Question

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

Answer 1

Answer:

First, we'll assume that √a + √b is rational, where a and b are distinct primes  

√a + √b= x, where x is rational  

Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.  

(√a + √b)² = x²  

a + 2√(ab) + b = x²  

2√(ab) = x² - a - b

√(ab) = (x² - a - b) / 2  

Now x, x², a, b and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - a - b) / 2 is rational.  

But since a and b are both primes, then ab is not a perfect square and therefore √(ab) is not rational. But this is a contradiction. Original assumption must be wrong.  

So √a + √b is irrational, where p and q are distinct primes  

We can also show that √a + √b is irrational, where p and q are non-distinct primes, i.e. a= b

We use same method: Assume √a + √b is rational.  

√a + √b = x, where x is rational  

√a + √a = x  

2√a= x  

√a = x/2

Since both x and 2 are rational, and rational numbers are closed under division, then x/2 is rational. But since p is not a perfect square, then √a is not rational. But this is a contradiction. Original assumption must be wrong.  

So √a + √b is irrational, where p and q are non-distinct primes  

∴ √a + √b is irrational, where p and q are primes

Step-by-step explanation:

Please mark brainliest