Real Numbers - Class 10 Maths - Extra Questions

$$25 = 5 \times 5=5^2$$

$$40 = 2 \times 2\times  2 \times 5=2^3\times 5$$

$$\therefore$$ H.C.F. of $$25$$ and $$40$$ is $$5$$

Since,$${{\left( 7-4\sqrt{2} \right)}^{2}}$$

$$=49-32-56\sqrt{2}$$

$$=17-56\sqrt{2}$$

$$\sqrt{2}$$ is irrational ,so is $$17-56\sqrt{2}$$ and so is .$$\left( 7-4\sqrt{2} \right)$$

Consider given numbers,

$$\Rightarrow 180,252,354$$

Now,

$$ 180=2\times 2\times 3\times 3\times 5 $$

$$ 252=2\times 2\times 3\times 3\times 7 $$

$$ 354=2\times 3\times 59 $$

H.C.F of $$180,252,354$$ is _{=$$2\times 3=6$$}

$$\sqrt {23} $$

Since, 23 is not a perfect square,

$$\sqrt {23} $$ is an irrational number.

$${\textbf{Step - 1: Writing any pair of irrational number}}{\text{.}}$$      

                  $${\text{We will get an irrational number if two numbers do not cancel each other}}{\text{.}}$$                               

                   $${\text{As we know that the sum of irrational numbers }}$$

                   $${\text{can be both rational and irrational numbers,}}$$

                   $${\text{so carefully choose the pair of number}}{\text{.}}$$

                   $${\text{ We can write any pair of irrational with same sign}}$$

                   $$\sqrt 2  + 2\sqrt 2  = 3\sqrt 2 $$

$${\textbf{Hence, A pair of irrational numbers whose sum irrational is }}\sqrt 2  + 2\sqrt 2  = 3\sqrt 2 \;$$

Let us consider $$\sqrt{2}$$ be a rational number, then

$$\sqrt{2} = p/q$$, where ‘p’ and ‘q’ are integers, $$q \neq 0$$ and p, q have no common factors (except 1).

So,

$$2 = p^2 / q^2$$

$$p^2 = 2q^2$$ …. (1)

As we know, ‘$$2$$’ divides $$2q^2$$, so ‘$$2$$’ divides $$p^2$$ as well. Hence, ‘$$2$$’ is prime.

So $$2$$ divides p

Now, let $$p = 2k$$, where ‘k’ is an integer

Square on both sides, we get

$$p^2 = 4k^2$$

$$2q^2 = 4k^2$$ [Since, $$p^2 = 2q^2$$, from equation (1)]

$$q^2 = 2k^2$$

As we know, ‘$$2$$’ divides $$2k^2$$, so ‘$$2$$’ divides $$q^2$$ as well. But ‘$$2$$’ is prime.

So $$2$$ divides q

Thus, p and q have a common factor of $$2$$. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that $$\sqrt{2}$$ is not a rational number.

$$\sqrt{2}$$ is an irrational number.

Now, let us assume $$3 -\sqrt{2}$$ be a rational number, ‘r’

So, $$3 -\sqrt{2}= r$$

$$3 – r = \sqrt{2}$$

We know that, ‘r’ is rational, ‘$$3- r$$’ is rational, so ‘$$\sqrt{2}$$’ is also rational.

This contradicts the statement that $$\sqrt{2}$$ is irrational.

So, $$3- \sqrt{2}$$ is an irrational number.

Hence proved.

Let us consider $$1/\sqrt{11}$$ be a rational number, then

$$1/\sqrt{11} = p/q$$, where ‘p’ and ‘q’ are integers, $$q \neq 0$$ and p, q have no common factors (except 1).

So,

$$1/11 = p^2 / q^2$$

$$q^2 = 11p^2$$ …. (1)

As we know, ‘$$11$$’ divides $$11p^2$$, so ‘$$11$$’ divides $$q^2$$ as well. Hence, ‘$$11$$’ is prime.

So $$11$$ divides q

Now, let $$q = 11k$$, where ‘k’ is an integer

Square on both sides, we get

$$q^2 = 121k^2$$

$$11p^2 = 121k^2$$ [Since, $$q^2 = 11p^2$$, from equation (1)]

$$p^2 = 11k^2$$

As we know, ‘$$11$$’ divides $$11k^2$$, so ‘$$11$$’ divides $$p^2$$ as well. But ‘$$11$$’ is prime.

