Question

The value of \[\left( {6 + \sqrt {27} } \right) - \left( {3 + \sqrt 3 } \right) + \left( {1 - 2\sqrt 3 } \right)\] when simplified is:
A) Positive and irrational
B) Negative and rational
C) Positive and rational
D) Negative and irrational

Answer 1

Hint: Here in this question, given an expression of irrational numbers. First we can solve the given expression to the simplest form by using basic arithmetic operation. Analyzing the simplest form of the expression we can say that it will neither positive nor negative and neither rational nor irrational.

Complete step-by-step solution:
A rational number is a number that can be expressed as a fraction where both the numerator and the denominator in the fraction are integers. The denominator in a rational number cannot be zero.
Expressed as an equation, a rational number is a number \[\dfrac{a}{b}\], \[b \ne 0\]
where a and b are both integers.
This equation shows that all integers, finite decimals, and repeating decimals are rational numbers. In other words, most numbers are rational numbers.
The opposite of rational numbers are irrational numbers. In simple terms, irrational numbers are real numbers that can’t be written as a simple fraction.
Now consider the given expression
\[ \Rightarrow \,\,\,\left( {6 + \sqrt {27} } \right) - \left( {3 + \sqrt 3 } \right) + \left( {1 - 2\sqrt 3 } \right)\]
Remove all the parenthesis using sign convention, then
\[ \Rightarrow \,\,\,6 + \sqrt {27} - 3 - \sqrt 3 + 1 - 2\sqrt 3 \]
\[\sqrt {27} \] Can be written as \[\sqrt {27} = \sqrt {9 \times 3} = \sqrt {{3^2} \times 3} = \sqrt {{3^2}} \cdot \sqrt 3 = 3\sqrt 3 \], then
\[ \Rightarrow \,\,\,6 + 3\sqrt 3 - 3 - \sqrt 3 + 1 - 2\sqrt 3 \]
Isolate real and imaginary numbers
\[ \Rightarrow \,\,\,6 - 3 + 1 + 3\sqrt 3 - \sqrt 3 - 2\sqrt 3 \]
On simplification, we get
\[ \Rightarrow \,\,\,7 - 3 + \left( {3 - 1 - 2} \right)\sqrt 3 \]
\[ \Rightarrow \,\,\,4 + \left( 0 \right)\sqrt 3 \]
\[ \Rightarrow \,\,\,4\]
The number 4 is positive and is an integer, hence 4 is a rational number.
Because 4 can also be expressed as \[\dfrac{4}{1}\].
When expressed as 4, both the numerator and the denominator are integers. The denominator doesn’t equal 0.

Hence the correct answer is option (C) i.e,Positive and rational.

Note: In mathematics we have many kinds of numbers namely, natural numbers are the counting numbers, whole numbers are the natural numbers along with zero, integers are the whole numbers and the also include the negative of natural numbers, rational numbers are in the form of p by q form, irrational numbers and real numbers.