple of 10, which implies that p is also a multiple of 10.
Let p = 10k, where k is an integer.
Substituting this value of p in the equation 10q^2 = p^2, we get:
10q^2 = (10k)^2
10q^2 = 100k^2
q^2 = 10k^2
This means that q^2 is a multiple of 10, which implies that q is also a multiple of 10.
But we assumed that p and q have no common factors. This is a contradiction since both p and q are multiples of 10.
Therefore, our assumption that √2 * √5 is a rational number is false.
Conclusion:
Hence, we have proved that √2 * √5 is an irrational number.