So $$11$$ divides p

Thus, p and q have a common factor of $$11$$. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).

We can say that $$1/\sqrt{11}$$ is not a rational number.

$$1/\sqrt{11}$$ is an irrational number.

Hence proved.

$$\sqrt{2} + \sqrt{5}$$

Now, let us assume $$\sqrt{2} + \sqrt{5}$$ be a rational number, ‘r’

So, $$\sqrt{2} + \sqrt{5} = r$$

$$\sqrt{5} = r – \sqrt{2}$$

Square on both sides,

$$(\sqrt{5})2 = (r – \sqrt{2})2$$

$$5 = r^2 + (\sqrt{2})2 – 2r\sqrt{2}$$

$$5 = r^2 + 2 – 2\sqrt{2}r$$

$$5 – 2 = r^2 – 2\sqrt{2}r$$

$$r^2 – 3 = 2\sqrt{2}r$$

$$(r^2 – 3)/2r = \sqrt{2}$$

We know that ‘r’ is rational, ‘$$(r^2 – 3)/2r$$’ is rational, so ‘$$\sqrt{2}$$’ is also rational.

This contradicts the statement that $$\sqrt 2$$ is irrational.

So, $$\sqrt{2} +\sqrt{5}$$ is an irrational number.

$$2\sqrt{3} – 7$$

Now, let us assume $$2\sqrt{3} – 7$$ be a rational number, ‘r’

So,$$2\sqrt{3} – 7 = r$$

$$2\sqrt{3} = r + 7$$

$$\sqrt{3} = (r + 7)/2$$

We know that ‘r’ is rational, ‘$$(r + 7)/2$$’ is rational, so ‘$$\sqrt{3}$$’ is also rational.

This contradicts the statement that $$\sqrt{3}$$ is irrational.

So, $$2\sqrt{3} – 7$$ is an irrational number.

$$23/75$$23/75 hence NON-terminating.

denominator is NOT  multiple of $$2^{m}\times 5^{n}$$ hence NON-terminating.

It is a non-terminating repeating decimal.

Let us divide both numerator and denominator by $$3$$

$$6/15=(6\div 3)/(15 \div 3)$$

$$= 2/5$$

Since the denominator is $$5$$.

It is a terminating decimal.

$$3/\sqrt 3$$

Let us simplify,

By rationalizing, we get

$$3/\sqrt 3=3\sqrt{3}/(\sqrt 3 \times \sqrt 3)$$

=$$ 3\sqrt 3/3$$

$$= \sqrt 3$$

Hence, it is an irrational number

$$3 – 5\sqrt{3}$$

Now, let us assume $$3 – 5\sqrt{3}$$ be a rational number, ‘r’

So, $$3 – 5\sqrt{3} = r$$

$$3 – r = 5\sqrt{3}$$

$$(3 – r)/5 = \sqrt{3}$$

We know that ‘r’ is rational, ‘$$(3 – r)/5$$’ is rational, so ‘$$\sqrt{3}$$’ is also rational.

This contradicts the statement that $$\sqrt{3}$$ is irrational.

So, $$3 – 5\sqrt{3}$$ is an irrational number.

$$-\dfrac{2}{3} + \sqrt[3]{2}$$

Let us simplify,

$$-\dfrac{2}{3} + \sqrt[3]2 = -\dfrac{2}{3} + 2^{\dfrac{1}{3}}$$

Since, $$2$$ is not a perfect cube.

Hence it is an irrational number.

Let $$x = 345.0456456$$

Multiplying by $$10$$ on both sides, we get

$$10x = 3450.456456$$

Since there is three repeating digit after the decimal point,

Multiplying by $$1000$$ on both sides, we get

$$1000x = 3450456.456456$$…

Now, subtract both the values,

$$10000x – 10x = 3450456 – 345$$

$$9990x = 3450111$$

$$x = 3450111/9990$$

Since it is a non-terminating repeating decimal.

$$345.0\overline{456}$$ is a rational number.

