If x and y are positive integers, is x > y?
(1)\(x^2\) < y
(2) The square root of x is less than y
Since x > 0 and y > 0.
Statement 1)\(x^2 < y\)
x*x < y so x would definitely be less than y always.
ALTERNATIVELY:
Verifying using numbers and fractions.
Case 1: x =\(\frac{1}{2}\) and y > 1(2 OR 3 OR any value).
Case 2: x =\(\frac{1}{2}\) and 0 < y < 1 such that\(x^2\) < y always (y =\(\frac{2}{3}\) OR\(\frac{4}{9}\) )
...
(1)\(x^2\) < y
(2) The square root of x is less than y
Since x > 0 and y > 0.
Statement 1)\(x^2 < y\)
x*x < y so x would definitely be less than y always.
ALTERNATIVELY:
Verifying using numbers and fractions.
Case 1: x =\(\frac{1}{2}\) and y > 1(2 OR 3 OR any value).
Case 2: x =\(\frac{1}{2}\) and 0 < y < 1 such that\(x^2\) < y always (y =\(\frac{2}{3}\) OR\(\frac{4}{9}\) )
...
