Question

show that root a+root b is an irrational if root ab is an irrational number​

Answer 1

Answer:

√a+√b is an irrational number.

Step-by-step explanation:

Let √a+√b be a rational number. There exist two number p and q where q ≠0 and p,q are co prime i.e. p/q =√a+√b

Then,

(p/q)² = (√a+√b)² [squaring both sides]

=› p²/q =a²+b/q ————(¹)

Since p and q are co prime L.H.S. is always fractional and R.H.S. is always integer . If q =1,the equation (¹) is hold good but it was impossible that there was no number whose square is a² +b .

This is the contradiction to our assumption. Hence a+√b is an irrational number.

Answer 2

Answer:

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Step-by-step explanation:

Given, √(ab) is an irrational number.

Let √a + √b is rational.

So, √a + √b can be written as p/q form where q ≠ 0

=> √a + √b = p/q

Squaring on both side, we get

=> (√a + √b)2 = (p/q)2

=> a + b + 2√a * √b = p2 /q2

=> a + b + 2√(ab) = p2 /q2

=> 2√(ab) = p2 /q2 - (a + b)

=> √(ab) = {p2 /q2 - (a + b)}/2   ...........1

Since √a + √b is rational, So √a , √b is also rational.

But LHS of equation 1 is irrational.

which contradict our assumption.

Hence, √a + √b is an irrational number.