$$-\sqrt{(2/49)}, 3/200, \sqrt{(25/3)}, -\sqrt{(49/16)}$$

$$-\sqrt{(2/49)} = -\sqrt 2/7$$

$$3/200 = 3/200$$

$$\sqrt{(25/3)} = 5/\sqrt 3$$

$$-\sqrt{(49/16)} = -7/4$$

So,

$$-\sqrt{2}/7$$ and $$5/\sqrt 3$$ are irrational decimals.

$$3/200$$ and $$-7/4$$ are rational decimals.

$$3 – \sqrt{(7/25)}$$

Let us simplify,

$$3 – \sqrt{(7/25)} = 3 – \sqrt 7/\sqrt {25}$$

$$= 3 – \sqrt 7/5$$

Hence, $$3 – \sqrt 7/5$$ is an irrational number.

$$ \sqrt[3]3$$

We know that $$\sqrt[3]3 = 3^{1/3}$$

Let us consider $$3^{1/3} = p/q$$, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

$$3^{1/3} = p/q$$

$$3 = p^3/q^3$$

$$p^3 = 3q^3$$ ….. (1)

We know that $$3$$ divides $$3q^3$$ then $$3$$ divides $$p^3$$

So, $$3$$ divides p

Now, let us consider $$p = 3k$$, where k is an integer

Substitute the value of p in (1), we get

$$p^3 = 3q^3$$

$$(3k)^3 = 3q^3$$

$$9k^3 = 3q^3$$

$$3k^3 = q^3$$

We know that, $$3$$ divides $$9k^3$$ then $$3$$ divides $$q^3$$

So, $$3$$ divides q

Thus p and q have a common factor ‘3’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, $$∛3$$ is an irrational number.

$$(3 + \sqrt 5)^2$$

Let us simplify,

By using $$(a + b)^2 = a^2 + b^2 + 2ab$$

$$(3 + \sqrt 5)^2 = 3^2 + (\sqrt 5)^2 + 2.3.\sqrt 5$$

$$= 9 + 5 + 6\sqrt 5$$

$$= 14 + 6\sqrt 5$$

Hence, it is an irrational number.

$$ \sqrt[3]2$$

We know that $$\sqrt[3]2 = 2^{1/3}$$

Let us consider $$2^{1/3} = p/q$$, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

$$2^{1/3} = p/q$$

$$2 = p^3/q^3$$

$$p^3 = 2q^3$$ ….. (1)

We know that $$2$$ divides $$2q^3$$ then $$2$$ divides $$p^3$$

So, $$2$$ divides p

Now, let us consider $$p = 2k$$, where k is an integer

Substitute the value of p in (1), we get

$$p^3 = 2q^3$$

$$(2k)^3 = 2q^3$$

$$8k^3 = 2q^3$$

$$4k^3 = q^3$$

We know that, $$2$$ divides $$4k^3$$ then $$2$$ divides $$q^3$$

So, $$2$$ divides q

Thus p and q have a common factor ‘2’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, $$\sqrt[3]2$$ is an irrational number.

$$ (2/5 \sqrt 7)^2$$ 

Let us simplify,

$$(2/5 \sqrt 7)^2 = (2/5 \sqrt 7) \times (2/5 \sqrt 7)$$

$$= 4/ (25 \times 7)$$

$$= 4/175$$

Hence it is a rational number.

$$-2/7 \sqrt[3] 5$$ 

Let us simplify,

$$-2/7 \sqrt[3] 5 = -2/7 (5)^{1/3}$$

Since $$5$$ is not a perfect cube.

Hence it is an irrational number.

We know that $$\sqrt[4]5 = 5^{1/4}$$

Let us consider $$5^{1/4} = p/q$$, where p, q are integers, q>0.

p and q have no common factors (except 1).

So,

$$5^{1/4} = p/q$$

$$5 = p^4/q^4$$

$$P^4 = 5q^4$$ ….. (1)

We know that $$5$$ divides $$5q^4$$ then $$5$$ divides $$p^4$$

So, $$5$$ divides p

Now, let us consider $$p = 5k$$, where k is an integer

Substitute the value of p in (1), we get

$$P^4 = 5q^4$$

$$(5k)^4 = 5q^4$$

$$625k^4 = 5q^4$$

$$125k^4 = q^4$$

We know that, $$5$$ divides $$125k^4$$ then $$5$$ divides $$q^4$$

So, $$5$$ divides q

Thus p and q have a common factor ‘$$5$$’.

