If r 0, is r^(1/2) an integer?

Question

If r > 0, is  \sqrt{r}  an integer?

(1) r^2 is an integer. 
(2) r = m^2, where m is an integer.

Skywalker18 · Answer

(1) r^2 is an integer.

If r = 3 , r^2 = 3^2 = 9 is an integer
(3)^(1/2) is not an integer 
r= 4 , r^2 = 4^2 = 16 is an integer
(4)^(1/2) that is 2 is an integer 
Not sufficient

(2) r = m^2, where m is an integer.
=> (r)^(1/2) = m
Square root of r will be an integer .
Sufficient

Answer B

dayasantoshi · Answer

Given, r>0

1. r^2= Integer
Case 1: eg, r^2=25, r = +/-5 but √5 is not integer
Case 2: eg, r^2= 16,r = +/-4 and √4 = +/- 2 is an integer.
Hence Insufficient

2. r=m^2 where m = Int
 => √r = +/- m , which is an Int
Hence B.

BelalHossain046 · Answer

Statement 1: r^2 is an integer.
 If r^2=1, √r=√1=1 (Integer)
If r^2=2, √r=√2=1.41 (NOT Integer)
Insufficient.

Statement 2: r = m^2, where m is an integer.
By Square rooting both sides, it becomes....
=> (r)^(1/2) = m^2*(1/2)= m 
since m is an integer, (r)^(1/2) must be an integer.

bumpbot · Answer

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If r > 0, is r^(1/2) an integer?

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