Math stuff that hasn't been proven

[TylerH]

In elementary school or high school, we often use stuff that has never actually been proven (in that class). For example

- Pythagoras' theorem.
- Addition of natural numbers is associative.
- Every number can be uniquely (up to order) decomposed in prime factors.

Accepting such a things really annoyed me, I would always ask why something is true. The answer that most teachers gave me was "can you find an example where it doesn't work," sigh. I had to wait until university to actually see a proof for such a things...

So, were you ever annoyed that something wasn't proven in school?? And what would have liked to see a proof/reason of??

I'd like to see a proof for the 2nd and 3rd. They're just true, there is no proof other than: It's true, by definition. Can you even imagine what terrible effect that would have on a high school student asking why? They wouldn't get it, and would likely be offended by its resemblance to how parents tell their kids "just because" when they don't feel like explaining something.

I once had a high school algebra teacher who thought .999... != 1. I would have appreciated a proof of that.

Since we're basically looking back on elementary math with higher math prospective, I think this is related. I really wish my teachers would have taken the time to explain how the associative/commutative/identity/etc. properties of the arithmetic operators are generalized in higher math. The aforementioned teacher started down that road when I asked him why those properties are useful (other than their blatantly obvious application to elementary algebra). But, when I asked him what a set is, he gave up. (Which is rather annoying to think of now. How hard would have been to just say: "a collection of objects/stuff"?)

 

[Tosh5457]

In elementary school or high school, we often use stuff that has never actually been proven (in that class). For example

- Pythagoras' theorem.
- Addition of natural numbers is associative.
- Every number can be uniquely (up to order) decomposed in prime factors.

Accepting such a things really annoyed me, I would always ask why something is true. The answer that most teachers gave me was "can you find an example where it doesn't work," sigh. I had to wait until university to actually see a proof for such a things...

So, were you ever annoyed that something wasn't proven in school?? And what would have liked to see a proof/reason of??

Not really, but I always felt I didn't understand something well if I didn't have at least an intuition of how to deduce some math stuff. What annoyed me were some formulas that we used in physics without deducing them, but that doesn't annoy me anymore, because, who cares?

 

[Hurkyl]

I'm not saying that you can't learn about logarithms. Looking at that definition, you won't even know that it is a logarithm. Why not define it as a logarithm with base e?

I did define it to be the logarithm to the base e.



This process of coming up with the rigorous definition for transcendental functions usually comes after you have already learned the basic facts -- in particular I already know the equation $\log x = \int_1^x dt/t$.

But, even if I was learning about transcendental functions for the first time, all of the basic properties of $\log$, $\exp$, and real exponentiation will be presented around the same time. Pedagogically, choosing one of a myriad of facts to be given the title of "definition" (if any are chosen at all) should be done to simplify the exposition. (unless, of course, you're writing for an audience that reads far too much into the word "definition")


Anyways, to define $\log(x)$ as the inverse of $\exp(x)$ requires you to define $\exp(x)$. To define it as the inverse of $e^x$ requires you to define $e$ and to define $x^y$.

 

[disregardthat]

Any discussion regarding whether log(x) should be defined as the integral of 1/x or as the inverse function of e^x is pure preference-but a preference based on what? I can't tell.

 

[Hurkyl]

Any discussion regarding whether log(x) should be defined as the integral of 1/x or as the inverse function of e^x is pure preference-but a preference based on what? I can't tell.

The one I'm familiar with is sheer simplicity, as I already mentioned.

If we're operating at this level of rigor and technical detail, then there are five things we need to do:

• Define $\log(x)$
• Define $\exp(x)$
• Define $e$
• Define $x^y$
• Derive all of the usual identities satisfied by these three operations

By far the simplest programme I know of is to start with $\log(x) = \int_1^x dt/t$, use change-of-variable to derive $\log(xy) = \log(x) + \log(y)$, define $\exp = \log^{-1}$, $e = \exp(1)$, $x^y = \exp(y \log(x))$, and then uses these to derive everything else.

What are the alternatives? I've mentioned the starting points of the main ones I recall. Do you need to see the sketch of how I recall that they proceed?

 

[disregardthat]

Another one (using our knowledge of absolutely converging power series) is to define exp(x) = 1+x+x^2/2 + ..., and define e = exp(1). It is simple algebra + binomial theorem to show that exp(x)exp(y) = exp(x+y). From here it is obvious that exp is positive and increasing, and we can define log = exp^-1, and it follows easily that log(xy) = log(x) + log(y).