This contradicts the statement, p and q have no common factor (except 1).

Hence, $$\sqrt[4]5$$ is an irrational number.

$$(2 – \sqrt 3) (2 + \sqrt 3)$$

Let us simplify,

By using the formula,

$$(a + b) (a – b) = (a)^2 -(b)^2$$

$$(2 – \sqrt 3) (2 + \sqrt 3) = (2)^2 – (\sqrt 3)^2$$

$$= 4 – 3$$

$$= 1$$

Hence, it is a rational number.

The Euclidean Algorithm for finding HCF $$(A,B)$$ is as follows:

If $$A = 0$$ then HCF $$(A,B)=B,$$ since the HCF $$(0,B)=B,$$ and we can stop.  

If $$B = 0$$ then HCF $$(A,B)=A,$$ since the HCF $$(A,0)=A,$$ and we can stop.  

Write $$A$$ in quotient remainder form $$(A = BQ + R)$$

Find HCF $$(B,R)$$ using the Euclidean Algorithm since 

HCF $$(A,B) = HCF(B,R)$$

Here, HCF of $$441$$ and $$567$$ can be found as follows:-

$$567=441\times 1+126$$

$$\Rightarrow$$ $$441=126\times 3+63$$

$$\Rightarrow$$ $$126=63\times 2 + 0$$

Since remainder is $$0$$, therefore, 

H.C.F of $$\left( 441,567 \right) $$ is $$=63$$

Now H.C.F of $$63$$ and $$693$$ is

$$693=63\times11 +0$$

Therefore, H.C.F of $$\left( 63,693 \right) =63$$

Thus, H.C.F of $$\left( 441,567,693 \right) =63.$$   

Assume that $$\sqrt 5  + \sqrt 3 $$ is rational. So,

$$\sqrt 5  + \sqrt 3  = \dfrac{p}{q}$$

$$5 + 3 + 2\sqrt {15}  = \dfrac{{{p^2}}}{{{q^2}}}$$

$$\sqrt {15}  = \dfrac{1}{2}\left( {\dfrac{{{p^2}}}{{{q^2}}} - 8} \right)$$

Since the RHS of the above equation is rational but $$\sqrt {15} $$ is an irrational number, so the assumption is wrong.

Therefore, $$\sqrt 5  + \sqrt 3 $$ is irrational.

 Since 12576 > 4052

12576 = 4052 × 3 + 420

 Since the remainder 420 ≠ 0

4052 = 420 × 9 + 272

Consider the new divisor 420 and the new remainder 272

420 = 272 × 1 + 148

Consider the new divisor 272 and the new remainder 148

272 = 148 × 1 + 124

Consider the new divisor 148 and the new remainder 124

148 = 124 × 1 + 24

Consider the new divisor 124 and the new remainder 24

124 = 24 × 5 + 4

Consider the new divisor 24 and the new remainder 4

24 = 4 × 6 + 0

The remainder has now become zero, so procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Assume that $$3 + 2\sqrt 5 $$ is rational. Then,

$$3 + 2\sqrt 5  = \dfrac{a}{b}$$

$$\sqrt 5  = \dfrac{{a - 3b}}{{2b}}$$

This gives that $$\dfrac{{a - 3b}}{{2b}}$$ is rational but $$\sqrt 5 $$ is an irrational number, therefore, the assumption is wrong.

Hence, $$3 + 2\sqrt 5 $$ is irrational.

Assume that $$\sqrt 5 $$ be a rational number. So,

$$\sqrt 5  = \dfrac{p}{q}$$, where $$p$$ and $$q$$ are co prime.

$$5 = \dfrac{{{p^2}}}{{{q^2}}}$$

$$5{q^2} = {p^2}$$                        (1)

This shows that $${p^2}$$ is divisible by 5, then $$p$$ is divisible by 5, then for any positive integer $$c$$, it can be said that $$p = 5c$$, $${p^2} = 25{c^2}$$.

Then equation (1) can be written as,

$$5{q^2} = 25{c^2}$$

$${q^2} = 5{c^2}$$

This gives that $$q$$ is divisible by 5.

