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
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
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?
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?
OA is Dkvcpk wrote:No Problem.. is the OA D?nikhilkatira wrote: Sorry guys..my mistake
its (2n+4)/2.
√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???
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!
