I have a question, take is a real function. If increases, then increases ( is )? Does this statement hold for the converse and contrapositive?

TheNihilist
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 08122010 14:28
Last edited by TheNihilist; 13122010 at 19:38. 
conorstone
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 08122010 14:29
pESS oFF wEANER!"

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 08122010 14:31
(Original post by TheNihilist)
I have a question, take f:R> R is a real function. If f increases, then f^2 increases? Does this statement hold for the converse and contrapositive?
The statement holds if and only if the contrapositive holds, so you only need to worry about the converse. 
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 08122010 16:32
(Original post by nuodai)
Can you think of any examples where it's not true?
The statement holds if and only if the contrapositive holds, so you only need to worry about the converse.
Contrapositive when if f^2 is decreasing, then f is decreasing right? 
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 08122010 16:45
(Original post by TheNihilist)
so f(x1) < f(x2). Then f^2(x1) < f^2(x2)?Contrapositive when if f^2 is decreasing, then f is decreasing right?
Statement: if f is increasing then f² is increasing
Contrapositive: if f² isn't increasing then f isn't increasing
Converse: if f² is increasing then f is increasingLast edited by nuodai; 08122010 at 16:46. 
TheNihilist
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 12122010 00:31
(Original post by nuodai)
That's not true. For example, but . This example should give you a hint at some sort of counterexample.
The contrapositive wouldn't say anything about it being decreasing. It would be about it "not being increasing". A function which isn't increasing isn't necessarily decreasing (e.g. is neither). These are the statements you have:
Statement: if f is increasing then f² is increasing
Contrapositive: if f² isn't increasing then f isn't increasing
Converse: if f² is increasing then f is increasing 
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 12122010 12:31
(Original post by TheNihilist)
Ah thanks, that cleared up things for me. I was looking around the internet and some put . I thought was squared, but which is right?Last edited by nuodai; 12122010 at 12:33. 
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 12122010 12:59
Personally my preference is to write and for, well, . I try to avoid writing plain , unless putting in all the implied arguments gets messy. You can sometimes figure out which is meant by looking at the type of the function  if it's, say, , then doesn't even make sense, so necessarily means . If it's , then makes sense, but what does mean? (Yes, usually it means the scalar product . But you may not be working in a context where that makes sense.)

TheNihilist
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 12122010 14:40
(Original post by Zhen Lin)
Personally my preference is to write and for, well, . I try to avoid writing plain , unless putting in all the implied arguments gets messy. You can sometimes figure out which is meant by looking at the type of the function  if it's, say, , then doesn't even make sense, so necessarily means . If it's , then makes sense, but what does mean? (Yes, usually it means the scalar product . But you may not be working in a context where that makes sense.)
I'm thinking if is indeed , then what I was thinking maybe I could do:
Take as our increasing function,defined on , then .
Hence .
Differentiate to get , which is always going to be positive and increasing since for all which is a real number.
So what is it going to be for the converse and contrapositive statements?Last edited by TheNihilist; 12122010 at 15:10. 
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 12122010 15:36
(Original post by TheNihilist)
What if it's ? Should I use both as squared, and ? doesn't make sense in my opinion.
I'm thinking if is indeed , then what I was thinking maybe I could do:
Take as our increasing function,defined on , then .
Hence .
Differentiate to get , which is always going to be positive and increasing since for all which is a real number.
So what is it going to be for the converse and contrapositive statements?
For what it's worth, I'd write for . If you had compositions, say then I'd write ; and for I'd write .Last edited by nuodai; 12122010 at 15:38. 
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 12122010 15:44
(Original post by nuodai)
For what it's worth, I'd write for . If you had compositions, say then I'd write ; and for I'd write . 
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 12122010 16:49
(Original post by Zhen Lin)
Interesting. I've considered doing something like that before, but the purist in me says that for the map is just plain wrong and we should not need to disambiguate. (Along the same lines, writing out the circles for composition should be unnecessary, yet my aesthetic preference is to write them out. I should probably resolve this contradiction at some point...)
I try not to write or the like, because it's not always clear whether it means or . It's not too much of a bother normally, but when I'm studying spaces of functions under various operations I sometimes find it confusing if it's not made obvious. 
TheNihilist
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 13122010 19:19
Just got told this morning that for this question, is . Now it's half the trouble for me.
So far the statement doesn't hold eg:.
Now I working towards the converse of the statement. If is increasing, then increasing.
What I got so far is let (g(x)=f^2(x)=x^(2n+1)) for , which is obviously and increasing function since (g'(x)=(2n+1)x^(2n)>0) for all .
Then . Differentiate , I get.
But then there is a problem. If I take n=0, then . I cannot take ve x values, only x.Last edited by TheNihilist; 13122010 at 19:24. 
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 13122010 19:23
(Original post by TheNihilist)
Just got told this morning that for this question, is . Now it's half the trouble for me.
So far the statement doesn't hold eg:.
Now I working towards the converse of the statement. If is increasing, then increasing.
What I got so far is let for , which is obviously and increasing function since for all .
Then . Differentiate , I get.
But then there is a problem. If I take n=0, then . I cannot take ve x values, only x.
If is an increasing function for some function , then is necessarily increasing?
Equivalently, if isn't increasing, then must also not be increasing?Last edited by nuodai; 13122010 at 19:25. 
TheNihilist
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 14122010 15:36
My friend said she got converse as false and contrapositive statement as true. Hmm...
For the converse counterexample, let , then it's increasing for but decreasing for . Then the converse doesn't hold.Last edited by TheNihilist; 14122010 at 16:01. 
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 14122010 15:41
not all the time no

