48−−√<36−−√

Question

is this statement true?

Answer 1

To determine if the statement \( \sqrt{48} < \sqrt{36} \) is true, we can evaluate both square roots:

1. Calculate \( \sqrt{48} \): \[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \approx 4 \times 1.732 \approx 6.928 \]

2. Calculate \( \sqrt{36} \): \[ \sqrt{36} = 6 \]

Now we compare the two results:

• \( \sqrt{48} \approx 6.928 \)
• \( \sqrt{36} = 6 \)

Since \( 6.928 > 6 \), the statement \( \sqrt{48} < \sqrt{36} \) is false.

Answer 2

or, since sqrt(x) is monotone increasing,
48 > 36, so √48 > √36

Answer 3

Yes, that's correct! The square root function \( \sqrt{x} \) is indeed a monotone increasing function, which means that if \( x > y \), then \( \sqrt{x} > \sqrt{y} \).

In this case, since \( 48 > 36 \), it follows that:

\[ \sqrt{48} > \sqrt{36} \]

This is another way to show that the original statement \( \sqrt{48} < \sqrt{36} \) is false. Your reasoning is completely valid!