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  1. #1
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    Post proof (mathematical proof?)

    Hi again,

    This time I'd like to ask for your opinion on a "mathematical" proof I wrote as an answer to a problem from my teacher:

    Problem: Be f(x) continuous for every x, passes through the point (1,2) (f(1) = 2), it is known that for every x there is
    f(x) x.
    prove that f(x) > x for every x.

    My answer:
    It is given that f(1)=2 which means that f(x)> x at least at some point.
    According to the intermediate value theorem, for f(x) < x to exist
    f(x) = x has to exist at least once in the interval (-oo,oo).
    BUT it is given that
    f(x) x for every x.
    Therefore f(x) > x , always.




    Would that work as a "mathematical" proof for co
    llege etc..?

    Thanks.

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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    Hi again,

    This time I'd like to ask for your opinion on a "mathematical" proof I wrote as an answer to a problem from my teacher:

    Problem: Be f(x) continuous for every x, passes through the point (1,2) (f(1) = 2), it is known that for every x there is
    f(x) x.
    prove that f(x) > x for every x.

    My answer:
    It is given that f(1)=2 which means that f(x)> x at least at some point.
    According to the intermediate value theorem, for f(x) < x to exist
    f(x) = x has to exist at least once in the interval (-oo,oo).

    This is NOT a true statement. The "intermediate value theorem" does NOT say "f(x)= x has to exist at least once".
    You appear to be missing a statement. Did you intend this to be a "proof by contradiction" and say "suppose, for some x, [tex]f(x)\le x[tex]"?

    BUT it is given that
    f(x) x for every x.
    Therefore f(x) > x , always.




    Would that work as a "mathematical" proof for co
    llege etc..?

    Thanks.

    Thanks from ryu1
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  3. #3
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    Re: proof (mathematical proof?)

    Quote Originally Posted by HallsofIvy View Post
    This is NOT a true statement. The "intermediate value theorem" does NOT say "f(x)= x has to exist at least once".
    You appear to be missing a statement. Did you intend this to be a "proof by contradiction" and say "suppose, for some x, [tex]f(x)\le x[tex]"?
    How to prove it?
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  4. #4
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    Re: proof (mathematical proof?)

    It is a correct idea to apply the IVT. But in order to apply the first version of the IVT as is it stated in Wikipedia, you need to give the following:

    (1) a closed interval [a, b],
    (2) a continuous function f from [a, b] to real numbers, and
    (3) a number u between f(a) and f(b).

    Until you present those objects, you cannot say that you apply the IVT.
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  5. #5
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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    Problem: Be f(x) continuous for every x, passes through the point (1,2) (f(1) = 2), it is known that for every x there is
    f(x) x.
    prove that f(x) > x for every x.
    Let h(x)=f(x)-x. Is h an continuous function? How and/or why?
    h(1)=~?

    Suppose that \exists t such that f(t)<t.

    What does that tell you about h(t)~?

    Now use the IVT.
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  6. #6
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    Re: proof (mathematical proof?)

    Quote Originally Posted by Plato View Post
    Let h(x)=f(x)-x. Is h an continuous function? How and/or why?
    h(1)=~?

    Suppose that \exists t such that f(t)<t.

    What does that tell you about h(t)~?

    Now use the IVT.
    Let's see...
    h(x) is continuous because f(x) is continuous and x is also continuous. so the difference is also continuous.
    h(1) =f(1)-1 = 1

    in the case f(t) < t that tells me h(t) < h(x). because i get h(t) = f(t) - t (and t is bigger than f(t)), also h(t) is always negative(?).

    For the IVT i have a hard time...not sure how to apply it or to what. how does f(x) != x come in ?
    Thank you.
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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    For the IVT i have a hard time...not sure how to apply it or to what. how does f(x) != x come in ?
    So far so good. Do we know that for some value between t~\&~1 say s, such that h(s)=0~? WHY?
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  8. #8
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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    in the case f(t) < t that tells me h(t) < h(x). because i get h(t) = f(t) - t (and t is bigger than f(t)), also h(t) is always negative(?).
    The fact that f(t) < t should tell you that h(t) = f(t) - t < 0, not that h(t) < h(x). The latter statement is wrong because, for one, x is not even defined there. In contrast, the use of t is OK because we are proving the statement "For all x, f(x) > x" by contradiction. In doing so, we assume the negation, i.e., "There exists an x such that f(x) <= x" with the goal of deriving a contradiction. Since at least one such x exists, we say, "Let's take one such x and call it a t". (It would be better to call it an x.) Then f(t) <= t by this assumption, but x is a new name, and it is not clear what it means in "h(t) < h(x)".

