Question

Show that root 2 is irrational number

Answer 1

ᴀɴSᴡᴇƦ :-

ʟᴇᴛ ᴜs ᴀssᴜᴍᴇ ᴛʜᴀᴛ √2 ɪs ᴀ ʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ.

ᴛʜᴇʀᴇғᴏʀᴇ,

$\frac{ \sqrt{2} }{1} = \frac{a}{b}$

ᴡʜᴇʀᴇ ᴀ ᴀɴᴅ ʙ ᴀʀᴇ ɪɴᴛᴇɢᴇʀs ᴀɴᴅ ʙ ɪs ɴᴏᴛ ᴇǫᴜᴀʟs ᴛᴏ 0.

ɴᴏᴡ,

$\sqrt{2 } b = a$

Sǫᴜᴀʀɪɴɢ ʙᴏᴛʜ sɪᴅᴇs.

$2 {b}^{2} = {a}^{2} \\ \\ {b}^{2} = \frac{ {a}^{2} }{2}$

ɪғ 2 ᴅɪᴠɪᴅᴇs

${a}^{2}$

ɪᴛ ᴀʟsᴏ ᴅɪᴠɪᴅᴇs ᴀ

ɴᴏᴡ,

$a = 2c$

Sǫᴜᴀʀɪɴɢ ʙᴏᴛʜ sɪᴅᴇs,

${a}^{2} = 4 {c}^{2} \\ \\ 2 {b}^{2} = 4 {c}^{2} \\ \\ {b}^{2} = 2 {c }^{2} \\ \\ {c}^{2} = \frac{ {b}^{2} }{2}$

ɪғ 2 ᴅɪᴠɪᴅᴇs

${b}^{2}$

2 ᴀʟsᴏ ᴅɪᴠɪᴅᴇs ʙ

ᴛʜᴇʀᴇғᴏʀᴇ, ᴛʜɪs ɪs ᴀ ᴄᴏɴᴛʀᴀᴅɪᴄᴛɪᴏɴ,

ʜᴇɴᴄᴇ,

√2 ɪs ᴀɴ ɪʀʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ,

ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs ʏᴏᴜ.

${\huge{\pink{\ddot{\smile}}}}$

Answer 2

ᴀɴSᴡᴇƦ :-

ʟᴇᴛ ᴜs ᴀssᴜᴍᴇ ᴛʜᴀᴛ √2 ɪs ᴀ ʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ.

ᴛʜᴇʀᴇғᴏʀᴇ,

$\frac{ \sqrt{2} }{1} = \frac{a}{b}[tex]</p><p>1</p><p>2</p><p> </p><p> </p><p> </p><p> = </p><p>b</p><p>a</p><p> </p><p> </p><p></p><p>ᴡʜᴇʀᴇ ᴀ ᴀɴᴅ ʙ ᴀʀᴇ ɪɴᴛᴇɢᴇʀs ᴀɴᴅ ʙ ɪs ɴᴏᴛ ᴇǫᴜᴀʟs ᴛᴏ 0.</p><p></p><p>ɴᴏᴡ,</p><p></p><p>[tex]\begin{lgathered}\sqrt{2 } b = a \\\end{lgathered}$

2

b=a

Sǫᴜᴀʀɪɴɢ ʙᴏᴛʜ sɪᴅᴇs.

$\begin{lgathered}2 {b}^{2} = {a}^{2} \\ \\ {b}^{2} = \frac{ {a}^{2} }{2}\end{lgathered}$

ɪғ 2 ᴅɪᴠɪᴅᴇs

${a}^{2}$a

2

ɪᴛ ᴀʟsᴏ ᴅɪᴠɪᴅᴇs ᴀ

ɴᴏᴡ,

a = 2ca=2c

Sǫᴜᴀʀɪɴɢ ʙᴏᴛʜ sɪᴅᴇs,

$\begin{lgathered}{a}^{2} = 4 {c}^{2} \\ \\ 2 {b}^{2} = 4 {c}^{2} \\ \\ {b}^{2} = 2 {c }^{2} \\ \\ {c}^{2} = \frac{ {b}^{2} }{2} \\ \\\end{lgathered}$

ɪғ 2 ᴅɪᴠɪᴅᴇs

${b}^{2}$b

2

2 ᴀʟsᴏ ᴅɪᴠɪᴅᴇs ʙ

ᴛʜᴇʀᴇғᴏʀᴇ, ᴛʜɪs ɪs ᴀ ᴄᴏɴᴛʀᴀᴅɪᴄᴛɪᴏɴ,

ʜᴇɴᴄᴇ,

√2 ɪs ᴀɴ ɪʀʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ,

ʜᴏᴘᴇ ɪᴛ ʜᴇʟᴘs ʏᴏᴜ.

${\huge{\pink{\ddot{\smile}}}}$