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by kvcpk » Wed Jul 28, 2010 11:05 pm
nikhilkatira wrote:If n = p/q (p and q are nonzero integers), is n an integer?

1. n^2 is an integer.
2. 2n+4/2 is an integer
n = p/q

1. n^2 is an integer.
n^2 is an integer and n is not integer occurs when n is square root of intger.
Example when n is sqrt(2).
But, sqrt(2) cannot be expressed in the form p/q, because it is irrational.
Hence n should be an integer.
SUFF

2. 2n+4/2 is an integer

2n+4 = 2k
2n = 2k-4
n=k-2
K is integer. Hence k-2 is also integer.
SUFF

pick D
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by outreach » Thu Jul 29, 2010 1:30 am
stmt2

2n+4/2 = 2n+2
n can be 1/2,3/2,2 etc
so it should be insuff

where am i wrong?



kvcpk wrote:
nikhilkatira wrote:If n = p/q (p and q are nonzero integers), is n an integer?

1. n^2 is an integer.
2. 2n+4/2 is an integer
n = p/q

1. n^2 is an integer.
n^2 is an integer and n is not integer occurs when n is square root of intger.
Example when n is sqrt(2).
But, sqrt(2) cannot be expressed in the form p/q, because it is irrational.
Hence n should be an integer.
SUFF

2. 2n+4/2 is an integer

2n+4 = 2k
2n = 2k-4
n=k-2
K is integer. Hence k-2 is also integer.
SUFF

pick D
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by selango » Thu Jul 29, 2010 1:58 am
n=p/q

stmt1,

n^2 is an integer.

p^2/q^2 is an integer.In order to be an integer p^2 and q^2 must have common prime factors so that p^2 is divisble by q^2.

Suff

stmt2,

2n+4/2 is an integer

I think it's (2n+4)/2 is an integer

n+2 is an integer.So n must be surely an integer.

Suff.

Pick D
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by kvcpk » Thu Jul 29, 2010 1:59 am
outreach wrote:stmt2

2n+4/2 = 2n+2
n can be 1/2,3/2,2 etc
so it should be insuff

where am i wrong?
I think it is (2n+4)/2.
You took it as 2n+(4/2).
I am not sure which one the intended statement was.
Nikhil - Can you confirm?
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by nikhilkatira » Thu Jul 29, 2010 2:31 am
kvcpk wrote:
outreach wrote:stmt2

2n+4/2 = 2n+2
n can be 1/2,3/2,2 etc
so it should be insuff

where am i wrong?
I think it is (2n+4)/2.
You took it as 2n+(4/2).
I am not sure which one the intended statement was.
Nikhil - Can you confirm?
Sorry guys..my mistake
its (2n+4)/2.
Best,
Nikhil H. Katira
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by kvcpk » Thu Jul 29, 2010 2:50 am
nikhilkatira wrote: Sorry guys..my mistake
its (2n+4)/2.
No Problem.. is the OA D?
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by nikhilkatira » Thu Jul 29, 2010 4:57 am
kvcpk wrote:
nikhilkatira wrote: Sorry guys..my mistake
its (2n+4)/2.
No Problem.. is the OA D?
OA is D
Best,
Nikhil H. Katira
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by ymach3 » Thu Nov 18, 2010 12:44 am
hi , got a doubt for stmt 1

n^2=2 then n=sq root(2)=1.414=1414/1000=p/q..

Is'nt this correct???
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by Rahul@gurome » Thu Nov 18, 2010 4:15 am
ymach3 wrote:...
n^2=2 then n=sq root(2)=1.414=1414/1000=p/q..

Is'nt this correct???
√2 is an irrational number and by definition an irrational number cannot be expressed as the ratio of two integers. We can easily prove that √2 is an irrational number and the proof is an interesting one! :)

Proof: If √2 is rational, then it has the form p/q for integers p, q not both even.
Then, p = (√2)q => p² = 2q², hence p is even
Say, p = 2m. Thus 4m² = 2q² => 2m² = q², hence q is also even, a contradiction.


In fact, √2 = 1.414213562... ≈ 1.414 (When rounded to 3 places after decimal)
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by kyle8285 » Thu Nov 18, 2010 12:35 pm
Rahul,

Is it then sufficient to say that any number that is not a perfect square (has an integer square root) is irrational and therefore cannot be expressed as the ratio of two integers? For example, sq root(5), sq root(12), etc. Thank you!
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by Rahul@gurome » Thu Nov 18, 2010 1:09 pm
kyle8285 wrote:Rahul,

Is it then sufficient to say that any number that is not a perfect square (has an integer square root) is irrational and therefore cannot be expressed as the ratio of two integers? For example, sq root(5), sq root(12), etc. Thank you!
Yes. But the generalized proof is different and a bit complicated and not necessary too.
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