    Note, by the way, that the negation of f(x) > x is f(x) <= x, not f(x) < x, but if there exists an x such that f(x) = x, then we immediately get a contradiction, without the IVT.

    So we have according to the plan in post #4:

    (1) an interval [1, t],
    (2) a continuous function h(x) = f(x) - x, and
    (3) a number u = 0 between h(1) = 2 > 0 and h(t) < 0.

    Therefore, we can apply the IVT.
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    Re: proof (mathematical proof?)

    Quote Originally Posted by Plato View Post
    So far so good. Do we know that for some value between t~\&~1 say s, such that h(s)=0~? WHY?

    Yes according to the IVT, the function h is continuous and in the interval [t,1] it changes sign.
    h(t) = minus
    h(1) = plus

    So there is a value s such that h(s) = 0

    ...*reading post #8*....

    So we can say that because theres an interval [t,1] (why it's not [1,t] ?) in which h changes signs.
    That happens in the value u at which the function h(u) = 0
    So we can say that h(u) = f(u) - u = 0
    Therefore f(u)-u = 0
    So f(u) = u
    So we conclude that the number at which the function f(x) = 0 is x.
    But if f(x) != x then we get a contradiction because f(x) != x.

    Is that a proper way to put it?

    u (from emakarov) = s (from Plato) for that matter, right?

    Thanks, hope I concluded correctly in this one.
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  10. #10
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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    Yes according to the IVT, the function h is continuous and in the interval [t,1] it changes sign.
    h(t) = minus
    h(1) = plus
    All these facts are not according to the IVT; rather, they are prerequisites to the IVT and have to be shown before one applies the IVT.

    Quote Originally Posted by ryu1 View Post
    So we can say that because theres an interval [t,1] (why it's not [1,t] ?) in which h changes signs.
    I should have noted that t can be anywhere with respect to 1, so the interval is either [1, t] ot [t, 1]; it does not matter.

    Quote Originally Posted by ryu1 View Post
    That happens in the value u at which the function h(u) = 0
    So we can say that h(u) = f(u) - u = 0
    Therefore f(u)-u = 0
    So f(u) = u
    So we conclude that the number at which the function f(x) = 0 is x.
    But if f(x) != x then we get a contradiction because f(x) != x.
    This is some plausible text, but I am not sure you understand the sequence of logical steps. Namely, I am not convinced you (1) fulfilled all prerequisites of the IVT, (2) applied the IVT, and (3) drew an appropriate inference from the conclusion of the IVT.

    Again:

    Quote Originally Posted by ryu1 View Post
    So we can say that because theres an interval [t,1] (why it's not [1,t] ?) in which h changes signs.
    What happens because there's an interval? Why does h change sign inside it?

    Quote Originally Posted by ryu1 View Post
    That happens in the value u at which the function h(u) = 0
    What exactly is u, by definition? What guarantees that such u exists?

    Quote Originally Posted by ryu1 View Post
    So we conclude that the number at which the function f(x) = 0 is x.
    But if f(x) != x then we get a contradiction because f(x) != x.
    What is x? The value u is what the IVT gives us (it says, "...there exists a number u such that..."). Every name you use has to be introduced. If I mention Jessings in a conversation with you and none of your acquaintances is called Jessings, would you have an idea about whom I am talking? The same thing happens when you use x out of the blue, without properly introducing or defining it.

    Quote Originally Posted by ryu1 View Post
    u (from emakarov) = s (from Plato) for that matter, right?
    Yes.

    I suggest you rewrite the whole proof in a clear way.
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    Lightbulb Re: proof (mathematical proof?)

    Quote Originally Posted by emakarov View Post
    All these facts are not according to the IVT; rather, they are prerequisites to the IVT and have to be shown before one applies the IVT.

    I suggest you rewrite the whole proof in a clear way.
    Hows this:

    f(1) = 2, given.
    Let h(x) = f(x) - x, continuous for all x.
    h(1) = 1 > 0
    Let t be the value for f(t) < t.
    h(t) = f(t) - t < 0

    Let's assume the negation that there is some number x such that h(x) = 0, in the interval [1, t] (has to be in this order [1,t] right? because 1 < t).

    According to the IVT that number x will be the value where h(x) changes signs:
    h(x) = f(x) - x = 0
    f(x) = x. Contradiction with the condition f(x) != x.

    We found that the value x in which f(x) = x is the value where the function f(x) has to pass through to change signs, but according to the condition f(x) != x, f(x) will never change signs, therefore f(x) > x for all x.


