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Show that { root(n+1) + root(n-1)}  is an irrational number where  n E N.

How is its done...???

Suppose there exists a positive integer n for which  is a rational number.

, where p and q positive integers and q ≠ 0.

From (1) and (2),

⇒ n + 1 and n – 1 perfect square of positive integers.

Now, (n + 1) – (n – 1) = 2, which is not possible since any two perfect squares differ by atleast 3.

Hence, there is no positive integer n for which  is a rational number. 

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 i didn't understood d last thing !!!! wats "n E N " ????///

  • -6

Me also don't  understood that thing only...thats why i asked....

My teacher gave us a zerox copy of sme pages..nd have to complete that....bt what does it me i too dont know...

dat middle sign was just like  capital  " E ".......

  • -9

Suppose a = Ön+1 + Ön-1 is a rational number.

a - Ön-1 = Ön+1

a2 2aÖn-1 + n-1 = n+1 [ squaring on both the sides ]

2aÖn-1 = a2 2

Ön-1 = a2 2/ 2a is rational.

Similarly, Ön+1 = a2 + 2/ 2a is rational.

If Ön+1 = p and Ön-1 = q, then p, q are positive integers because p and q are rational and n Î N.

+1 = p2 and n-1 = q2 [ squaring on both the sides ]

p2 - q2 =2

(p-q)(p+q) = 2

But factors of 2 are 1 and 2 only

p + q = 2 and p q = 1

Now as p q ¹ 0 , p and q are distinct positive integers.

Hence the minimumvalue of p + q should be 2 + 1 = 3 .

So p + q ¹ 2

Hence our supposition that Ön+1 + Ön-1 is a rational number is wrong.

For any n Î N, Ön+1 + Ön-1 is an irrational number.

  • -8

Suppose a = Ön+1 + Ön-1 is a rational number.

a - Ön-1 = Ön+1

a2 2aÖn-1 + n-1 = n+1 [ squaring on both the sides ]

2aÖn-1 = a2 2

Ön-1 = a2 2/ 2a is rational.

Similarly, Ön+1 = a2 + 2/ 2a is rational.

If Ön+1 = p and Ön-1 = q, then p, q are positive integers because p and q are rational and n Î N.

+1 = p2 and n-1 = q2 [ squaring on both the sides ]

p2 - q2 =2

(p-q)(p+q) = 2

But factors of 2 are 1 and 2 only

Now as p q ¹ 0 , p and q are distinct positive integers.

Hence the minimumvalue of p + q should be 2 + 1 = 3 .

So p + q ¹ 2

p + q = 2 and p q = 1

Now as p q ¹ 0 , p and q are distinct positive integers.

Hence the minimumvalue of p + q should be 2 + 1 = 3 .

So p + q ¹ 2

Hence our supposition that Ön+1 + Ön-1 is a rational number is wrong.

For any n Î N, Ön+1 + Ön-1 is an irrational number.

  • -6

Suppose a = root(n+1) + root(n-1) is rational number.

a - root(n-1) = root(n+1)

a2-2aroot(n-1) + n-1 = n+1 [ By squaring on both the sides ]

2aroot(n-1) = a2- 2 / 2a is rational.

simillarly , root)n+1) = a2+ 2 / 2a is rational.

If root(n+1) = p and root(n-1) =q then p, q are positive integers because p and q are rational and n belongs to natural number.

n+1 = p2 and n-1 = q2 [ n minus 1 = q square]

p2 - q2= 2 [ p square minus q square = 2 ]

(p-q)(p+q) =2

but factors of 2 are 2 and 1 only.

p + q = 2 and p-q =1 { p minus q =1}

Now as p - q { p minus q} is not equal to zero p and q are distinct positive integers,

Hence the minimum value of p+q is 2 + 1 = 3 .

So p + q is not equal to 2.

Hence our supposition that root(n+1) + root(n-1) is rational is wrong.

Thus for any n belongs to N root(n+1) + root(n-1) is an irrational number.

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