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This seems like a very simple question, the answer to which should be obvious to me. So I won't be offended if this question is closed. But I'm having trouble figuring out the answer.

I would like to write a function that returns True if a list contains one or more numbers greater than 1. Why does MemberQ[{2, 1}, # > 1 &] return False? Thanks.

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    $\begingroup$ Select[{2, 1}, # > 1 &] != {} $\endgroup$ – Dr. belisarius Oct 17 '12 at 16:13
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    $\begingroup$ Catch[Scan[If[# > 1, Throw[True]] &, {1, 2, 6, 8}]] $\endgroup$ – Dr. belisarius Oct 17 '12 at 16:19
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    $\begingroup$ Simply put, the second argument of MemberQ[] should be a pattern, not a test. What does MatchQ[2, # > 1 &] return? $\endgroup$ – J. M. is away♦ Oct 17 '12 at 16:28
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    $\begingroup$ The comment by @J.M. answers your question; if you insist on using MemberQ you can do MemberQ[{2, 1}, _?(# > 1 &)] to transform a test into a pattern. $\endgroup$ – b.gates.you.know.what Oct 17 '12 at 16:37
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    $\begingroup$ The simplest way is clearly ! IntegerQ@Log2[1 + FromDigits[UnitStep[1 - {1, 2}]~Prepend~1, 2]] $\endgroup$ – Rojo Oct 17 '12 at 18:44
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As J. M. kindly points out, I must specify form as a pattern, not a test. Thus, as b.gatessucks kindly points out, I must instead use:

MemberQ[{2, 1}, _?(# > 1 &)]

or one of the other suggestions given as comments to the original post.

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