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 14122010 15:54
(Original post by TheNihilist)
My friend said she got converse as false and contrapositive statement as true. Hmm...
For the converse counterexample, let , then it's increasing for but decreasing for . Then the converse doesn't hold.
Basically we have 4 statements:
1. f is increasing => f² is increasing
2. f² is not increasing => f is not increasing
3. f² is increasing => f is increasing
4. f isn't increasing => f² isn't increasing
(1) and (2) are always either both true or both false, since (2) is the contrapositive of (1) and vice versa. Also, (3) and (4) are always either both true or false, for the same reason. (3) is the converse of (1). We've shown that (1) is false earlier in the thread, and your example here has shown that (3) is false (assuming we're only looking at the region , since x² is decreasing on !) and so (2) and (4) are both necessarily false.
The contrapositive for the converse (i.e. (4)) is "if f isn't increasing then f² isn't increasing". Well, as you've just shown, isn't increasing on the interval , but is increasing on this interval, so this is false. 
TheNihilist
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 14122010 16:16
(Original post by nuodai)
I'm a bit muddled with which contrapositive and converse we're talking about here (and also I'm guessing you meant ?).
Basically we have 4 statements:
1. f is increasing => f² is increasing
2. f² is not increasing => f is not increasing
3. f² is increasing => f is increasing
4. f isn't increasing => f² isn't increasing
(1) and (2) are always either both true or both false, since (2) is the contrapositive of (1) and vice versa. Also, (3) and (4) are always either both true or false, for the same reason. (3) is the converse of (1). We've shown that (1) is false earlier in the thread, and your example here has shown that (3) is false (assuming we're only looking at the region , since x² is decreasing on !) and so (2) and (4) are both necessarily false.
The contrapositive for the converse (i.e. (4)) is "if f isn't increasing then f² isn't increasing". Well, as you've just shown, isn't increasing on the interval , but is increasing on this interval, so this is false.
So to summarise what you are saying:
1. If is increasing, then is increasing. False as given in a previous example.
2. If is increasing, then is increasing. False as given in a previous example.
3. If is not increasing, then is not increasing. False since statement 1 is false. (counterexample would be similar to counterexample in 1?)
Does it mean all are false statements? 
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 14122010 16:17
(Original post by TheNihilist)
Ah, I was meant to say . This is now corrected.
So to summarise what you are saying:
1. If is increasing, then is increasing. False as given in a previous example.
2. If is increasing, then is increasing. False as given in a previous example.
3. If is not increasing, then is not increasing. False since statement 1 is false. (counterexample would be similar to counterexample in 1?)
Does it mean all are false statements?Last edited by nuodai; 14122010 at 16:18.
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