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.

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) 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 college etc..?

Thanks.

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.

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]"?

Quote:

BUT it is given that

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?

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.

Re: proof (mathematical proof?)

Quote:

Originally Posted by

**ryu1**

Let $\displaystyle h(x)=f(x)-x$. Is $\displaystyle h$ an continuous function? How and/or why?

$\displaystyle h(1)=~?$

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

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

Now use the IVT.

Re: proof (mathematical proof?)

Quote:

Originally Posted by

**Plato** Let $\displaystyle h(x)=f(x)-x$. Is $\displaystyle h$ an continuous function? How and/or why?

$\displaystyle h(1)=~?$

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

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

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.

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 $\displaystyle t~\&~1$ say $\displaystyle s$, such that $\displaystyle h(s)=0~?$ WHY?

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.

Re: proof (mathematical proof?)

Quote:

Originally Posted by

**Plato** So far so good. Do we know that for some value between $\displaystyle t~\&~1$ say $\displaystyle s$, such that $\displaystyle h(s)=0~?$ 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.

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.

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

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.

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?

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.

These are general comments.

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

The **IVT** grantees that there is a point $\displaystyle c$ **between** $\displaystyle a~\&~b$ such that $\displaystyle g(c)=0$.

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

In your proof it does not matter. All you need to know that $\displaystyle h(t)=f(t)-t=0$.

That is a contradiction which proves the statement, $\displaystyle \forall x,f(x)>x$

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.