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    (Original post by metaltron)
    To summarise, this is the proof that everybody is stressing about:

     x^2 > x \Leftrightarrow x > 1 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.
    Precisely
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    I'd love for DFranklin or ghostwalker to comment on this thread, just so I can LOL at them being corrected/argued with by an A-Level and IB student :lol:

    When Giveme45 said he'd "studied maths beyond university level" I was confused how someone with a degree in maths could be saying half of what he's posting, then I realised he meant he read a pamphlet on a topic in maths.
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    (Original post by metaltron)
    To summarise, this is the proof that everybody is stressing about:

     x^2 > x \Leftrightarrow x > 1 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.
    As has been said in quite a few of the above posts, this is incorrect logic (generally speaking). What you've said is that, if P is true and you're trying to show that P => Q, then showing Q => P (which is what you do when you get down to x-1>0 from x^2>x) proves that Q is true - but this doesn't make sense unless the argument is reversible. This means that you still need to show that P => Q and so the argument is circular.

    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.
    I agree that it has got out of hand, and it will be closed very shortly but you didn't appear to understand what myself (and other posters) were getting it - I think it was more a case of how your explained yourself which led to the confusion.
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    (Original post by Noble.)
    I'd love for DFranklin or ghostwalker to comment on this thread, just so I can LOL at them being corrected/argued with by an A-Level and IB student :lol:

    When Giveme45 said he's "studied maths beyond university level" I was confused how someone with a degree in maths could be saying half of what he's posting, then I realised he meant was he read a pamphlet on a topic in maths.
    +1

    The fact that a thread about such a simple proof has got out of control to such an extent is beyond belief :lol:
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    (Original post by Farhan.Hanif93)
    As has been said in quite a few of the above posts, this is incorrect logic (generally speaking). What you've said is that, if P is true and you're trying to show that P => Q, then showing Q => P (which is what you do when you get down to x-1>0 from x^2>x) proves that Q is true - but this doesn't make sense unless the argument is reversible. This means that you still need to show that P => Q and so the argument is circular.


    I agree that it has got out of hand, and it will be closed very shortly but you didn't appear to understand what myself (and other posters) were getting it - I think it was more a case of how your explained yourself which led to the confusion.
    x > 1 is given to you, you're not trying to prove it, it is a fact. Using this fact you can show that x^2 > x leads to x > 1 which is a fact.

    What you're saying is by starting with x>1, which you're saying is unproven, you can then go and prove x^2 > x, which is of course true.
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    (Original post by metaltron)
    x > 1 is given to you, you're not trying to prove it, it is a fact. Using this fact you can show that x^2 > x leads to x > 1 which is a fact.

    What you're saying is by starting with x>1, which you're saying is unproven, you can then go and prove x^2 > x, which is of course true.
    I've not said anywhere in the previous post that x>1 is unproven. In fact, my interpretation of what you did started with the truth of P (which is the x>1 condition). And I've explained why it doesn't work - if you can tell me what I've said wrongly in that post then we'll continue this discussion; otherwise we'll just continue to go in a circle.
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    (Original post by Farhan.Hanif93)
    I've not said anywhere in the previous post that x>1 is unproven. In fact, my interpretation of what you did started with the truth of P (which is the x>1 condition). And I've explained why it doesn't work - if you can tell me what I've said wrongly in that post then we'll continue this discussion; otherwise we'll just continue to go in a circle.
    What I think you've said wrongly is this:

    x > 1 is a fact. If we start with x^2 > x using a sequence of logical and correct statements, using x > 1 as a fact, we arrive at x >1, which is of course true. Each of these steps is reversible and there is nothing wrong with my proof.
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    I have made no assumptions about x, it is given that x > 1, so where is my proof incorrect?
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    (Original post by metaltron)
    I have made no assumptions about x, it is given that x > 1, so where is my proof incorrect?
    Farhan and I have already pointed this out.

    Given the assumption that x>1 then you can infer that x^2 > x.

    Given the assumption that x^2 > x you cannot infer that x >1

    The two statements "x>1" and "x^2 > x" are not logically equivalent.

    What you're saying is "If I assume x^2>x AND x>1, then I can show x>1" which is trivially true because for any 2 propositions P and Q, (P \cap Q) \Rightarrow Q
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    (Original post by davros)
    Farhan and I have already pointed this out.

    Given the assumption that x>1 then you can infer that x^2 > x.

    Given the assumption that x^2 > x you cannot infer that x >1

    The two statements "x>1" and "x^2 > x" are not logically equivalent.

    What you're saying is "If I assume x^2>x AND x>1, then I can show x>1" which is trivially true because for any 2 propositions P and Q, (P \cap Q) \Rightarrow Q
    No I did not say that x^2>x. I said x > 1, and started with the possibly false statement x^2 > x. This leads to x > 1, given that x > 1. At this point x^2>x could still be false. However, x > 1 is true and therefore x^2 > x. It is your proof in reverse.
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    (Original post by metaltron)
    No I did not say that x^2>x. I said x > 1, and started with the possibly false statement x^2 > x. This leads to x > 1, given that x > 1. At this point x^2>x could still be false. However, x > 1 is true and therefore x^2 > x. It is your proof in reverse.
    Good grief, how many ways do you need someone to explain it to you?
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    (Original post by metaltron)
    No I did not say that x^2>x. I ... started with the possibly false statement x^2 > x.
    Do you seriously not see the logical flaw in what you've written there?
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    (Original post by Noble.)
    Good grief, how many ways do you need someone to explain it to you?
    I don't need somebody to explain it to me, I need one of you to backtrack, and if you're so sure your right that my proof is wrong, then this is such a joke the thread might as well be closed. If somebody can tell me where I made the assumption that x^2 > x then please do but all attempts so far have been utter ....
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    (Original post by davros)
    Do you seriously not see the logical flaw in what you've written there?
    Sometimes in induction you might start at the (k+1) stage and try to work back to the inductive step, even though the (k+1) step could possibly be wrong. This is what I have done here, worked back from x^2 > x to show that it is true.
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    (Original post by metaltron)
    Sometimes in induction you might start at the (k+1) stage and try to work back to the inductive step, even though the (k+1) step could possibly be wrong. This is what I have done here, worked back from x^2 > x to show that it is true.
    Firstly, this isn't induction - that has a precise form that relates to propositions about natural numbers.

    Secondly, you've now explicitly stated that you've assumed x^2 > x in order to prove it, which is what I was trying to tell you.

    I think this thread should now be closed - you're not taking in anything that people are telling you, and I think we've all got better things to do
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    (Original post by davros)
    Firstly, this isn't induction - that has a precise form that relates to propositions about natural numbers.

    Secondly, you've now explicitly stated that you've assumed x^2 > x in order to prove it, which is what I was trying to tell you.

    I think this thread should now be closed - you're not taking in anything that people are telling you, and I think we've all got better things to do
    But a few posts ago I explicitly said that x^2 > x could be false.
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    (Original post by metaltron)
    Sometimes in induction you might start at the (k+1) stage and try to work back to the inductive step, even though the (k+1) step could possibly be wrong. This is what I have done here, worked back from x^2 > x to show that it is true.
    Erm... you always do that in induction, but the way induction works is entirely different.
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    Thread closed.
 
 
 
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