Show that √2 ; √3; √5 can not be terms of a single A.P.
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Show that √2 ; √3; √5 can not be terms of a single A.P.

[From: ] [author: ] [Date: 13-08-11] [Hit: ]
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)-2LHS isa rational number [since k and nare integers]butRHS isan irrationalnumber,thus the equality fails and our initial assumptionthatthese numbers are terms of an A.P. is wrongi.......
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);

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Assume they are,
then from √2 to √3 is p steps of d,
√3 to √5 is q steps of d,
where p and q are integers

(√3 - √2)/(√5 - √3) = p/q = rational

multiply LHS to and bottom by complex conjugate (√5 + √3),
which must result in an integer denominator.

(√3 - √2)(√5 + √3) = rational

√3√5 + 9 - √2√5 - √2√3 = rational

√3√5 - √2√5 - √2√3 = rational

Each LHS term is irrational.

Now, starting with all irrational numbers, and then adding or subtracting, we will only get a rational result in the special case of where the same irrational number is present in both parts eg
(3 - √2) + √2
But that is not the case here. Thus the LHS is irrational and we have arrived at a contradiction, thus disproving the premise.

-
Forgive my desire for clarification, what is "single A.P." ?

-
I'm not sure if I'm right.
√3 - √2 ≠ √5 - √3
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