Question

prove thatroot3,root11,root17are irrational number​

Answer 1

Answer:

HERE IS YOUR ANSWER:

Step by step explanation:

1.Let us assume that root 3 is rational. 

Root 3 = a/b where a and b are integers and coprimes.

Root 3 * b = a

Square LHS and RHS

3b2 = a2

b2 = a2/3

Therefore 3 divides a2 and 3 divides a.

Now take ,

a = 3c

Square ,

a2 = 9c2

3b2 = 9c2

b2/3 = c2

Therefore 3 divides b2 and b.

2.By the method of contradiction..

Let √11 be rational , then there should exist √11=p/q ,where p & q are coprime and q≠0(by the definition of rational number). So,

√11= p/q

On squaring both side, we get,

11= p²/q² or,

11q² = p². …………….eqñ (i)

Since , 11q² = p² so ,11 divides p² & 11 divides p

Let 11 divides p for some integer c ,

so ,

p= 11c

On putting this value in eqñ(i) we get,

11q²= 121p²

or, q²= 11p²

So, 11 divides q² for p²

Therefore 11 divides q.

So we get 11 as a common factor of p & q but we assumpt that p & q are coprime so it contradicts our statement. Our supposition is wrong and √11 is irrational.

3.suppose root 17 is rational number

let root 17 =p/q

( p/q is positive integer )

17 =p²/q²

17q²=p²

let p = 17 n ( n is positive rational number )

17q²=17²N²

q=17N²

let q=17m(m is positive integer)

That's contradiction

root 17 is irrational no.

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Answer 2

refer to the attachments