Inequality proof

  • Thread starter phospho
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  • #1
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Homework Statement


Prove

If ## 0 \leq a < b ## and ## 0 \leq c < d ## then ## ac < bd ##

The Attempt at a Solution



not sure how to even start on this,

was thinking if a = 0 or c = 0, then ac = 0, but bd > 0 (which is given) so bd > ac
however this seems like I'm cheating because they give you bd > ac, so I don't think it's valid
 

Answers and Replies

  • #2
verty
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They don't give you ac < bd, that is what you must show.

What you have done so far is correct but you still have to show it is true when 0 < a, 0 < c.

Hint: you were probably given the rule that, for 0 < c, a < b → ac < bc. Use it.
 
  • #3
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They don't give you ac < bd, that is what you must show.

What you have done so far is correct but you still have to show it is true when 0 < a, 0 < c.

Hint: you were probably given the rule that, for 0 < c, a < b → ac < bc. Use it.
Yes we were given that, but I couldn't proceed how to use it. I used ac < bd to get to where I am right now.
 
  • #4
tiny-tim
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I used ac < bd to get to where I am right now.
no, you can't use ac < bd !!

use verty's :smile: hint

(second hint: if you could prove ac < bc, what would you still need to get ac < bd ? :wink:)
 
  • #5
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I'm sorry, I'm trying but I just don't see how to proceed.

For the proof of c>0 and a<b prove ac<bc it was very simple, all I had to use was (b-a).c > 0, but when there are two inequality signs I don't know how to deal with it. Is multiplying by b or d to each of the inequalities allowed? For instance, if 0<=a<b and 0 <=c<d is it true that 0<=bc < bd?
 
  • #6
tiny-tim
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hi phospho! :smile:
(second hint: if you could prove ac < bc, what would you still need to get ac < bd ? :wink:)
… is it true that 0<=bc < bd?
that's right! … if you can prove ac < bc and bc < bd,

then obviously you have ac < bc < bd, ie ac < bd

sooo … now prove ac < bc and bc < bd (separately) :wink:
 

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