    Did I missed anything? didn't used u at all there, is it a problem?


    I would never think of the idea of making a new function h(x) = f(x) - x it's pretty neat but how would you think about it normally?
    and then making a new variable t such that f(t) < t so that h(t) will be < 0.... brilliant
    Last edited by ryu1; March 24th 2013 at 04:22 PM.
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  12. #12
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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    Hows this:

    f(1) = 2, given.
    Let h(x) = f(x) - x, continuous for all x.
    h(1) = 1 > 0
    Correct so far.

    Quote Originally Posted by ryu1 View Post
    Let t be the value for f(t) < t.
    When you say this, you need to know why such t exists. (Also, if you say, "the value" rather than "some value", you need to know that such t is unique.) You have not proved or assumed it so far.

    It is at this point that you need to start an argument by contradiction. You need to show that h(x) > 0 for all x (this would imply the required fact that f(x) > x for all x). The negation of "For all x, h(x) > 0" is "There exists an x such that h(x) ≤ 0". So you say, "Towards contradiction, assume that there exists some t such that h(t) ≤ 0. If h(t) = 0, then f(t) = t, which contradicts the assumption that f(x) ≠ x for all x".

    Quote Originally Posted by ryu1 View Post
    h(t) = f(t) - t < 0

    Let's assume the negation that there is some number x such that h(x) = 0, in the interval [1, t] (has to be in this order [1,t] right? because 1 < t).
    The negation of what? If the negation is h(x) = 0 for some x, then the original fact, of which it is a negation, is that h(x) ≠ 0, i.e., f(x) ≠ x, for all x? But this is not what you need to prove; you need to show that f(x) > x for all x. Also, you need to start an argument by contradiction earler, as I said above.

    That t for which you assumed, towards contradiction, that h(t) ≤ 0, can be equal to, less than or greater than 1; the rest of the proof does not change. So it is better to refer to "[1, t] or [t, 1]" or "the closed interval between 1 and t".

    Quote Originally Posted by ryu1 View Post
    According to the IVT that number x will be the value where h(x) changes signs:
    h(x) = f(x) - x = 0
    f(x) = x. Contradiction with the condition f(x) != x.
    First, the IVT says nothing about changing signs. For example, below is the graph of g(x) = x(x - 1)2.



    We have g(-1) = -4, g(2) = 2 and g(x) is continuous. By the IVT applied to [-1, 2], g(x) and y = 0, there exists an x in [-1, 2] such that g(x) = 0. This x can very well be x = 1 where the function does not change sign.

    More importantly, you invoke the IVT to get an x such that h(x) = 0, but above you have already assumed that such x exists. To do it correctly, you assume that there exists a t such that h(t) < 0 (the case h(t) = 0 has already been considered) and apply the IVT to the closed segment between 1 and t (since h(1) > 0 and h(t) < 0, i.e., 0 lies between h(1) and h(t)), the continuous function h(x) and y = 0. The IVT gives you an x such that h(x) = 0, which contradicts the assumption. Thus, whether h(t) < 0 or h(t) = 0, you get a contradition, which proves that the assumption "There exists a t such that h(t) ≤ 0" is false and therefore the original claim "For all x, h(x) > 0" is true.

    Quote Originally Posted by ryu1 View Post
    didn't used u at all there, is it a problem?
    The name of a variable is irrelevant. Saying "For all u, P(u)" and "For all s, P(s)" for some property P is the same thing.
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    Re: proof (mathematical proof?)

    Quote Originally Posted by emakarov View Post

    More importantly, you invoke the IVT to get an x such that h(x) = 0, but above you have already assumed that such x exists. To do it correctly, you assume that there exists a t such that h(t) < 0 (the case h(t) = 0 has already been considered) and apply the IVT to the closed segment between 1 and t (since h(1) > 0 and h(t) < 0, i.e., 0 lies between h(1) and h(t)), the continuous function h(x) and y = 0. The IVT gives you an x such that h(x) = 0, which contradicts the assumption. Thus, whether h(t) < 0 or h(t) = 0, you get a contradiction, which proves that the assumption "There exists a t such that h(t) ≤ 0" is false and therefore the original claim "For all x, h(x) > 0" is true.
    I read it couple of times and I cant understand this part, can you explain it in a simpler fashion please? like how would you write it for real (like some parts you wrote at previous sections of your post i.e ""Towards contradiction, assume that there exists some t such that h(t) ≤ 0. If h(t) = 0, then f(t) = t, which contradicts the assumption that f(x) ≠ x for all x". ")

