Question

Prove that sqaure root of 3 is an irrational number..​

Answer 1

Answer:

Let us assume to the contrary that v3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q 0.

V3 = p/q

3 = p?/q? (Squaring on both the sides)

3q? = p?.

..(1)

This means that 3 divides p2. This means that 3 divides p because each factor should appear two times for the square to

exist.

So we have p = 3r

where r is some integer.

p? = 9r2.

from equation (1) and (2) 3q? = 9r2

q3 r2

0.6KB/s *

Where q 2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co prime.

Consequently, p/q is not a rational number. This demonstrates that v3 is an irrational number.

$hope \: it \: help \: you$

Answer 2

Aɴsᴡᴇʀ:

Lᴇᴛ ᴜs ᴀssᴜᴍᴇ ᴛᴏ ᴛʜᴇ ᴄᴏɴᴛʀᴀʀʏ ᴛʜᴀᴛ ᴠ3 ɪs ᴀ ʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ.

Iᴛ ᴄᴀɴ ʙᴇ ᴇxᴘʀᴇssᴇᴅ ɪɴ ᴛʜᴇ ғᴏʀᴍ ᴏғ ᴘ/ϙ

ᴡʜᴇʀᴇ ᴘ ᴀɴᴅ ϙ ᴀʀᴇ ᴄᴏ-ᴘʀɪᴍᴇs ᴀɴᴅ ϙ 0.

V3 = ᴘ/ϙ

3 = ᴘ?/ϙ? (Sϙᴜᴀʀɪɴɢ ᴏɴ ʙᴏᴛʜ ᴛʜᴇ sɪᴅᴇs)

3ϙ? = ᴘ?.

..(1)

Tʜɪs ᴍᴇᴀɴs ᴛʜᴀᴛ 3 ᴅɪᴠɪᴅᴇs ᴘ2. Tʜɪs ᴍᴇᴀɴs ᴛʜᴀᴛ 3 ᴅɪᴠɪᴅᴇs ᴘ ʙᴇᴄᴀᴜsᴇ ᴇᴀᴄʜ ғᴀᴄᴛᴏʀ sʜᴏᴜʟᴅ ᴀᴘᴘᴇᴀʀ ᴛᴡᴏ ᴛɪᴍᴇs ғᴏʀ ᴛʜᴇ sϙᴜᴀʀᴇ ᴛᴏ

ᴇxɪsᴛ.

Sᴏ ᴡᴇ ʜᴀᴠᴇ ᴘ = 3ʀ

ᴡʜᴇʀᴇ ʀ ɪs sᴏᴍᴇ ɪɴᴛᴇɢᴇʀ.

ᴘ? = 9ʀ2.

ғʀᴏᴍ ᴇϙᴜᴀᴛɪᴏɴ (1) ᴀɴᴅ (2) 3ϙ? = 9ʀ2

ϙ3 ʀ2

0.6KB/s *

Wʜᴇʀᴇ ϙ 2 ɪs ᴍᴜʟᴛɪᴘʟʏ ᴏғ 3 ᴀɴᴅ ᴀʟsᴏ ϙ ɪs ᴍᴜʟᴛɪᴘʟᴇ ᴏғ 3.

Tʜᴇɴ ᴘ, ϙ ʜᴀᴠᴇ ᴀ ᴄᴏᴍᴍᴏɴ ғᴀᴄᴛᴏʀ ᴏғ 3. Tʜɪs ʀᴜɴs ᴄᴏɴᴛʀᴀʀʏ ᴛᴏ ᴛʜᴇɪʀ ʙᴇɪɴɢ ᴄᴏ ᴘʀɪᴍᴇ.

Cᴏɴsᴇϙᴜᴇɴᴛʟʏ, ᴘ/ϙ ɪs ɴᴏᴛ ᴀ ʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ. Tʜɪs ᴅᴇᴍᴏɴsᴛʀᴀᴛᴇs ᴛʜᴀᴛ ᴠ3 ɪs ᴀɴ ɪʀʀᴀᴛɪᴏɴᴀʟ ɴᴜᴍʙᴇʀ.