Free Number Systems 01 Practice Test - 9th Grade 

Natural numbers are all the counting numbers starting from 1, i.e. {1, 2, 3, ...}
Therefore, statement: '1 is smallest natural number'; is correct.

Whole numbers are all counting numbers and zero, i.e. {0, 1, 2, 3, ...}
Therefore, statement: '1 is smallest whole number'; is wrong.

Integers include the set of natural numbers, their negatives and 0; i.e. {...-3, -2, -1, 0, 1, 2, 3, ...}. Thus, -1 is the greatest negative integer, not the smallest one.

0 is included among integers.
Hence, the statement is correct.

Given: perpendicular sides of a right angled triangle are $\sqrt{6}$ and $\sqrt{3}$.

Using Pythagoras theorem,

Hypotenuse = $\sqrt{{\sqrt{6}}^2+{\sqrt{3}}^2}$

                    = $\sqrt{6+3}$

                   = $\sqrt{9}$  cm

                  =  3 cm

We know the identity:
$a^2-b^2=(a+b)(a-b)$

$(3+\sqrt{5})(3-\sqrt{5})=(3^2-(\sqrt{5}{)}^2=9-5=4$
 
Hence, $(3+\sqrt{5})(3-\sqrt{5})=4$, which is a rational number.

$2\sqrt{3}\times 3\sqrt{2}=2\times 3\times \sqrt{2\times 3}=6\sqrt{6}$
The presence of factor $\sqrt{6}$ makes the expression irrational.
 

$6\times 6$ = 36, a perfect square, hence $\sqrt{36}$ is not irrational .
Also 25 is the perfect square of 5, hence $\sqrt{25}$ is rational.

$4^2=16$ & $9^2=81$, hence all the numbers between  $\sqrt{16}$ and $\sqrt{81}$  lie between 4 and 9.

We can see that  $\sqrt{17}$, $\sqrt{31}$   are irrational numbers  between $\sqrt{16}$ and $\sqrt{81}$, and they lie between 4 and 9.

To convert 0.45 into a fraction:
$0.45=\frac{45}{100}$ 
[Multiplying 0.45 with $\frac{100}{100}$]

Dividing the numerator and the denominator by 5, we get
$\frac{45}{100}=\frac{9}{20}$ 

$\therefore 0.45=\frac{9}{20}$

${\left(\frac{7}{9}\right)}^3=\frac{7\times 7\times 7}{9\times 9\times 7}\Rightarrow {\left(\frac{7}{9}\right)}^3=\frac{343}{729}$

Given:
${\left(\sqrt{1024}\right)}^3=8^x$

$\Rightarrow \left(\sqrt{(}{2}^{10}\right){)}^3=8^x$.   

$\Rightarrow \left(\sqrt{(}\right(2^5{)}^2){)}^3=8^x$.  {as ${\left(a^m\right)}^n$=$a^{mn}$}

${\Rightarrow \left(2^5\right)}^3={\left(2^3\right)}^x$

${\Rightarrow \left(2^3\right)}^5={\left(2^3\right)}^x$

$\Rightarrow (8{)}^5$= $(8{)}^x$

We know that, if $a^n$= $a^m$ then, $n=m$.

Hence,
$\Rightarrow x=5$

Rationalizing the expression we get,
$\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}+\sqrt{7}}{(\sqrt{5}-\sqrt{7})(\sqrt{5}+\sqrt{7)}}$ 

=$\frac{\sqrt{5}+\sqrt{7}}{(\sqrt{5}{)}^2-(\sqrt{7}{)}^2}$

=$\frac{\sqrt{5}+\sqrt{7}}{5-7}$
=$\frac{-(\sqrt{5}+\sqrt{7})}{2}$

We know that $\sqrt[n]{na}\times \sqrt[n]{nb}=\sqrt[n]{nab}$

$\text{So,}\sqrt[n]{n2}\times \sqrt[n]{n3}=\sqrt[n]{n2\times 3}=\sqrt[n]{n6}$