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August 28th, 2018, 11:55 AM  #1 
Member Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0  Prove that at least one number must be negative. abc = 1000
So I am having this question in my math textbook that says: Assume $\displaystyle abc = 1000$ Prove that at least one number must be negative. So the only proof method that is availible for the chapter is to do contrary evidence (the opposite to prove that it can't be true). So do it become this then? Prove that more than one number must be positive. Do I need to find a abstract example or can I show a concrete example that doesn't work? Case 1: a and b is positive and c is negative, then abc will be a negative number. Case 2: a, b and c is positive, then abc will be a positive number. If I am correct I need to have that more than one number must be positive. If so, I only have two cases. So this seems on the right track to me, but I haven't really proven anything, I guess. So I need to know how I prove my cases because right now they are just statements. So how can I continue with that? 
August 28th, 2018, 12:19 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,501 Thanks: 1372 
I think it's easier than you make it. Assume that all 3 numbers are nonnegative. Then $a b c$ is nonnegative. This is a property of multiplication of real numbers. We are told to assume $abc = 1000 < 0$, i.e. $abc$ is negative. Thus we have a contradiction between our two assumptions. We are told to assume $abc$ is negative so it must be that all 3 numbers are not nonnegative and thus at least 1 is negative. This can all be stated in a lot fewer words than above. 
August 28th, 2018, 12:28 PM  #3 
Member Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0 
I see. Seeing it from that perspective made the problem much easier and less text to write too 

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1000, abc, negative, number, prove 
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