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# Triangles

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With the third-side triangle property, the general term is a-b<c<a+b. Correct? Well, to make the sentence work, a has to be greater than b because you can't have a negative length. with this _b a | / c crude triangle, when I say the inequality stated above, do I need to state that a > b? Here is an example question: Set a domain for c in terms of a and b. (For the triangle above)
tcache Contact Me ----------- AH! MY BRAIN IS GOING TO SELF-DETONATE! -- Yours Truly (Jan, 2003) [edited by - Tazel on March 24, 2003 8:47:17 AM] [edited by - Tazel on March 24, 2003 8:48:11 AM]

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less-than signs don't show up right if you type them like that, the browser thinks it's html. You need to type &lt, which gives you a real '<'.

quote:
what he tried to write:
With the third-side triangle property, the general term is a-b<c<a+b. Correct? Well, to make the sentence work, a has to be greater than b because you can't have a negative length.

Anyway, I don't think a-b is really a length, so it can be negative. And even if it is negative, the inequality holds true.

[edited by - Bagel Man on March 24, 2003 3:03:14 AM]

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Well when you''re doing a triangle, it has to be positive. Just think of it as a length. It can''t be negative or 0 because there is something there which takes out zero and a line / length can''t be negative.

tcache
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You just said it yourself, shouldn''t that be a length? Probably absolute value of a-b, not just a-b. Or think of it as, subtract the smaller from the larger . (a-b if a is larger, and b-a if b is larger).

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quote:
Original post by Tazel
Well when you''re doing a triangle, it has to be positive. Just think of it as a length. It can''t be negative or 0 because there is something there which takes out zero and a line / length can''t be negative.

His point is that a-b doesn''t represent anything geometrically, so it can be a negative number. I agree with that.

Cédric

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if a and b are lengths, they are bigger than 0. You can fill in the equation and see that it''s correct.
for both a and b negative it doesn''t hold.

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Here''s an example

4(a)/\6(b)
--
X(c)

According to my original inequality, it would be 4-6<X<4+6 which would be -2 to 10. Now, X can''t be negative...? right? So how would I fix this?

tcache
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AH! MY BRAIN IS GOING TO SELF-DETONATE! -- Yours Truly (Jan, 2003)

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X can''t be negative, but it is still comprised between -2 and 10. We''re also sure that it is above 0, but
0 < X implies -2 < X
So there''s no problem.

As someone else pointed out, you can use the absolute value and be done with your problem

|a-b|
Cédric

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Sigh...Ok....

tcache
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AH! MY BRAIN IS GOING TO SELF-DETONATE! -- Yours Truly (Jan, 2003)

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