Squaring Any Real

INDIRECT PROOF

x^2 < x

x^2 - x < 0

x(x-1)<0

0<x<1

so, the statement is FALSE when 0<x<1

Technically neither. A more mathematically accurate way to word it would be "Is this true for all real numbers x?"

The square of a positive fraction will be less than the original number.

$x^{2} \ge x$ is false for all the Real Numbers x such that $0<x<1$ and true for all other Real Numbers.

Its only true if x is greater than 1

if x is 0.5 the x2 is 0.25 so x2 is smaller than x

Fractions!!

It is not true for all real numbers less than 1

0 and 1 squared do not increase

If x is defined as a quotient then the numbers' squared product will be less than the number itself. The only definition for x that will make the proposition true is when x is greater than or equal to 1 or less than or equal to 0. Since x^2 must be greater than or equal to x.

Very obvious. 1/2 * 1/2 is 1/4.

x^2 is not always greater than or equal to x... It is true only when x=0 or x is any real number greater than or equal to 1. But when 0<x<1, the statement is false. For example, if x=0.5, then x^2=(0.5)^2=0.25 which is less than 0.5

x^2>or=x divide both sides by x.

x>1

above step is not true for all real numbers.

therefore original statement is false

x^2 isn't always >x

.5^2 = .25, and .25 is less than .5

The statement is false for

              0<x<1

Hence the answer is false

Given, x^2≥x Or, x^2-x≥0 Or, x(x-1) ≥0 Now,This implies that x ϵ (-∞,0]U[1,∞),i.e,for real numbers between 0 and 1 will not satisfy the given inequality. Thus the answer will be false.

x^2-x>=0 then derivation, solution of derivation, then three intervals (-infinite; 0) [0;1/2] (1/2;infinite)

Or you could do x=1

(1)^2=1

If 0<x<1, then x^2 is actually smaller than x. So it is not true for ALL real numbers x.

It's true if x>1 or if -1>x>0.

It will be smaller when X is a fraction number