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