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stmt2 is not sufficient. sqrt(y) < x but y can be greater than, equal to or smaller than x.

I don't seem to understand this a bit more elaborate explanation will help

if x is between 0 and 1 then sqrt(x) will be greater than x. However, if x > 1 then sqrt(x) will always be smaller than x. We can check this with any real numbers.

With this logic, if sqrt(x) > y and since x > sqrt(x) as x > 1, hence x > y.

1) sqrt x>y x>y^2 x has to be bigger than y since x is bigger then y*y

2) sqrt y<x y<x^2 x^2 > y in this case x is not neccessarily bigger than y

for instance x=2, y=3 and x^2=4, y=3, true in this case but x=3, y=3 means x^2=9. x^2 is bigger than y, but x is not bigger than y so false in this case

How do you rule out the second statement ? I understood the first statement in your first post I had asked explanation for the second one alone.

Thanks

Ok. take y = 4, x = 5. In this case sqrt(y) < x and also y < x. Take another example where y = 16, x = 5. In this case, sqrt(y) < x but y > x. Third example, y = 4, x = 4. Here, sqrt(y) < x, but y = x.

As a rule, if y > 1 then y > sqrt(y) In the given question x > sqrt(y). Hence, the two cannot be equated. We cannot definitely say that y > x or y < x or y = x.

How do you rule out the second statement ? I understood the first statement in your first post I had asked explanation for the second one alone.

Thanks

Ok. take y = 4, x = 5. In this case sqrt(y) < x and also y < x. Take another example where y = 16, x = 5. In this case, sqrt(y) < x but y > x. Third example, y = 4, x = 4. Here, sqrt(y) < x, but y = x.

As a rule, if y > 1 then y > sqrt(y) In the given question x > sqrt(y). Hence, the two cannot be equated. We cannot definitely say that y > x or y < x or y = x.

yes, examples in this case is the easiest & fastest way to rule out. Try 9 for y and 4 for x

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