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Simple inequalities equation


 
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PostPosted: Mon Jan 08, 2007 6:01 pm    Post subject: Simple inequalities equation Reply with quote

1/k > 0. I would think that hence 1> K- But this is wrong. Why???

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Argen
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PostPosted: Wed Jan 10, 2007 11:55 am    Post subject: Reply with quote

When k is greater than 1, 1/k is less than 1 but still greater than 0. If 1/k>0 then 0<k.
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PostPosted: Wed Jan 10, 2007 8:37 pm    Post subject: Reply with quote

Oops! Ignore this post - what I explained doesn't apply to this problem, as I should have seen if I'd re-read what I typed! (Though the first paragraph, below, is still true in general, so I left it.) See later post for clarification.

Be very careful when manipulating variables in inequalities. When you multiply or divide an inequality by a negative number, you have to reverse the sign. If you are multiplying or dividing by a variable and you don't know whether that variable is positive or negative, you have two scenarios:

[Removed this math so no one thinks it's valid - I was totally wrong!]

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Last edited by Stacey Koprince on Thu Jan 11, 2007 2:57 pm; edited 3 times in total
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PostPosted: Thu Jan 11, 2007 1:39 am    Post subject: Reply with quote

Hi Stacey, could you explain what you mean by if k is negative then 1<k. If k is negative then k<0<1. Also, if 1/k>0, then k can't be negative, right?
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PostPosted: Thu Jan 11, 2007 2:01 am    Post subject: Reply with quote

Oops! Teach me to respond after a long day at work and while I'm trying to cook dinner at the same time. Smile

I wasn't paying careful attention to the actual expression given and tried to explain something that doesn't even actually apply here. Ignore what I said before!

Yes, if you're given 1/k > 0, then k must represent a positive number. To solve for k, take the reciprocal of both sides and get k>0.

Alternatively you can test some numbers for k and see what works. k could equal 1, 2, 3, any positive integer. k could also equal 1/2, 1/3, or any positive fraction between zero and one. k can't equal zero (can't divide by zero) and it can't be negative (b/c the expression won't be true.

If you're not sure how to solve it algebraically and go the "try some numbers" route, it's important to try positive integers, positive fractions between zero and one, zero, one, negative numbers, and possibly negative fractions between zero and one (though this last one usually only comes into play on really hard questions). Each of those categories has specific / unique attributes when performing various operations. So when I initially narrowed it down to positive, I knew I had two categories of positive numbers to test: integers and fractions bet. zero and one.

Sorry about that! I'll go edit the old post so people don't read it and get messed up!

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