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Which number is irrational, an integer, and a real number?

A. 0
B. [tex]\(\frac{-4}{2}\)[/tex]
C. There is no such number.
D. [tex]\(\sqrt{4}\)[/tex]

Sagot :

GinnyAnsweridn
To determine which number is irrational, an integer, and a real number, let’s analyze each of the options given:

A. [tex]\(0\)[/tex]:
- [tex]\(0\)[/tex] is an integer.
- [tex]\(0\)[/tex] is a rational number because it can be expressed as [tex]\( \frac{0}{1} \)[/tex].
- Therefore, [tex]\(0\)[/tex] is not an irrational number.

B. [tex]\( \frac{-4}{2} \)[/tex]:
- Simplifying [tex]\( \frac{-4}{2} \)[/tex] gives us [tex]\(-2\)[/tex].
- [tex]\(-2\)[/tex] is an integer.
- [tex]\(-2\)[/tex] is a rational number because it can be expressed as [tex]\( \frac{-2}{1} \)[/tex].
- Therefore, [tex]\(-2\)[/tex] is not an irrational number.

C. There is no such number.
- An irrational number cannot be an integer because integers by definition are rational numbers.
- An integer is always rational and cannot be both irrational and an integer at the same time.
- Therefore, Option C seems to be a valid conclusion.

D. [tex]\( \sqrt{4} \)[/tex]:
- Simplifying [tex]\( \sqrt{4} \)[/tex] gives us [tex]\(2\)[/tex].
- [tex]\(2\)[/tex] is an integer.
- [tex]\(2\)[/tex] is a rational number because it can be expressed as [tex]\( \frac{2}{1} \)[/tex].
- Therefore, [tex]\(2\)[/tex] is not an irrational number.

Based on the analysis of all the options, we can conclude that:

C. There is no such number.

is the correct answer because an integer cannot be an irrational number by definition.
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