I think that's how it should be interpreted, but i must admit I did a doubletake when I read it (and the following posts) for the first time!(Original post by Noble.)
Yeah, true, but the way I read it was that you're starting from x > 1 and supposed to show x^2 > x from that.

davros
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 29032013 17:45

metaltron
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 29032013 17:51
(Original post by davros)
I think that's how it should be interpreted, but i must admit I did a doubletake when I read it (and the following posts) for the first time!
x > 1 is a fact. Using this we can divide by x to get x > 1 which is true so x^2 > x is true. Or the other way round x > 1 can be multiplied by x as x>1>0, so x^2 > x. 
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 29032013 17:52
(Original post by Noble.)
Yeah, true, but the way I read it was that you're starting from x > 1 and supposed to show x^2 > x from that.(Original post by metaltron)
I was starting to give in before you came along!(Original post by Noble.)
Well I'm sorry! If a question states "Show that if P then Q" if you start with Q and show P you've not answered the question. The way I've read the OP is that you're given x>1 (P) and supposed to show x^2 > x (Q). But this is a very good example of why you need to be unambiguous in mathematics.
You're square root proof, I haven't looked at yet, but what it seems to be is that if untrue, it probably is due to ambiguities of incorrect mathematical operations concerning the square of both sides of the inequality. I will check it if you insist. 
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 29032013 17:54
(Original post by Giveme45)
What you don't understand, Noble, is that it does not matter from what point you start to prove this equation, as long as all steps there on will be mathematically correct. If a question starts with "Show that if P then Q", you still would have answered the question starting from Q. It would even be considered a correct and formal proof if you would show that if not Q then not P. According to you this would also be a "wrong start" to the question and be "bad practice". But infact, this is referred to as Reductio ad absurdum. You can read about this here http://en.wikipedia.org/wiki/Reductio_ad_absurdum.
You're square root proof, I haven't looked at yet, but what it seems to be is that if untrue, it probably is due to ambiguities of incorrect mathematical operations concerning the square of both sides of the inequality. I will check it if you insist.
Also, proof by contradiction only works... when there is a contradiction . There is no contradiction here. 
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 29032013 17:58
(Original post by Noble.)
Are you just trolling or being an idiot. If you are asked "Show if P then Q" and you start with Q, you have no answered the question  this is proof 101 . You obviously aren't studying maths at University, because noone out of the first week would be posting such nonsense.
I even linked you wikipedia article which shows that "Show if P then Q" can even be answered by stating that if NOT Q then NOT P.
How can an individual be so stubborn and full of judgement and irrational conclusions?
Please, if you don't have any reasonable argument to add to this and want to tell me about bad practice or proof 101 instead. Then go for it but I will not answer. This is embarassing and you are just making a fool of yourself. 
metaltron
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 29032013 18:03
(Original post by Giveme45)
No insults. No reason to. I am not trolling. I have studied maths beyond university level. "Show if P then Q" does to NO EXTENT state that you have to start with P to prove Q. You COULD EVEN start with 1+1=2 to prove that Q under the assumption that P. Do you have any arguments to support your claim about "proof 101"? No.
I even linked you wikipedia article which shows that "Show if P then Q" can even be answered by stating that if NOT Q then NOT P.
How can an individual be so stubborn and full of judgement and irrational conclusions?
Please, if you don't have any reasonable argument to add to this and want to tell me about bad practice or proof 101 instead. Then go for it but I will not answer. This is embarassing and you are just making a fool of yourself. 
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 29032013 18:05
(Original post by Noble.)
Also, proof by contradiction only works... when there is a contradiction . There is no contradiction here.
(Original post by metaltron)
Exactly, I think Proof 101 needs to go into Room 101. 
davros
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 29032013 18:21
(Original post by metaltron)
I don't even understand how it does make a difference. This is the way I see it is:
x > 1 is a fact. Using this we can divide by x to get x > 1 which is true so x^2 > x is true. Or the other way round x > 1 can be multiplied by x as x>1>0, so x^2 > x.