At last we define x^y = exp(ylog(x)) for positive x.

(Note that showing that exp(x) = exp(1)^x never came into question here (showing that this coincides with x^n for integer n is trivial)).

It also follows easily from the definition that exp(x)' = exp(x).

EDIT: I can see you have mentioned this before.

 

[JonF]

Yes, but why is the ratio a constant?? That seems nontrivial to me...

I remember asking this question in my high school geometry course and was told “what makes you think it wouldn’t be?”

 

[disregardthat]

pi can be calculated for a circle of radius r, and finding it's a constant independent of r answers the question.

 

[Kindayr]

I'd like to see a proof for the 2nd.

The abstract way that I've learned in class is:

First show that $(N, s(n), 1)$ to be a Peano space.

By the Recursion Theorem we can define for each $m\in N$ the function $f_m : N\to N$ as follows: $f_m (1)=s(m)$, and $f_m (s(n))=s(f_m(n)),\, \forall n\in N$. We then define a binary operation on N, which we call $addition$ and denote by "$+$". For each $(m,n)\in P\times P$, we define $m+n$ to be $f_m (n)$. We also call $m+n$ the $sum$ of $m$ and $n$.

Note we have then: $f_m (1)=s(m)=m+1$ and $m+(n+1)=f_m (s(n))=s(f_m (n))=(f_m (n))+1=(m+n)+1$

(We're almost there)

Claim: $k+(m+n)=(k+m)+n,\,\,\forall k,m,n\in N$ (Associativity of Addition)

$Proof.$ We will use induction on $n$. Let $P(n):=k+(m+n)=(k+m)+n$. $P(1)$ is true by definition. So suppose $P(n)$ is true, we must show $P(n)\implies P(n+1)$.

So:
$k+[m+(n+1)]=k+[(m+n)+1]=[k+(m+n)]+1=[(k+m)+n]+1=(k+m)+(n+1).$

Where the first, second, and fourth equality are by definition, and the third is the inductive step.
-----

By showing $N$ to be a Peano space, you can prove that induction 'works' on it, and can therefore show the operation of addition to be associative $\forall n\in N$ by the previous proof.

Obviously this way is too abstract and complicated for elementary or high school. So the most intuitive way would be to just explain what induction is and use it.

EDIT: Definition of a Peano space can be found https://www.physicsforums.com/showthread.php?t=518053", (within PF). The proof that there exist's a Peano space can be typed up if anyone is interested, as well as how $(N, s(n)=n+1, 1)$ is a Peano space; which would be sufficient to hold the background to these definitions and claim.

References: https://www.amazon.com/dp/0763727334/?tag=pfamazon01-20

 

[stringy]

If all we are after is some intuition (nothing rigorous) behind the definition

$$e = \lim_{x\to \infty} \left( 1 + \frac{1}{x}\right)^x,$$

here's a nice little argument...

Our goal and motivation: find a base $a$ such that $d/dx (a^x) = a^x.$

From the definition, we have

$$a^x = \lim_{h\to 0} \frac{ a^{x+h} - a^x}{h}.$$

We'll let x=1, so

$$a = \lim_{h\to 0} \frac{ a^{1+h} - a}{h}.$$

For small h,

$$a \approx \frac{ a^{1+h} - a}{h},$$

or

$$a^{1+h} \approx a(1+h).$$

But this simplifies to

$$a \approx (1+h)^{1/h}.$$

This approximation becomes more and more accurate as h goes to zero. Equivalently, put h=1/x and send x to infinity. We get

$$a = \lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x.$$

Put e=a. I would have liked to have seen something like this when I was first learning calculus with the promise that it'll all be made more rigorous later on.

 

[disregardthat]

The problem with this kind of argument is that a^(1+h) is approximately equal to a(1+2h) for small h as well.

 

[stringy]

It's not meant to be an ironclad argument. Of course you can poke holes in it. But you're missing the point. If a Calc 1 student points at the definition and asks, "Why?," you could show them this. If they ask a lot of questions, then it's a good segue into discussing their future enrollment in your university's analysis course.

 

[Dr. Seafood]

I thought that was pretty awesome, stringy.

