Question

prove that 2+root3/5 is an irrational number given that root 3 is irrational number?? please fully explain this question​

Answer 1

Answer:

Let it be rational number

therefore it can be written in form of a and b where a and b are co-prime numbers.

2√3=5a/b

5a/b is rational number as it is of the form p/q which is a rational number.

but we know that √3 is irrational number so our assumption is wrong.

2√3/5 is irrational.

Step-by-step explanation:

here we used contradiction method..

Answer 2

Answer :

hola please put 3 instead of 5 and 5 instead of 3

Explanation:

first we have toproove that √5 is an irrational no.

that is by following

Let us suppose that√5 is an rational no.

√5=p/q {where p and q is coprime and q=× 0}

√5q=p

on squaring both sides

$5 {q}^{2} = {p}^{2} \: \: eq1$

${q}^{2} = {p}^{2} \div 5$

{as 5 divides p(square), 5 will also divide p}

p÷5=r (where r is remainder)

on squaring both sides

${p}^{2} \div 25 = {r}^{2}$

${p}^{2} = {r}^{2} \div 25$

putting value of p from eq1

$5 {q}^{2} = 25 {r}^{2}$

dividing it we get

${q}^{2} \div 5 = {r}^{2}$

{as 5 divides q(square), 5 will also divide q}

our assumption gets wrong as 5 divides both p and q

therefore √5 is an irrstional no.

then we will solve like that so that √5 be equal to an rational no.

I am explaining you 2+√5÷3 so please change it accordingly

$\sqrt{5 } = 3 - 2$

$\sqrt{5} = 1$

as an irrational no can't be equal to an rational no so

$2 + \sqrt{5} \div 3$

is an irrational no

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