If x>1 then x>0 so we can multiply the inequality x>1 by x on both sides while preserving the direction of the inequality to get x^2 > x as required. This is a proof.
However, you can't "prove" that second inequality by assuming it to be true and then dividing it by something. What you can do is do the division as if it were true and then recognize that if you started from this end position you could construct the proof as given in my previous paragraph. 
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 29032013 18:23
(Original post by davros)
Now I think you're confusing things slightly.
If x>1 then x>0 so we can multiply the inequality x>1 by x on both sides while preserving the direction of the inequality to get x^2 > x as required. This is a proof.
However, you can't "prove" that second inequality by assuming it to be true and then dividing it by something. What you can do is do the division as if it were true and then recognize that if you started from this end position you could construct the proof as given in my previous paragraph.
Read the whole conversation if you wish to understand more about it, feel free to ask questions 
metaltron
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 29032013 18:27
(Original post by davros)
Now I think you're confusing things slightly.
If x>1 then x>0 so we can multiply the inequality x>1 by x on both sides while preserving the direction of the inequality to get x^2 > x as required. This is a proof.
However, you can't "prove" that second inequality by assuming it to be true and then dividing it by something. What you can do is do the division as if it were true and then recognize that if you started from this end position you could construct the proof as given in my previous paragraph.
as x > 1 > 0.
is true, so:
if x >1.
This shows that you don't have to start with x > 1, you can start with x^2 > x. Also, I'm not saying that anybody else's way is wrong, I am just disputing with some people who are saying this method is incorrect. 
ThatPerson
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 29032013 18:28
(Original post by Giveme45)
No insults. No reason to. I am not trolling. I have studied maths beyond university level. "Show if P then Q" does to NO EXTENT state that you have to start with P to prove Q. You COULD EVEN start with 1+1=2 to prove that Q under the assumption that P. Do you have any arguments to support your claim about "proof 101"? No.
I even linked you wikipedia article which shows that "Show if P then Q" can even be answered by stating that if NOT Q then NOT P.
How can an individual be so stubborn and full of judgement and irrational conclusions?
Please, if you don't have any reasonable argument to add to this and want to tell me about bad practice or proof 101 instead. Then go for it but I will not answer. This is embarassing and you are just making a fool of yourself.
http://www.thestudentroom.co.uk/show...782&highlight= 
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 29032013 18:30
(Original post by ThatPerson)
Yet apparently a few days ago you were an IB student?
http://www.thestudentroom.co.uk/show...782&highlight=
I am an IB student, but does that mean that I haven't studied maths beyond universtiy level? Is it impossible for a person to pursue a particular area of interest (in my case maths), and study/research this beyond what is taught in that person's school lesson's?
Full of assumptions this thread. 
Farhan.Hanif93
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 29032013 18:31
(Original post by Giveme45)
No insults. No reason to. I am not trolling. I have studied maths beyond university level. "Show if P then Q" does to NO EXTENT state that you have to start with P to prove Q. You COULD EVEN start with 1+1=2 to prove that Q under the assumption that P. Do you have any arguments to support your claim about "proof 101"? No.
He makes a fair point in saying that starting with Q is generally bad practice because it usually requires you to end up proving something stronger (namely the reverse implication too). I agree with you that most proofs are correct on this thread but there's a very fine line between proving it the way you've done, and suggesting that double implications are valid here (as Metaltron suggested earlier in the thread  my only qualms with his posts were the potential to confuse and nothing more.
I even linked you wikipedia article which shows that "Show if P then Q" can even be answered by stating that if NOT Q then NOT P.
How can an individual be so stubborn and full of judgement and irrational conclusions?
This is embarassing and you are just making a fool of yourself.
