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17 Apr 2005, 12:57
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N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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If r > 0 is sqrt(r) an integer
I r^2 is an integer
II r = m^2 where m is an integer
OA = B
I dont understand why I is also not a sufficient answer i was thinking it wud be D
thanks
I r^2 is an integer
II r = m^2 where m is an integer
OA = B
I dont understand why I is also not a sufficient answer i was thinking it wud be D
thanks
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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17 Apr 2005, 13:14
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really tricky
for statement 1 you know that r^2 is an integer (suppose 4), so r is integer, in this case 2, but sqrt[2] is not an integer...
at the same time, r^2 could be 16, and r 4, and sqrt[4]=2, which is an integer...
for statement 1 you know that r^2 is an integer (suppose 4), so r is integer, in this case 2, but sqrt[2] is not an integer...
at the same time, r^2 could be 16, and r 4, and sqrt[4]=2, which is an integer...
18 Apr 2005, 03:58
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B
(1) imagine r^2 = 3, r is not an integer
(2) we know in B that r = m^2 where m is an integer so r is an integer (a square of an integer is always an integer) and root(r) is definitely an integer because root(r) = m
(1) imagine r^2 = 3, r is not an integer
(2) we know in B that r = m^2 where m is an integer so r is an integer (a square of an integer is always an integer) and root(r) is definitely an integer because root(r) = m
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
