Question

Classify the following numbers as rational or irrational.

(i) √2 ​

Answer 1

Answer:

let √2 is rational number

√2= a/b ( A and B are two integers whose

sum and hcf is 1 )

√2/b = a

( squaring on both side)

(√2/b)^2 = (a)^2

2/b^2 = a^2

so 2 divides by a^2

2 divide by a

we can write a=2c. ( where c is an integer)

2b^2 = 2c^2

2b^2 = 4c^2

b^2 = 4c^2/2

b^2. = 2c^2

2 divides by b^2

2 divide. by b

So 2 is common factor of both

So our assumption is wrong

so √2 is irrational number

Answer 2

Step-by-step explanation:

To prove that root 2 is irrational.

$let \: us \: suppose \: that \: \sqrt{2 } \: is \: rational$

$then \: \sqrt{2} = \frac{p}{q} (where \: p \: and \: q \: are \: integers \: \\ having \: no \: common \: factors \: other \: \\ than \: 1....................(1)$

$squaring \: both \: sides \: we \: get \\ = {( \sqrt{2} )}^{2} = { (\frac{p}{q}) }^{2} \\ = 2 = \frac{ {p}^{2} }{ {q}^{2} } \\ = 2 {q}^{2} = {p}^{2} .........................(2) \\ = 2 \: divides \: {a}^{2} \\ = 2 \: divides \: a..........................(3) \\ \\ let \: r \: be \: an \: integer \: such \: that \: \\ p = 2r \\ putting \: p = 2r \: i n\: eq.(2) \: we \: get \\ 2 {q}^{2} = {(2r)}^{2} \\ = 2 {q}^{2} = 4 {r}^{2} \\ = {q}^{2} = 2 {r}^{2} \\ = {r}^{2} = \frac{ {q}^{2} }{2} \\ = 2 \: divides \: {q}^{2} \\ = 2 \: divides \: q......................(4) \\ from \: eq.(3) \: and \: (4) \: we \: conclude \: \\ that \: 2 \: is \: the \: other \: common \: factor \: \\ of \: p\: and \: q \: which \: contradicts \: eq.(1).$

hence root 2 is irrational.

similarily we can prove for other irrational numbers.

Thank you........

Hope it helps ..........