It is always a pain to deal with these "maths university level pros" which through their stubborness confuse others. You said you almost "gave in". In the future, dont give in to these kind of idiots, if your argument makes complete sense, which it did +Rep 
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 29032013 18:36
(Original post by Farhan.Hanif93)
Yet, in that post you quoted, Noble made no mention of starting with P… He just said that you shouldn't start with Q (and before you bring up proof by contradiction; that's starting with 'not Q' and that's definitely not the same thing as starting with Q).
He makes a fair point in saying that starting with Q is generally bad practice because it usually requires you to end up proving something stronger (namely the reverse implication too). I agree with you that most proofs are correct on this thread but there's a very fine line between proving it the way you've done, and suggesting that double implications are valid here (as Metaltron suggested earlier in the thread  my only qualms with his posts were the potential to confuse and nothing more.
I'm sure he knows that.
Any further offensive comments will be met with warnings. Please keep your posts firmly on the topic at hand.
If he sure knows of my quotation of the wikipedia article then how come he still makes an "offensive comment" based on his misunderstanding on what is reductio by absurdum: "Also, proof by contradiction only works... when there is a contradiction . There is no contradiction here."
How is it acceptable that him, a 4000+ post member allowed to make fun of me and insult me with no location of evidence in his statements, but as soon as I respond, WITH EVIDENCE, that I get warned? 
ThatPerson
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 29032013 18:37
(Original post by Giveme45)
I am an IB student, but does that mean that I haven't studied maths beyond universtiy level? Full of assumptions this thread.
In the context you posted, it seemed like formal study, such as a Masters, etc.
I too have studied a tiny amount of undergraduate maths (not formally), but I'm no expert, and I think there is a distinct difference between an amateurish interest, and formal academic study. 
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 29032013 18:39
(Original post by ThatPerson)
Yes but an other assumption would be that, if you have studied maths beyond university level, then that means you know what you are talking about.
In the context you posted, it seemed like formal study, such as a Masters, etc.
I too have studied a tiny amount of undergraduate maths, but I'm no expert, and I think there is a distinct difference between an amateurish interest, and formal academic study.
As you say, claims and statements should not be looked at by their probability of validity in regard to who made the statement. So it is completely irrelevant if I have pursued a formal academic study or not as long as my claims are based on solid evidence. 
davros
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 29032013 18:41
(Original post by metaltron)
This was my proof:
as x > 1 > 0.
is true, so:
if x >1.
This shows that you don't have to start with x > 1, you can start with x^2 > x. Also, I'm not saying that anybody else's way is wrong, I am just disputing with some people who are saying this method is incorrect.
At this point you would like to divide by x to conclude that x1>0 and hence x>1, but you can't perform that action without assuming that x>0. However, when you stated that x>0 you inferred this from the statement that x>1, which is the thing you're trying to prove! 
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 29032013 18:46
(Original post by davros)
You can't start with x^2 > x without making a further assumption about x.
At this point you would like to divide by x to conclude that x1>0 and hence x>1, but you can't perform that action without assuming that x>0. However, when you stated that x>0 you inferred this from the statement that x>1, which is the thing you're trying to prove! 
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 29032013 18:51
(Original post by metaltron)
No you're not trying to prove it, you're given that x > 1.
If you are given x>1 then you can deduce directly that x^2 > x as Farhan pointed out on the first page of this thread.
If you just start with the statement x^2 > x then you can't infer that x > 1 (without making a further assumption).
I think we're going in a circle here so I'm off to have some tea 
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 29032013 18:53
To summarise, this is the proof that everybody is stressing about:
as x > 1 > 0.
The aspect of this that seems to be confusing some members of this forum is the fact that the final statement is x>1, which makes it look like I am trying to prove x>1. THIS IS NOT THE CASE. This is just a true statement, so the original assumption is true.
I'd also like to add that this discussion has got out of hand and that I'm sure the OP has stopped watching this thread for ages now. What has upset me is that some of the more prestigious members have used there power to try and quash my argument without explaining properly what they mean resulting in a circular argument.
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