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a) A finite decimal

b) 1.4121

c) Non-Terminating, Recurring

d) Non-Terminating, Non-Recurring

Answer
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If a number is rational, either it has a finite decimal value or, it may have a non-terminating, recurring decimal value.

And if the number is an irrational number, the decimal value of that number is non-terminating and non-recurring.

A rational number can be written in the form of a simplified fraction as $\dfrac{p}{q};q\ne 0$ but irrational numbers cannot be written as a fractional form.

To check if $\sqrt{2}$ is a rational or irrational number:

Let us assume that $\sqrt{2}$ is a rational number.

So, we can write $\sqrt{2}=\dfrac{p}{q}......(1)$

Now, square both sides of the equation (1), we get:

$\begin{align}

& 2=\left( \dfrac{{{p}^{2}}}{{{q}^{2}}} \right) \\

& \Rightarrow {{p}^{2}}=2\times {{q}^{2}}......(2) \\

\end{align}$

Equation (2) implies that $2=\dfrac{{{p}^{2}}}{{{q}^{2}}}$must be an even number.

Also, if the square of a number is an even number, that means the number itself is even.

Therefore, p is an even number.

Since all the even numbers are multiples of 2, therefore, p is a multiple of 2.

Similarly, if p is multiple of 2, it means ${{p}^{2}}$is a multiple of ${{2}^{2}}$, i.e. 4.

Now, we have:

$2=\dfrac{{{p}^{2}}}{{{q}^{2}}}$ and ${{p}^{2}}$as multiple 4

So, for the equation to be valid, ${{q}^{2}}$needs to be multiple of 2

But if both the numbers would be even numbers, we would be able to simplify the fraction.

But we initially said that the fraction is already in its simplified form.

Let us say that it was not simplified. So, if we try to simplify further, we again get p and q as even numbers, and the process repeats.

Hence, our assumption that $\sqrt{2}$ is a rational number is totally wrong

Therefore, $\sqrt{2}$ is an irrational number. That’s why it will have non-terminating and non-recurring decimal values.

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