Which number produces an irrational number when multiplied by [tex]\frac{1}{4}[/tex]?

A. [tex]\sqrt{12}[/tex]

B. [tex]-\sqrt{36}[/tex]

C. [tex]\frac{4}{3}[/tex]

D. [tex]0.444444 \ldots[/tex]



Answer :

To determine which of the given numbers results in an irrational number when multiplied by [tex]\(\frac{1}{4}\)[/tex], we need to analyze each option carefully.

1. Option A: [tex]\(\sqrt{12}\)[/tex]

[tex]\(\sqrt{12}\)[/tex] is the square root of 12. Since 12 is not a perfect square, [tex]\(\sqrt{12}\)[/tex] is an irrational number. When we multiply an irrational number by a rational number (in this case, [tex]\(\frac{1}{4}\)[/tex]), the result remains irrational.

Therefore, [tex]\(\frac{1}{4} \times \sqrt{12}\)[/tex] is irrational.

2. Option B: [tex]\(-\sqrt{36}\)[/tex]

[tex]\(\sqrt{36}\)[/tex] is the square root of 36. Since 36 is a perfect square, [tex]\(\sqrt{36}\)[/tex] is equal to 6, which is a rational number. Thus, [tex]\(-\sqrt{36} = -6\)[/tex] is also a rational number. When a rational number (in this case, -6) is multiplied by another rational number ([tex]\(\frac{1}{4}\)[/tex]), the result is still rational.

Therefore, [tex]\(\frac{1}{4} \times -6\)[/tex] is rational.

3. Option C: [tex]\(\frac{4}{3}\)[/tex]

[tex]\(\frac{4}{3}\)[/tex] is already in its simplest form as a rational number. Multiplying two rational numbers results in another rational number.

Therefore, [tex]\(\frac{1}{4} \times \frac{4}{3} = \frac{4}{12} = \frac{1}{3}\)[/tex] is rational.

4. Option D: [tex]\(0.444444 \ldots\)[/tex]

The repeating decimal [tex]\(0.444444 \ldots\)[/tex] can be expressed as the fraction [tex]\(\frac{4}{9}\)[/tex], making it a rational number. When we multiply two rational numbers, the result is rational.

Therefore, [tex]\(\frac{1}{4} \times 0.444444 \ldots = \frac{1}{4} \times \frac{4}{9} = \frac{4}{36} = \frac{1}{9}\)[/tex] is rational.

From these analyses, we see that among the options provided, the only number that, when multiplied by [tex]\(\frac{1}{4}\)[/tex], results in an irrational number is:

A. [tex]\(\sqrt{12}\)[/tex]