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Which number is irrational?

A. [tex]$\frac{\sqrt[3]{-27}}{3}$[/tex]
B. [tex]$\frac{25}{9}$[/tex]
C. [tex]$\frac{\pi}{\sqrt[3]{27}}$[/tex]
D. [tex]$\overline{12}$[/tex]


Sagot :

Let's determine which of the given numbers is irrational. An irrational number cannot be expressed as the quotient [tex]$ \frac{a}{b} $[/tex] of two integers [tex]\( a \)[/tex] and [tex]\( b \)[/tex], where [tex]\( b \neq 0 \)[/tex].

### Option 1: [tex]\( \frac{\sqrt[3]{-27}}{3} \)[/tex]

First, let's simplify [tex]\( \sqrt[3]{-27} \)[/tex]:
[tex]\[ \sqrt[3]{-27} = -3 \][/tex]

So,
[tex]\[ \frac{\sqrt[3]{-27}}{3} = \frac{-3}{3} = -1 \][/tex]

The number [tex]\( -1 \)[/tex] is a rational number because it can be expressed as [tex]\( \frac{-1}{1} \)[/tex].

### Option 2: [tex]\( \frac{25}{9} \)[/tex]

The number [tex]\( \frac{25}{9} \)[/tex] is a ratio of two integers. Thus, it is a rational number.

### Option 3: [tex]\( \frac{\pi}{\sqrt[3]{27}} \)[/tex]

First, let's simplify [tex]\( \sqrt[3]{27} \)[/tex]:
[tex]\[ \sqrt[3]{27} = 3 \][/tex]

So,
[tex]\[ \frac{\pi}{\sqrt[3]{27}} = \frac{\pi}{3} \][/tex]

The number [tex]\(\pi\)[/tex] is a well-known irrational number. The quotient of an irrational number [tex]\(\pi\)[/tex] and a non-zero integer 3 remains irrational.
Therefore, [tex]\( \frac{\pi}{3} \)[/tex] is an irrational number.

### Option 4: [tex]\( \overline{12} \)[/tex]

The notation [tex]\( \overline{12} \)[/tex] typically represents the repeating decimal [tex]\( 12.121212\ldots \)[/tex]. Repeating decimals are rational numbers because they can be expressed as a ratio of two integers. For example,
[tex]\[ \overline{12} = \frac{1212}{99} = 12.121212\ldots \][/tex]

Thus, [tex]\( \overline{12} \)[/tex] is a rational number.

### Conclusion

The only number among the given options that is irrational is:
[tex]\[ \frac{\pi}{\sqrt[3]{27}} = \frac{\pi}{3} \][/tex]

So, the irrational number is [tex]\( \frac{\pi}{\sqrt[3]{27}} \)[/tex].