Prove: Let x and y be positive real numbers. If x ≤ y, then √x ≤ √y.

Question

asked Jan 27, 2020 61.1k views

1 Answer

Sastrija · Answer 1 · 2020-02-01T18:15:46+0000

Suppose that
x,y> 0 and that
$x\leq y$ . We can subtract
from both sides to obtain
$y-x \geq 0$ . Recall that
(a+b)(a-b)=a^2-b^2 . Now we can replace
and
in this identity with
√(y) and
√(x) respectively and we get
(√(y)-√(x))(√(y)+√(x))=y-x . From this it follows that
$(√(y)-√(x))(√(y)+√(x)) \geq 0$ . Since
√(y)+√(x) > 0 , we can divide by it both sides of the inequality without altering its direction and end up with
$√(y)-√(x) \geq 0$ . Now we just need to add
√(x) to both sides and conclude that
$√(y) \geq √(x)$ , which finishes our proof.