Free Number Systems 03 Practice Test - 9th Grade 

To check if a number lies between any two numbers, we first convert the numbers into the decimal form. $\frac{3}{4}=0.75$ ,$\frac{1}{2}=0.5$ and we have to check if $0.75$ lies between $0.5$ and $1$ and plot the numbers on the number line as follows  :

Therefore, we can say that $\frac{3}{4}$ lies between $\frac{1}{2}$ and 1.

Integers are constituted by natural numbers, their negatives and 0. Removing the negative numbers from integers would leave us with the whole numbers. Therefore, whole numbers include 0 as well as all the positive integers.

The product of two irrational numbers will be either  a rational or an irrational number.

Consider the following example: $(\sqrt{3}+\sqrt{2})\times (\sqrt{3}-\sqrt{2})=1$. 
Both $(\sqrt{3}+\sqrt{2})$ and $(\sqrt{3}-\sqrt{2})$ are irrational numbers and their product is 1. 
1 is an integer and since, all integers are rational numbers as well, we can infer that product of two irrational numbers may be an integer, a rational number or an irrational number and they are all real numbers.

When we multiply the irrational numbers $\sqrt{2}$ and $\sqrt{3}$, we get $\sqrt{6}$, which is also an irrational number. 

Irrational numbers are defined as those numbers which have a non-terminating and non-repeating decimal expansion and hence cannot be represented in the $\frac{p}{q}$ form. So, here we see that out of the given options the number 5.2731687143725186..... has a non-terminating and non-repeating decimal expansion as clearly one can interpret that there are no patterns formed in the decimal expansion and the number goes on expanding arbitrarily.

In $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, the denominator has to be rationalised for simplification. So , 

$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$\times$$\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$
$=\frac{(\sqrt{5}-\sqrt{3}{)}^2}{{\sqrt{5}}^2-{\sqrt{3}}^2}$
$=\frac{1}{2}$[${\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2-2\sqrt{15}$]
$=4-\sqrt{15}$ 
$\text{So, a=4 and b=1.}$
$\therefore$a + b = 4 + 1 = 5

The integers include positive integers i.e. 1, 2, 3, ... , negative integers i.e. -1, -2, -3, ... and zero. Hence, zero is neither a positive nor a negative integer. 

A common tendency to solve this question is to apply the algebraic identity $(a^2-b^2)=(a+b)(a-b)$. But by observation, we can see that $(\sqrt{2}-\sqrt{2})$, which is one of the factors, will result in 0, thus making the final value of the whole expression as 0 i.e.
 $(\sqrt{2}+\sqrt{2})(\sqrt{2}-\sqrt{2})$
$=(\sqrt{2}+\sqrt{2})\times \left(0\right)$
$=0$

Let 'x' = 14.287628762876 ------- (i)

then 10000x = 142876.287628762876 --------(ii)

Subtracting (i) from (ii) we get

10000x - x = 142876.287628762876  - 14.287628762876

9999x = 142862

 then 'x' = $\frac{142862}{9999}$  

p = 142862, q = 9999

So, p-q = 132863

For any positive real number '$a$' and integers '$m$' and '$n$', we define
1. $a^m\times a^n=a^{m+n}$
2. ${\left(a^m\right)}^n=a^{mn}$
3. $(ab{)}^m=a^m{b}^m$  and if  $a=b$, the expression becomes  $(a\times a{)}^m=(a^2{)}^m=a^{2m}$.

​​​​​​So, using first rule, we have
${14}^{\frac{1}{7}+\frac{1}{7}}={14}^{\frac{2}{7}}$
Now, using second rule,
$({14}^2{)}^{\frac{1}{7}}$= ${14}^{\frac{2}{7}}$
Finally, using third rule,
($14\times 14{)}^{\frac{1}{7}}={14}^{\frac{2}{7}}$
Hence, we see that all the three results are same.