Answer:

♧︎︎︎ ǫᴜᴇsᴛɪᴏɴ ♧︎︎

• Prove that 3 + 2√5 is an irrational number

♧︎︎︎ ᴀɴsᴡᴇʀ ♧︎︎︎

➪ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ᴘʀᴏᴠᴇ 3+2√5 ɪs ɪʀʀᴀᴛɪᴏɴᴀʟ

ʟᴇᴛ ᴜs ᴀssᴜᴍᴇ ᴛʜᴇ ᴏᴘᴘᴏsɪᴛᴇ

ᴛʜᴀᴛ ɪs 3+2√5 ɪs ʀᴀᴛɪᴏɴᴀʟ

ʜᴇɴᴄᴇ,

3+2√5=a/b

2√5 =a/b -3

2√5 =a-3b/b

√5=1/2 × a-3b/b

here a-3b /2b is rational number

but √5 is irrational

sɪɴᴄᴇ,ʀᴀᴛɪᴏɴᴀʟ ≠ɪʀʀᴀᴛɪᴏɴᴀʟ

ᴛʜɪs ɪs ᴀ ᴄᴏɴᴛʀᴅɪᴄᴛɪᴏɴ

ᴛʜᴇʀᴇ ғᴏʀᴇ ,ᴏᴜʀ ᴀssᴜᴍᴘᴛɪᴏɴ ɪs ɪɴᴄᴏʀʀᴇᴄᴛ

ʜᴇɴᴄᴇ 3+2√5 ɪs ɪʀʀᴀᴛɪᴏɴᴀʟ

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ

ɪ ʜᴏᴘᴇ ᴛʜɪs ʜᴇʟᴘs ᴜʜʜ

ᴍᴀʀᴋ.ᴀs ʙʀᴀɪɴʟᴇᴀsᴛ

[tex]let \: take \: that \:  3+2 \sqrt{5} \: is \\ an \: rational \: number[/tex]

[tex] \\ [/tex]

[tex]so, \: we \: can \: write \: the\\ answer \: as[/tex]

[tex]3 + 2 \sqrt{5} = \frac{a}{b} \\ [/tex]

[tex]Here \:  a and \: b  \: use \: two \: \\ c \: oprime \: number \: and b≠0[/tex]

[tex]2 \sqrt{5} = \frac{a}{b} - 3 \\ \\ 2 \sqrt{5} = \frac{a - 3b}{b} \\ \\ \sqrt{5} = \frac{a - 3b}{2b} [/tex]

[tex] \\ [/tex]

[tex]Here  \: a \:  and \:  b  \: are \: \\ integer \: so \: \frac{a - 3b}{2b} \: \\ is \\ a \: rational number \: so \:   \\ \sqrt{5}   \: should \: be \: rational \\ number \: but \: \: \sqrt{5}   \: is \\ a irrational \: number \: so \: it \: is \\ contradict[/tex]

[tex] \\ [/tex]

[tex]Hence \: 3 + 2 \sqrt{5} \: is \:\\ an \: irrational[/tex]