Question

Prove that root 2, root 3 and root 5 cannot be the terms of any

a.P

Answer 1

Step-by-step explanation:

you meant : can these three terms be in an A.P

Let us prove this by contradiction, assume such an A.P exists and root(2) is the first term of that

A.P.

then if root(3) is in that A.P.,

root(3)=root(2)+n.d; where n is a Natural number denoting root(3) is (n+1)th term of the A.P;

and d is the common difference of that A.P.

implies:

n.d=root(3)-root(2)-----eq(1)

similarly,

root(5)=root(2)+k.d ,where k is also a Natural no. and root(5) is (k+1)th term of the A.P.

implies:

k.d=root(5)-root(2)------eq(2)

eq(2)/eq(1) implies: [since d !=0 in our case,we can perform this division]

k/n= ( root(5)-root(2) ) / ( root(3)-root(2) )

on simplification:

k/n= root(15)-root(6)+root(10)-2

LHS is a rational number [since k and n are integers] but

RHS is an irrational number, thus the equality fails and our initial assumption

that these numbers are terms of an A.P. is wrong

i.e) for any d belonging to the set of Real numbers, you cannot define two positions n and k

from root(2) in an A.P. to accommodate root(3) and root(5);

Answer 2

Let's assume that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ can be in an A.P.

∴ $\sqrt{3} = \sqrt{2} + (n-1)d$ ----------(1)

and,

$\sqrt{5} = \sqrt{2} + (k-1)d$ -----------(2)

Re-arranging (1):

$\sqrt{3} - \sqrt{2} = (n-1)d$

Re-arranging (2):

$\sqrt{5} - \sqrt{2} = (k-1)d$

Subtracting (1) from (2):

$\sqrt{5} - \sqrt{3} = (k-n)d$

The sum of $\sqrt{5}$ and $(-\sqrt{3)}$ will be irrational.

$(k-n)d$ will be rational as the product of two rational numbers is always rational.

Thus, LHS ≠ RHS.

So our assumption must be wrong. Thus, $\sqrt{2} , \sqrt{3}$ and $\sqrt{5}$ can't form an A.P.