Prove that, if a, b, c and d be positive rationals such that, a+√b=c+√d, then either a=c and b=d or b and d are squares of rationals.
If a=c, then a+√b=c+√d⇒√b=√d⇒b=d
So, let a≠c. Then, there exists a positive rational number x such that a=c+x.
Now,
⇒a+√b=c+√d
⇒c+x+√b=c+√d [∵a=c+x]
⇒x+√b=√d
⇒(x+√b)2=(√d)2
⇒x2+2√bx+b=d
⇒√b=d−x2−b2x
⇒√b is rational [∴d,x,b are rationals∴d−x2−b22xis rational]
⇒ b is the square of a rational number.
From(i), we have
√d=x+√b
⇒ √d is rational
⇒ d is the square of a rational number.
Hence, either a=c and b=d or b and d are the squares of rationals.