Question

prove that 5+root3 is irrational ​

Answer 1

Answer:

hope this will be helpful to u

$let \: us \: assume \: that \: 5 + \sqrt{3} \: is \: rational \: number \\ 5 + \sqrt{3} = p \div q \: \: \: \: \: \: q = 0 \\ \sqrt{3 } = p \div q \: - 5 \\ \sqrt{3} = p \div q - 5q \: \\ therefore \: p \div q - 5q \: is \: rational \: number \\ but \: \sqrt{3} \: is \: irrational \: number \\ this \: leads \: t o \: contradiction \\ our \: assumption \: was \: wrong \\ hence \: is \: 5 + \sqrt{3} \: is \: irrational \: number$

Answer 2

Answer:

it is a irrational number

Step-by-step explanation:

assume that 5+root isa rational no

5+root=p÷q(p,qbelongs to Z ( integer) q not equal to 0)

send root3to another side

now it is like as

5=p÷q-root 3

squaring on both sides

25=(p÷q-√3)square

25=(p÷q)square+3-2√3p÷q

now send -2√3p÷q to left side it is as

and also send 25 to right side

2√3p÷q=(p÷q)square+3-25

2√3p÷q=(p÷q) square-22

now do lcm on right side

• it is as. 2√3p÷q=(p÷q) square-22(q)square/(q)square. now cancel q which is on left side denominator with Q square on right side denominator

now sent 2p÷q to right side it is as

√3=(p÷q)square -22(q)square/2pq

note that lHS 1 is not =to RHS 1

on which is a contradiction that what our assumption is wrong i.e 5+√3 isa rational no

hence we proved that 5+√3 is irrational number