So, $$p$$ and $$q$$ has a common factor 5 which is a contradiction to the assumption that they are co prime.

Hence, $$\sqrt 5 $$ is an irrational number.

Assume that $$\sqrt 2  + \sqrt 3 $$ is a rational number. Then,

$$\sqrt 2  + \sqrt 3  = \frac{p}{q}$$

Where $$p$$ and $$q$$ are co-prime numbers and $$q \ne 0$$.

Squaring both sides,

$$2 + 3 + 2\sqrt 3  = \frac{{{p^2}}}{{{q^2}}}$$

$$5 + 2\sqrt 3  = \frac{{{p^2}}}{{{q^2}}}$$

$$2\sqrt 3  = \frac{{{p^2}}}{{{q^2}}} - 5$$

$$2\sqrt 3  = \frac{{{p^2} - 5{q^2}}}{{{q^2}}}$$

$$\sqrt 3  = \frac{{{p^2} - 5{q^2}}}{{2{q^2}}}$$

It can be observed that $$\frac{{{p^2} - 5{q^2}}}{{2{q^2}}}$$ is a rational number but $$\sqrt 3 $$ is an irrational number which cannot be possible.

Therefore, the assumption is wrong.

Hence, $$\sqrt 2  + \sqrt 3 $$ is an irrational number.

Consider that $$3\sqrt 2 $$ is a rational number. Then,

$$3\sqrt 2  = \frac{p}{q}$$

Where $$p$$ and $$q$$ are co-prime numbers and $$q \ne 0$$.

Squaring both sides,

$$18 = \frac{{{p^2}}}{{{q^2}}}$$

$$18{q^2} = {p^2}$$                   (1)

This shows that $${p^2}$$ is divisible by 18 and thus $$p$$ is divisible by 18. Therefore,

$$p = 18k$$

$${p^2} = 324{k^2}$$                  (2)

From equation (1) and (2),

$$18{q^2} = 324{k^2}$$

$${q^2} = 18{k^2}$$

It can be observed that $$p$$ and $$q$$ both are divisible by 18 which is a contradiction to the fact that $$p$$ and $$q$$ are co-primes.

Therefore, the assumption is wrong.

Hence, $$3\sqrt 2 $$ is an irrational number.

Consider that $$\sqrt 3 $$ is a rational number. Then,

$$\sqrt 3  = \frac{p}{q}$$

Where $$p$$ and $$q$$ are co-prime numbers and $$q \ne 0$$.

Squaring both sides,

$$3 = \frac{{{p^2}}}{{{q^2}}}$$

$$3{q^2} = {p^2}$$                   (1)

This shows that $${p^2}$$ is divisible by 3 and thus $$p$$ is divisible by 3. Therefore,

$$p = 3k$$

$${p^2} = 9{k^2}$$                  (2)

From equation (1) and (2),

$$3{q^2} = 9{k^2}$$

$${q^2} = 3{k^2}$$

It can be observed that $$p$$ and $$q$$ both are divisible by 3 which is a contradiction to the fact that $$p$$ and $$q$$ are co-primes.

Therefore, the assumption is wrong.

Hence, $$\sqrt 3 $$ is an irrational number.

Assume that $$5 - \sqrt 3 $$ is a rational number. Then it can be expressed as a fraction,

$$5 - \sqrt 3  = \frac{p}{q}$$

$$5 - \frac{p}{q} = \sqrt 3 $$

Where $$p$$ and $$q$$ are co-prime numbers and $$q \ne 0$$.

$$\dfrac{{5q - p}}{q} = \sqrt 3 $$

It can be observed that $$\dfrac{{5q - p}}{q}$$ is a rational number because adding or subtracting two rational numbers gives a rational number but $$\sqrt 3 $$ is an irrational number which cannot be possible.

Therefore, the assumption is wrong.

Hence, $$5 - \sqrt 3 $$ is an irrational number.

$$17,23,29$$ are prime numbers.they have only two factors i.e 1 and itself
$$f(17)={1,17}$$
$$f(23)={1,23}$$
$$f(29)={1,29}$$

hcf of prime numbers =1
hcf$$(17,23,29)=1$$

lcm of prime numbers= product of the numbers

hcf$$(8,9,25)=1$$(1 is common factor of each number)
$$lcm =17*29*23=11339$$