Finding the counterexample?

Question

I got a couple problems i need help with. So here it is.

"Show the conjecture is false by finding a counterexample."

The square root of a number x is always less than x.

&

If m is a nonzero integer, then m+2(over m) is always greater than 1.

so i have to prove these wrong. how do i do that?

Answer 1

sqrt(.25) = .5

.5 > .25

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let m = -1

(m+2) / m = (-1 + 2)/(-1) = - 1/2 < 1

Answer 2

Example: x=0.5 ➞ x²=0.25

Generally, if 0<x<1, x²<x.

Example: m=-1 ➞ (m+2)/m = -1 < 1

Answer 3

sqrt 1=1

so that disproves the fact that 1 has to be greater than the sqrt of 1. Same thing stands for 0.

(m+2)/m =1 if m in INFINITELY LARGE

Since adding 2 to a BIG number will do nothing.

Answer 4

the trivial counterexamples are √ 1 and m = negative integer.{ if m = positive then statement is valid & if x > 1 the 1st is also true}