Revised 2013-05-18 Introduction to this website website TOC website index blog Back to top of Understanding Math chapter

When mathematicians consider a mathematical object, they are typically interested in two different aspects of it:

I want a **conceptual understanding** of the object.

- How do I
**think**about it? - What
**properties**does it have? - How is it
**different**from other math objects? - How can I understand it so that I can see possible
**applications**?

- How do I find a
**value**of the object (if that makes sense)? - How to I tell how
**big**it is (in some sense of big)? - How do I
**determine in an efficient way**what properties it has?

Proofs can have a conceptual side and a computational side too.

- A
**conceptual**proof helps you understand why the statement is true. - A
**computational**or**symbolic**proof may be easier to check systematically to see if it is correct, and to automate using some suitable computer program.

- Conceptual presentations are commonly geometric, but they don't have to be.
- If you understand some technicalities really well, you might call something conceptual that looks like a horrible bunch of symbolic manipulations to someone else.
- Whether something is conceptual or not depends on what concepts you understand!

Here is a simple example that shows the distinction between concept and computation. This identity \[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\] holds for all real numbers. (In fact it holds in any commutative ring.)

This proof shows an explicit series of steps that verify the identity using basic laws of algebra: \[\begin{align*} (a-b)(a+b)&= (a-b)a+(a-b)b \;\;\; \text{distributive law} \\ &= {{a}^{2}}-ba+ab-{{b}^{2}} \;\;\; \text{distributive law applied to each term separately} \\ &= {{a}^{2}}-ab+ab-{{b}^{2}} \;\;\; \text{commutative law} \\ &= {{a}^{2}}+0-{{b}^{2}} \\ &= {{a}^{2}}-{{b}^{2}} \end{align*}\]

These laws hold in every commutative ring, so the identity holds in every commutative ring.

This diagram shows why the identity is true geometrically (for $b\le a$).

Boxes 1 and 2 are congruent, so have the same area. Therefore the area of boxes 1 and 3 is the same as the area of boxes 2 and 3. But the first one has area $(a+b)(a-b)$ and the second one has area $a^2-b^2$, so necessarily $a^2-b^2=(a+b)(a-b)$.

This conceptual proof requires no algebraic laws at all. On the other hand, it is not easy to see how to generalize it to a commutative ring. Think about drawing a picture that shows that\[{{a}^{2}}-{{b}^{2}}=(a-b)(a+b)\]
for all *complex* numbers $a$ and $b$.

The idea of "conceptual proof" depends on your experience. If you are not familiar with basic geometric facts, the preceding geometric proof may not be conceptual!

Here is another example that shows how "conceptual" depends on what you know.

For all real numbers $x$ $y$, and $z$, if $x\le z$, then either $x\le y$ or $y\le z$.

This is a standard trick for proving that "$P$ or $Q$", namely prove directly that if $P$ is false then $Q$ has to be true.

If $x\leq y$ is not true, then it must be true that $y\lt x$. We are given that $x\leq z$, so now we know that $y\lt x$ and $x\leq z$. It follows from that fact that the ordering of the reals is transitive that $y\leq z$.

The number $y$ has to be in one of three intervals in this picture of (part of) the real line. (The left interval goes off to infinity to the left and the right interval goes off to infinity to the right. The middle interval has finite length.)

If y is in the left interval or the middle interval, then $y\le z$. If it is in the middle or right interval, then $x\le y$.

The transitive law says that if $x\gt y$ and $y \gt z$, then $x \gt z$. The contrapositive of this statement is: If $x$ is not greater than $z$ then *either* $x$ is not greater than $y$ *or* $y$ is not greater than $z$ (this uses a DeMorgan Law). But for any real numbers $r$ any $s$, saying that $r$ is not greater than $s$ is the same as saying that $r\le s$. So, rewording the contrapositive, we get: if $x\le z$, then either $x\le y$ or $y\le z$, as was to be proved.

If you have some experience with mathematical logic, you might react to the logical proof (as I did) this way:

The statement in the theorem is nothing but the contrapositive of the transitive law!

When I realized that, I felt that I had acquired a new insight, so I would call this statement conceptual. The geometric proof gives you a *different* insight.

