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Hi. I couldn't solve a proof problem, can you help me about this?
the question is:
thanks for any help
The Contrapositive is this:
If a statement is
then the contrapositive is
We have also
, that is, if one is true then both are true. If one is false then both are false.So if the statement is this: "1<x or x<-1" implies "|x|>1"
It's contrapositive is hence: A, "|x|≤1" implies "-1≤x≤1"
Now if you can prove either one of them, then the other will follow immediately.
In this case, it is far more easy to prove the contrapositive first, then let the original statement follow.
"|x|≤1" implies "-1≤x≤1" is a well known and very useful absolute value identity. Unfortunately I can't help you in proving it though, since I learnt it a while ago.
While doing the proof make sure to note in your proof that |x| must be positive, just so the marker knows you know you've considered all the cases.
Last edited by Identity (2007-10-06 16:12:00)
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There are three ranges of values x can be assigned.
(i) x < -1. For example, x=-2, x=-5/2, x=-√2, x=-√5 etc.
(ii) x > 1. For example, x=2, x=5/2, x=√2, x=√5 etc.
(iii)-1≤x≤1. For example, x=-1/2, -1/5,0, 1/5, 1/2 etc.
In the three cases above, |x|>1 is true for both (i) and (ii).
However, |x|≤1 for the third range of values.
Therefore, in order to satisfy the equation |x|>1, either x<-1 or x>1.
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
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i'm not good at proving something, can you explain more to me? sorry
So if the statement is this: "1<x or x<-1" implies "|x|>1"
It's contrapositive is hence: A, "|x|≤1" implies "-1≤x≤1".
You got it the wrong way round!! The original statement is "|x| > 1" implies "x > 1 or x −1".
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i can't understand how to proof by cases (it says i need to use proof by cases)
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