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Which of the following gives all of the sets that contain √9?

a.the set of all irrational numbers
b. the set of all natural numbers, the set of all whole numbers, and the set of all integers
c.the set of all integers, the set of all rational numbers, and the set of all real numbers
d.the set of all natural numbers, the set of all whole numbers, the set of all integers, the set of all rational numbers, and the set of all real numbers

User Joe Gatt
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2 Answers

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I dont think its b because one of the square roots 9 (-3) is not a natural number

I think its choice c
User Rebol Tutorial
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1 vote

Answer: c.the set of all integers, the set of all rational numbers, and the set of all real numbers

Explanation:

The given number is
√(9).

We know that
(3)^2=9 and
(-3)^2=9

Therefore,
√(9)=\pm3

Therefore, the value of
√(9) does not belongs to the set of natural numbers since -3 does not belongs to it.

Since 3,-3 belongs to the set of integers.

Thus, the value of
√(9) belongs to the set of integers the set of all rational numbers, and the set of all real numbers
because the set of integers is contained in the set of rationals and the set of rational is contained in the set of real numbers.

User Nate Kohl
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