- Let $m$ and $n$ be positive integers. Then there are unique integers $q$ and $r$ for which $m=qn+r$ and $r\lt n$. In this case $q$ is the
**integer quotient**when $m$ is divided by $n$ and $r$ is the**remainder**. - For example, if $m=17$ and $n=5$, then $q=3$ and $r=2$ because $17=3\times 5+2$. In other words, "five goes into $17$ three times with remainder two."
- An integer $d$ is a
**common divisor**of $m$ and $n$ if $\frac{m}{d}$ and $\frac{n}{d}$ are both integers. In this case, for $m$, $m=\frac{m}{d}d+0$, so $q=\frac{m}{d}$ and $r=0$, and similarly for $n$ - For example, $5$ is a common divisor of $35$ and $45$ since $\frac{35}{5}$ and $\frac{45}{5}$ are both integers.
- Examples: the set of (positive) common divisors of $35$ and $45$ is $\{1,5\}$ and the set of positive common divisors of $12$ and $18$ is $\{1,2,3,6\}$.
- The
**greatest common divisor**is the**maximum**of the integers in the set of common divisors of $m$ and $n$. - So $\text{GCD}(35,45)=5$ and $\text{GCD}(12,18)=6$. Note that also $\text{GCD}(35,5)=5$ and $\text{GCD}(5,45)=5$.

This theorem states a property of the GCD of two integers.

**Let $m$ and $n$ be positive integers, and let $r$ be the quotient when $m$ is divided by $n$. Then**
\[\text{GCD} (m,n)= \text{GCD} (n,r)\]

**Lemma: An integer $d$ divides both $m$ and $n$ if and only if it divides both $n$ and $r$.**

- First suppose $d$ divides both $m$ and $n$, then $\frac{m}{d}$ and $\frac{n}{d}$ are both integers.
- Therefore, $\frac{qn}{d}$, which is $q\frac{n}{d}$ (the product of two integers), is an integer (this is a rabbit), so $\frac{r}{d}=\frac{(m-qn)}{d}=\frac{m}{d}-q\frac{n}{d}$.
- $\frac{m}{d}-q\frac{n}{d}$ is the difference of two integers, so $\frac{r}{d}$ is an integer.
- Therefore $d$ divides $r$.
- Now suppose $d$ divides both $n$ and $r$. Then $\frac{n}{d}$ and $\frac{r}{d}$ are both integers.
- Therefore $\frac{m}{d}=\frac{qn+r}{d}=q\frac{n}{d}+\frac{r}{d}$ which is the sum of two integers. So $\frac{m}{d}$ is an integer.
- Therefore $d$ divides $m$.

The Lemma means that* the set of common divisors of $m$ and $n$ is the same as the set of common divisors of $n$ and $r$*.

Since the GCD is the *maximum* of the set of common divisors and a (finite) set of integers has a unique maximum, $m$ and $n$ have the same GCD as $n$ and $r$.

The statement in the preceding paragraph is the conceptual part of the proof. It makes some students go nuts for awhile, then suddenly they understand it and think it is obvious. (This happens to all mathematicians from time to time. See ratchet).

The **derivative of a function** $f$ is another function $f'$ whose value at $a$ is the *slope of the tangent line to $f$ at $a$*. This picture shows the tangent line to $y=x^3-x$ at the point $x=-0.7$.

The slope of the tangent line at $x=-0.7$ is about $0.47$. You could in fact calculate that value, to one decimal place anyway, by drawing a very careful picture of the graph of $y=x^3-x$, positioning the tangent line by eye and then measuring the run and rise with a ruler.

You can get a better estimate by looking at the **secant lines**. They are shown below for $x=-0.7$. The other point in each case is $x=-0.7+h$ for $h=0.2,\, 0.1,\,0.06,\,0.02,\,0.01$ in that order.

The slope of the secant line is \[\frac{f(x+h)-f(x)}{x-(x-h)}=\frac{f(x+h)-f(x)}{h}\] when $h$ is small. For example, for $x=-0.7$ and $h=0.01$ the value is about $0.4491$, and when $h=0.001$ it is $0.467901$.

The trouble with getting the *exact* value of the slope, which is when $h=0$, is that it looks like you must divide by zero. However, **a miracle occurs** when $f(x)=x^3-x$: The formula for the slope simplifies to $3x^2+h^2-1$, so the limit when $h\to0$ is $3x^2-1$. ** So now we have a formula for the derivative function for $x^3-x$.** If you have taken a little calculus, you may know that we can always get a formula for the derivative of a polynomial this way, and (with extra cleverness in some cases) for the derivatives of most other calculus-type functions.

**This example shows how mathematicians (in this case in the seventeenth century) started from the naive or pictorial idea of the derivative as giving the slope of the tangent line and created calculations that allowed us to have a formula for the derivative of many functions.**

There is more detail about this in my blog post The power of being naive, which has diagrams you can manipulate with your mouse.

- The discussion of square roots in Definitions.

This work is licensed under a Creative Commons Attribution-ShareAlike 2.5 License.