 

[stringy]

Thank you, but I can't take credit for it. I got that out of a GRE study guide. It was the same guide that started with things like basic analytic geometry, trig identities, and logarithms...and ended with stuff like Lebesgue measure, point-set topology, and group theory. Evidently I remember more from the first half than I do the second. Lol!

 

[HallsofIvy]

pi can be calculated for a circle of radius r, and finding it's a constant independent of r answers the question.

No, it doesn't. The only way to "calculate" pi from a circle is to actually measure the circumference and divide by the diameter. Doing that for any number of circles and finding that you always get the same thing does NOT show that is true for all circles.

 

[disregardthat]

No, it doesn't. The only way to "calculate" pi from a circle is to actually measure the circumference and divide by the diameter. Doing that for any number of circles and finding that you always get the same thing does NOT show that is true for all circles.

What are you talking about? What do you mean by "actually measure"?

Surely we can calculate (mathematically) the circumference by means of mathematical methods for a circle with radius r, must I show it to you?

A circle is, by the way, in this regard, a mathematical object.

 

[Fredrik]

I also find HallsofIvy's complaint a bit strange, but I have one of my own. Edit: Not anymore. See the next few posts after this one.

I would define the length of a differentiable curve $C:[a,b]\rightarrow\mathbb R^2$ as $$L(C)=\int_a^b\sqrt{C_1'(t)^2+C_2'(t)}\, dt.$$ A circle around (a,b) with radius r is the range of the curve C defined by \begin{align}
x(t) &=a+r\cos t\\
y(t) &=b+r\sin t\\
C(t) &=(x(t),y(t))
\end{align} for all $t\in[0,2\pi)$. So the circumference of that circle is $$L(C)=\int_0^{2\pi}\sqrt{x'(t)^2+y'(t)^2}\, dt=r\int_0^{2\pi}dt=2\pi r$$ Yes, we can see that $L(C)/2r$ is independent of a,b and r, but the argument looks circular. You might be able to fix it, but I don't immediately see how.

 

[I like Serena]

Your argument depends on cosine, sine, and pi, whose definitions should not be readily used.

A simple fix is using $y(x) = \sqrt{r^2 - x^2}$
$$L(C) = \int_{-r}^{+r} \sqrt{1 + y'(x)^2} dx$$

This integral measures half the circumference of a circle.
Calculate it and you will find $\pi r$.

 

[Dr. Seafood]

but the argument looks circular

I see what you did there

 

[Fredrik]

A simple fix is using $y(x) = \sqrt{r^2 - x^2}$

Cool. That approach looks correct and non-circular. \begin{align}
x(t) &=a+t\\
y(t) &=b+\sqrt{r^2-(x(t)-a)^2}=b+\sqrt{r^2-t^2}
\end{align} We want to show that L(C)/2r is independent of a,b and r.
\begin{align}
\frac{L(C)}{2r} &= \frac{1}{r}\int_{-r}^{r}\sqrt{x'(t)^2+y'(t)^2}\,dt =\frac{1}{r}\int_{-r}^{r}\sqrt{1+\frac{t^2}{r^2-t^2}}\,dt =\frac{1}{r}\int_{-r}^{r}\sqrt{\frac{r^2}{r^2-t^2}}\,dt \\
&=\frac{1}{r}\int_{-r}^{r}\frac{1}{\sqrt{1-\frac{t^2}{r^2}}}\,dt =\left[\text{Variable change: }s=\frac{t}{r}\right] = \int_{-1}^1\frac{1}{\sqrt{1-s^2}} ds
\end{align}That last integral is clearly independent of a,b and r, and Wolfram alpha says that it's equal to $\pi$.

 

[micromass]

Yes, but to calculate that last integral, we need cosines, sines and pi again. But we need to be careful to use these concepts here.

 

[Fredrik]

We don't need to do the calculation. We just need to see that it's independent of a,b and r.

 

[I like Serena]

Hmm, I think you need to add "a" in the limits of the integrals.
Otherwise that looks correct.

 

[Kindayr]

So would you just define $\pi = \int_{-1}^1 \frac{1}{\sqrt{1-s^2}} ds$ with $s=\frac{t}{r}$ and leave the approximation to someone else?

 

[I like Serena]

Yes, but to calculate that last integral, we need cosines, sines and pi again. But we need to be careful to use these concepts here.