    Thanks.

    trying again:

    Need to prove f(x) > x, for all x.
    f(x) > x = f(x) - x > 0
    Defining a new function h(x) = f(x) - x.
    Now we need to prove h(x) > 0 for all x.
    f(1) = 2 , given, therefore h(1) = 1 > 0
    Let's assume towards contradiction there is some t such that h(t) <= 0, if h(t) = 0 then h(t) = f(t) - t = 0 , f(t) = t which contradicts the condition f(x) != x.
    Now we still need to prove towards contradiction there is some t such that h(t) < 0.
    According to the IVT, in the closed segment [1,t] there is some x such that h(x) = 0, because h(1) > 0 and h(t) < 0.
    That x also contradicts the condition f(x) != x (again because h(x) = 0 = f(x) - x = 0 = f(x) = x).
    Thus we get a contradiction for h(t) <= 0, which means h(x) > 0 for all x, which proves f(x) > x for all x.


    I really hope this works, what you think?
    Last edited by ryu1; March 26th 2013 at 05:47 AM.
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  14. #14
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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    I read it couple of times and I cant understand this part, can you explain it in a simpler fashion please? like how would you write it for real (like some parts you wrote at previous sections of your post i.e ""Towards contradiction, assume that there exists some t such that h(t) ≤ 0. If h(t) = 0, then f(t) = t, which contradicts the assumption that f(x) ≠ x for all x". ")
    Need to prove f(x) > x, for all x.
    f(x) > x = f(x) - x > 0
    Defining a new function h(x) = f(x) - x.
    Now we need to prove h(x) > 0 for all x.
    f(1) = 2 , given, therefore h(1) = 1 > 0
    Let's assume towards contradiction there is some t such that h(t) <= 0, if h(t) = 0 then h(t) = f(t) - t = 0 , f(t) = t which contradicts the condition f(x) != x.
    Now we still need to prove towards contradiction there is some t such that h(t) < 0.
    According to the IVT, in the closed segment [1,t] there is some x such that h(x) = 0, because h(1) > 0 and h(t) < 0.
    That x also contradicts the condition f(x) != x (again because h(x) = 0 = f(x) - x = 0 = f(x) = x).
    Thus we get a contradiction for h(t) <= 0, which means h(x) > 0 for all x, which proves f(x) > x for all x.
    These are general comments.
    Suppose that g is a continuous function and a~\&~b are two points such that g(a)<0<g(b).
    The IVT grantees that there is a point c between a~\&~b such that g(c)=0.

    Note we do not know which of these is true: a<c\text{ or }a>c.

    In your proof it does not matter. All you need to know that h(t)=f(t)-t=0.
    That is a contradiction which proves the statement, \forall x,f(x)>x
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    Re: proof (mathematical proof?)

    Quote Originally Posted by ryu1 View Post
    Need to prove f(x) > x, for all x.
    f(x) > x = f(x) - x > 0
    Defining a new function h(x) = f(x) - x.
    Now we need to prove h(x) > 0 for all x.
    f(1) = 2 , given, therefore h(1) = 1 > 0
    Let's assume towards contradiction there is some t such that h(t) <= 0, if h(t) = 0 then h(t) = f(t) - t = 0 , f(t) = t which contradicts the condition f(x) != x.
    Now we still need to prove towards contradiction there is some t such that h(t) < 0.
    No, you don't need to prove it, you assumed it! More precisely, you assumed, towards contradiction, that h(t) <= 0. One option how this may happen is that h(t) = 0; the other is h(t) < 0. Both cases have to be examined (and shown to lead to a contradiction) under the same assumption, that there exists a t such that h(t) <= 0.

    I would make the following changes in bold.

    Quote Originally Posted by ryu1 View Post
    Let's assume towards contradiction there is some t such that h(t) <= 0. If h(t) = 0 then h(t) = f(t) - t = 0 , f(t) = t which contradicts the condition f(x) != x. If h(t) < 0, then, according to the IVT, in the closed segment between 1 and t there is some x such that h(x) = 0, because h(1) > 0 and h(t) < 0.
    I also replaced "the closed segment [1, t]" with "the closed segment between 1 and t" because t can be <= 1, as Plato said.

    One more remark is that = is usually used between mathematical objects (numbers, sets, matrices and so on) and not between propositions (something that is true or false). One usually writes ⇔ or "iff" between propositions, e.g., "f(x) > x ⇔ f(x) - x > 0".

    Otherwise, the proof looks good.
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