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The number $b$ is irrational. Which statement about $6 - b$ is true? $\newline$ Choices: $\newline$ (A) $6 - b$ is rational. $\newline$ (B) $6 - b$ is irrational. $\newline$ (C) $6 - b$ can be rational or irrational, depending on the value of $b$ .

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Properties of operations on rational and irrational numbers

Full solution

Q. The number

b

is irrational. Which statement about

6 - b

is true?

\newline

Choices:

\newline

(A)

6 - b

is rational.

\newline

(B)

6 - b

is irrational.

\newline

(C)

6 - b

can be rational or irrational, depending on the value of

b

.

Given Information: We are given that $b$ is an irrational number. By definition, an irrational number cannot be expressed as a fraction of two integers. Examples of irrational numbers include $\sqrt{2}$ , $\pi$ , and $e$ .
Nature of Expression: We need to determine the nature of the expression $6 - b$ . Since $6$ is a rational number (it can be expressed as $\frac{6}{1}$ , which is a fraction of two integers), we are subtracting an irrational number from a rational number.
Difference between Rational and Irrational Numbers: The difference between a rational number and an irrational number is always irrational. This is because if $6 - b$ were rational, then we could add $b$ to both sides of the equation $(6 - b) + b = c + b$ (where $c$ is a rational number), and we would get $6 = c + b$ , implying that an irrational number $b$ is equal to a rational number $6 - c$ , which is not possible.

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