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
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nikhilkatira
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kvcpk
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n = p/qnikhilkatira 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
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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outreach
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stmt2
2n+4/2 = 2n+2
n can be 1/2,3/2,2 etc
so it should be insuff
where am i wrong?
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:n = p/qnikhilkatira 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
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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selango
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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
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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kvcpk
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I think it is (2n+4)/2.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?
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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nikhilkatira
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Sorry guys..my mistakekvcpk wrote:I think it is (2n+4)/2.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?
You took it as 2n+(4/2).
I am not sure which one the intended statement was.
Nikhil - Can you confirm?
its (2n+4)/2.
Best,
Nikhil H. Katira
Nikhil H. Katira
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nikhilkatira
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OA is Dkvcpk wrote:No Problem.. is the OA D?nikhilkatira wrote: Sorry guys..my mistake
its (2n+4)/2.
Best,
Nikhil H. Katira
Nikhil H. Katira
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Rahul@gurome
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√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!ymach3 wrote:...
n^2=2 then n=sq root(2)=1.414=1414/1000=p/q..
Is'nt this correct???
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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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!
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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Rahul@gurome
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Yes. But the generalized proof is different and a bit complicated and not necessary too.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!
Rahul Lakhani
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Quant Expert
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