if r 0 is sqrt(r) an integer I) r^2 = integer II) r =

Question

if r > 0 is sqrt(r) an integer 

I) r^2 = integer 

II) r = m^2 where m = integer

vprabhala · Answer

I would say D.
I) r^2 = integer 
r>0 from the stem.. for eg. 1.5^2 will not give an integer ever
r can be either -ve or +ve . but we already know that r>0 so sufficient.

II) r = m^2 where m = integer 
 m is an integer so.. r is also an integer.
sufficient.

banerjeea_98 · Answer

"B" it is.

statement 1....if r = 9...r2 = 81....sqrt (9) = 3...but if r = 7....r2 = 49....sqrt(7) = not integer...insuff

statement 2: r = m2....for any value of integer m ....r = sqrt(m2) ===> integer...suff

vprabhala · Answer

oops. i missed the sqrt(r) part.. I must be dreaming.. yeah it is B..

thenine · Answer

I go B

I) sub in the number 3 for r...... 3^2 = 9. 9 is an integer, but the square root of 3 is not an integer

II) From this formula we know the square root of r = m. So if m is an integter, as defined, the II is sufficent.

if r > 0 is sqrt(r) an integer I) r^2 = integer II) r =

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