Noooo.
If you want you can define a special function called micromass(x) that happens to be the result of that integral (and that happens to work out to pi).

Oh, and not entirely by coincidence micromass(x) = arcsin(x).

So we found in a mathematically sound argument a definition for arcsin, as well as for pi! Yay!

 

[micromass]

Noooo.
If you want you can define a special function called micromass(x) that happens to be the result of that integral (and that happens to work out to pi).

Oh, and not entirely by coincidence micromass(x) = arcsin(x).

So we found in a mathematically sound argument a definition for arcsin, as well as for pi! Yay!

Yes, but how do you know that that integral even exists??

Math people are annoying, I know.

 

[Kindayr]

If you show $\frac{1}{\sqrt{1-x^2}}$ to be continuous between ]-1,1[, then its Riemann integrable. Would that work? Then just do the improper integral as $x\to 1$.

And it is continuous, so the improper integral should exist. I think at least, I could very well be wrong.

 

[HallsofIvy]

We don't need to do the calculation. We just need to see that it's independent of a,b and r.

No, you first have to show that integral exists- and it is an improper integral so that is non-trivial.

 

[I like Serena]

Yes, but how do you know that that integral even exists??

Math people are annoying, I know.

Ah, are you really forcing me to brush off my rusty integral theorems?
Reminds me of a glass bead necklace!

What about:
"A function on a compact interval [a,b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). "

And if you really want, we can adjust the boundaries of the integral to $\pm \frac 1 2 \sqrt 2$ to eliminate the singularities.

 

[KingNothing]

- Every number can be uniquely (up to order) decomposed in prime factors.

It's my impression that this is a matter of definition rather than proof: we've co-defined "prime numbers" along with this axiom in the same sense that we co-define the speed of light and the length of a meter.

That is, they are understood together.

 

[Kindayr]

It's my impression that this is a matter of definition rather than proof: we've co-defined "prime numbers" along with this axiom in the same sense that we co-define the speed of light and the length of a meter.

That is, they are understood together.

To my knowledge, the fundamental theorem of arithmetic isn't taken as an axiom and infact has a proof found http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic" .

 

[micromass]

Ah, are you really forcing me to brush off my rusty integral theorems?
Reminds me of a glass bead necklace!

What about:
"A function on a compact interval [a,b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). "

And if you really want, we can adjust the boundaries of the integral to $\pm \frac 1 2 \sqrt 2$ to eliminate the singularities.

But the function is not bounded. And the function can never be Riemann integrable since it has poles in 1 and -1. So you're working with improper Riemann integrals...

 

[Dr. Seafood]

Cool. That approach looks correct and non-circular.

Only semi-circular! HAHA LOLlkasfksdlkvcslamc

 

[I like Serena]

But the function is not bounded. And the function can never be Riemann integrable since it has poles in 1 and -1. So you're working with improper Riemann integrals...

All right, so let's adjust the boundaries of the integrals.
Let's only integrate the circumference of a quarter of a circle instead of half of a circle.
This is achieved by integration from $(a - \frac 1 2 r \sqrt 2)$ to $(a + \frac 1 2 r \sqrt 2)$.
The resulting integral becomes:
$$\int_{-\frac 1 2 \sqrt 2}^{+\frac 1 2 \sqrt 2} \frac 1 {\sqrt{1-s^2}} ds$$

This integral has no singular points and it is bounded.
Furthermore, it is still the same micromass(x) although we have clipped its extremities.

It will evaluate to $\frac \pi 2$.

 

[micromass]

All right, so let's adjust the boundaries of the integrals.
Let's only integrate the circumference of a quarter of a circle instead of half of a circle.
This is achieved by integration from $(a - \frac 1 2 r \sqrt 2)$ to $(a + \frac 1 2 r \sqrt 2)$.
The resulting integral becomes:
$$\int_{-\frac 1 2 \sqrt 2}^{+\frac 1 2 \sqrt 2} \frac 1 {\sqrt{1-s^2}} ds$$

This integral has no singular points and it is bounded.
Furthermore, it is still the same micromass(x) although we have clipped its extremities.

It will evaluate to $\frac \pi 2$.

Seems ok

\begin{Annoying-mathematician}
Of course, this only shows that the length of a quarter of a circle is independent of pi. Nothing is said about the entire circle.
\end{Annoying-mathematician}