Question

what is a counterexample to this claim? if x is any real number, then $x^2 \geq 1$. select the correct answer. $\circ$ $x = -2.5$ $\circ$ $x = -1.5$ $\circ$ $x = 0.5$ $\circ$ $x = 1.5$

Explanation:

Step1: Recall the claim

The claim is "If \( x \) is any real number, then \( x^2 \geq 1 \)". A counterexample is a value of \( x \) for which \( x^2 < 1 \).

Step2: Evaluate \( x = -2.5 \)

Calculate \( (-2.5)^2 \).
\( (-2.5)^2 = 6.25 \), and \( 6.25 \geq 1 \), so this is not a counterexample.

Step3: Evaluate \( x = -1.5 \)

Calculate \( (-1.5)^2 \).
\( (-1.5)^2 = 2.25 \), and \( 2.25 \geq 1 \), so this is not a counterexample.

Step4: Evaluate \( x = 0.5 \)

Calculate \( (0.5)^2 \).
\( (0.5)^2 = 0.25 \), and \( 0.25 < 1 \), so this satisfies \( x^2 < 1 \), making it a counterexample.

Step5: Evaluate \( x = 1.5 \)

Calculate \( (1.5)^2 \).
\( (1.5)^2 = 2.25 \), and \( 2.25 \geq 1 \), so this is not a counterexample.

Answer:

\( x = 0.5 \) (the option with \( x = 0.5 \))