# How to solve an inequality

I am trying to solve the following inequality:

Assuming[{R[k] >= 0, R[k + 1] = R[k] - 2^N Y, Y >= 0 && N >= 0},
Refine[Re[R[k + 1] >= 0]]


I want check whether R[k+1] >=0 is true or not, but I am getting Re[True < 0] and I don't know what this means.

R[k + 1] = R[k] - 2^N Y
Y >= 0
N >= 0
Simplify[R[k + 1] < 0, R[k] >= 0]


In this case i am getting R[k] < 2^k Y. I want true or false as result.

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I think you need to write R[k + 1] == R[k] - 2^N[k] Y in there and not R[k + 1] = R[k] - 2^N[k] Y i.e. use == and not = ? – Nasser Mar 4 '14 at 2:02
Thanks a lot Nasser for your response ,but after that modification getting following result..Re[R[1 + k] >= 0] – Tom Mar 4 '14 at 2:45
I attempted to edit your post for clarity, the second portion appears to contain a heavy dose of pseudo-code and would benefit from some editing on your part. – bobthechemist Mar 4 '14 at 2:49
what are you expecting? it is trivial to think of examples that will make R[k+1] less than, greater than or equal to zero. – george2079 Mar 4 '14 at 4:37
To more or less repeat what @george2079 stated, if you keep subtracting something of fixed positive size (2^n*y, in this case) then for some k you will have r[k+1]<=0. – Daniel Lichtblau Mar 4 '14 at 15:18