Question

.prove that if a,b,c and d are positive rational no. such that a+ √b = c+√d then either a=c & b=d or b&d are square of rational.

Answer 1

Given a + √b = c + √d
Case (i): Let a=c
⇒ a + √b = c + √d becomes
a + √b = a + √d
⇒ √b = √d
∴ b = d
Case (ii): Let a ≠ c
Let us take a = c + k where k is a rational number not equal to zero.
⇒ a + √b = c + √d becomes
(c + k) + √b = c + √d
⇒ k + √b = √d
Let us now square on both the sides,
⇒ (k + √b)2 = (√d)2
⇒ k2 + b + 2k√b = d
⇒ 2k√b = d – k2 – b
 
Notice that the RHS  is a rational number.
Hence √b is a rational number
This is possible only when b is square of a rational number.
Thus d is also square of a rational number as k + √b = √d.... Hope it helps

Answer 2

Given a + √b = c + √d Case (i): Let a=c ⇒ a + √b = c + √d becomes a + √b = a + √d⇒ √b = √d ∴ b = d Case (ii): Let a ≠ c Let us take a = c + k where k is a rational number not equal to zero. ⇒ a + √b = c + √d becomes (c + k) + √b = c + √d ⇒ k + √b = √d Let us now square on both the sides, ⇒ (k + √b)2 = (√d)2 ⇒ k2 + b + 2k√b = d ⇒ 2k√b = d – k2 – b  Notice that the RHS  is a rational number. Hence √b is a rational numberThis is possible only when b is square of a rational number. Thus d is also square of a rational number as k + √b = √d