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

Conceptual: What is it?

I want a conceptual understanding of the object.

Computational: How do I compute with it?


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

Conceptual and computational are not well defined ideas

Example: An algebraic identity

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.)

Computational proof

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.

Conceptual proof

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 $b$.

Example: Property of the ordering of the reals

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 about.


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

Geometric proof

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 $x\le y$.

Logical Proof

We know that for real numbers, if $x> y$ and $y > z$, then $x >z$ (this is the transitive law). 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:

Why, 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.

Another point: The logical proof works in a much more general setting: The statement is true in any totally ordered set. The geometric proof gives you no clue that that is true.

Example: Property of the greatest common divisor of two integers.


Let $m$ and $n$ be positive integers, and let $r=m\bmod n$. Then $\text{GCD} (m,n)= \text{GCD} (m,r)$.

Conceptual proof

Some definitions:

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

Proof: If $\frac{m}{d}$ and $\frac{n}{d}$ are both integers, then $\frac{qn}{d}$, which is $q\frac{n}{d}$, is an integer, so $\frac{r}{d}=\frac{(m-qn)}{d}=\frac{m}{d}-q\frac{n}{d}$, This is the difference of two integers, so is an integer.

If $\frac{n}{d}$ and $\frac{r}{d}$ are both integers, then $\frac{m}{d}=\frac{qn+r}{d}=q\frac{n}{d}+\frac{r}{d}$, which is the sum of two integers and so is an integer.

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 $r$. This statement 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).

Example: Derivatives

Conceptual definition

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 $x=-0.7$.

Concept to computation

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.

Computation to general formula

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.

Other examples

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