# proof (mathematical proof?)

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Mar 22nd 2013, 06:12 PM
ryu1
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) http://opal.openu.ac.il/pluginwiris/...634c4c14bd.png x.
prove that f(x) > x for every x.

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) http://opal.openu.ac.il/pluginwiris/...634c4c14bd.png x for every x.
Therefore f(x) > x , always.

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

Thanks.

• Mar 23rd 2013, 01:41 PM
HallsofIvy
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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) http://opal.openu.ac.il/pluginwiris/...634c4c14bd.png x.
prove that f(x) > x for every x.

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]"?

Quote:

BUT it is given that
Quote:

f(x) http://opal.openu.ac.il/pluginwiris/...634c4c14bd.png x for every x.
Therefore f(x) > x , always.

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

Thanks.

• Mar 24th 2013, 10:16 AM
ryu1
Re: proof (mathematical proof?)
Quote:

Originally Posted by HallsofIvy
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?
• Mar 24th 2013, 11:55 AM
emakarov
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.
• Mar 24th 2013, 12:32 PM
Plato
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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) http://opal.openu.ac.il/pluginwiris/...634c4c14bd.png 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).

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

Now use the IVT.
• Mar 24th 2013, 01:11 PM
ryu1
Re: proof (mathematical proof?)
Quote:

Originally Posted by Plato
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).

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.
• Mar 24th 2013, 01:20 PM
Plato
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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?
• Mar 24th 2013, 02:23 PM
emakarov
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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.
• Mar 24th 2013, 02:47 PM
ryu1
Re: proof (mathematical proof?)
Quote:

Originally Posted by Plato
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

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.
• Mar 24th 2013, 03:13 PM
emakarov
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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
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
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
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
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
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
u (from emakarov) = s (from Plato) for that matter, right?

Yes.

I suggest you rewrite the whole proof in a clear way.
• Mar 24th 2013, 04:18 PM
ryu1
Re: proof (mathematical proof?)
Quote:

Originally Posted by emakarov
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
• Mar 25th 2013, 03:51 AM
emakarov
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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
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
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
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.

https://s3.amazonaws.com/grapher/exports/5c1mtiyojl.png

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
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.
• Mar 26th 2013, 05:27 AM
ryu1
Re: proof (mathematical proof?)
Quote:

Originally Posted by emakarov

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?
• Mar 26th 2013, 06:45 AM
Plato
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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.

Suppose that $g$ is a continuous function and $a~\&~b$ are two points such that $g(a)<0.
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: $ac$.

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, $\forall x,f(x)>x$
• Mar 26th 2013, 11:24 AM
emakarov
Re: proof (mathematical proof?)
Quote:

Originally Posted by ryu1
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
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.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last