# Are proofs always synonymous with explanations?

As I have looked at more and more complicated mathematics I have often found formulas and theorems and so on where I understand the proofs for them but I still don't feel like I understand how they really work. Unless I am given a somewhat intuitive explanation, I tend to feel alienated. While I don't doubt the validity of the proofs, to me it often feels unsatisfying if I cannot explain concepts intuitively. Of course, some proofs go hand-in-hand with explanations (they are merely a way of setting out formally what we already know). More often though, it feels like I have to accept they 'just work' because they have been shown to be mathematically rigorous. Am I being too demanding, or is there a way to carry on this desire for intuitive explanations as I progress further?

To illustrate, here is an example of a proof that there are infinitely many primes (Euclid's Theorem) which also makes intuitive sense to me:

• Consider any finite list of prime numbers $$p_1,p_2,...,p_n$$
• Let $$P$$ equal the product of the list. $$P = p_1p_2...p_n$$
• Let $$q = P+1$$
• If $$q$$ is prime, then the list is incomplete
• If $$q$$ is composite and there were a number $$p_x$$ on the list that could divide it evenly, then $$p_x$$ would have to be able to divide $$P$$ and $$P+1$$, meaning it would also have to divide $$(P+1)-P=1$$. Since no prime goes into 1, no number $$p_x$$ meets this requirement. Therefore, we can conclude that if $$q$$ is composite, it must be divisible by a prime not on the list. Hence, no list is complete

I understand this in layman terms to mean if you multiply the list together and add 1, then every prime number in the list would be 'one off'. It seems the proof formalises this insight.

However, other proofs that involve lots of rearrangement don't seem to provide any idea of 'why' the theorems/formulas work. It just seems like with a lot of manipulation of an equation, the theorem/formula springs out. Example (proof of the Cosine Rule from brilliant.org):

Here, it almost seems as if we got lucky and because we ended up with the identity $$sin^2x + cos^2x \equiv 1$$, it all simplifies down and we end up with the Cosine Rule.

• Your proof for the infinite primes is not quite correct – fourier1234 Aug 28 at 20:05
• @fourier1234 I think I know where I went wrong, so I’ll edit it. If it is still incorrect, please notify me. – Joe Aug 28 at 20:16
• I think it's a matter of how much you can pile on abstract explanations until you human brain overloads and your eyes glaze over. Ideally, we like to believe the proof reader has a perfect brain and recognizes every argument with complete and thorough understanding. But... we're human. Your cosine proof can follow that from the Pyth. Th. by noting the inside right triangle has $b^2+c^2>(b-r)^2+h^2=a^2$. So the question is how much bigger. i.e. showing $[b^2-(b-r)^2]+[c^2-h^2]=2bc*\cos A$ which once we note $c^2=r^2+h^2$ (by PT) and $\cos A=\frac rh$ by definition, it follows pretty directly. – fleablood Aug 28 at 21:34

What I like to do when a proof is long and abstract, is I break it up into chunks that I can describe intuitively in 1 sentence. Then these sentences form the outline of the proof which is an explanation. The skill here is to decide how much detail to include in each sentence

For example I will do this for your cosine formula proof. And I will include enough detail as I need to make it intuitive enough to myself

• The goal is to show "Pythagorean theorem is almost true for all triangle, minus some error". In other words, that any side length can be expressed in terms of the other side lengths ($$b$$ and $$c$$), and the opposite angle ($$\alpha$$)
• First we can simplify the problem; split our triangle into 2 right angled triangles, which are familiar to work with
• On the right triangle, we can use the familiar Pythagorean theorem with $$a$$, $$h$$ and $$b-r$$
• But since $$h$$ and $$r$$ are shared by the left triangle, we can easily convert them using the familiar SOHCAHTOA to expressions involving $$b$$, $$c$$ and $$\alpha$$

For the law of cosines, we know that two sides and the included angle determine a triangle, so $$a$$ is definitely determined by $$b,$$ $$c,$$ and $$\alpha.$$

If you had no idea what formula related $$a$$ to $$b,$$ $$c,$$ and $$\alpha,$$ you could still try to find a formula by dividing the triangle into two right triangles as shown in the proof from brilliant.org. Once you have drawn the two right triangles it is just a matter of applying the Pythagorean Theorem twice in order to get some formula relating $$a$$ to $$b,$$ $$c,$$ and $$\alpha.$$ That much is guaranteed.

So you can easily get to the formula

$$a^2 = (c\sin\alpha)^2 + (b - c\cos\alpha)^2.$$

Here's where experience helps: this is not the only place where there is a fortunate opportunity to add $$\sin^2 + \cos^2$$ and get $$1$$ (or add $$k\sin^2 + k\cos^2$$ for some common factor $$k$$ and get $$k$$). So when we see a sine and a cosine of the same angle each inside a squared term, one obvious thing to try is to combine their squares in this way.

Or we might now see ahead that far, but we might try multiplying out the terms on the right side of the equation anyway, because who knows, sometimes you stumble over something that way:

$$a^2 = c^2\sin^2\alpha + b^2 - 2bc\cos\alpha + c^2\cos^2\alpha.$$

And now if it didn't occur to us before to look for an opportunity to simplify $$c^2\sin^2\alpha + c^2\cos^2\alpha,$$ it's a lot more obvious now that there is one.

This may not seem earth-shatteringly beautiful like the proof of the infinitude of primes, but then it is after all a much more pedestrian result: just a formula for computing one of the sides of a triangle knowing two others and the angle between them, or alternatively computing one of the angles knowing the three sides. The result is computational, so it's not so out of place for the proof to be computational. Much like proving that $$114 + 265 = 379,$$ we just do the arithmetic and we get an answer.

Too bad this proof only works for acute angles and we still have to do more work to show that the same formula happens to apply to obtuse angles as well.

The cosine proof can be made intuitive.

We know $$b > d= b-r$$ and $$c > h$$. And we know by the pythagorean th. and $$d$$ and $$h$$ are sides of a right triangle with hypotenuse $$a$$. SO $$d^2 + h^2 = a^2$$. So $$b^2 + c^2 > d^2 + h^2 = a^2$$. But how much bigger?

Draw a picture. If we let $$d = b-r$$ so $$b= d+r$$ then $$b^2 = (d+r)^2 = d^2 + 2*dr + r^2$$. We can see in our picture that $$b^2$$ is a square composed of two squares ($$d^2$$ and $$r^2$$) and two rectanges $$d\times r$$.

Now $$r$$ and $$h$$ are two sides of right triangle with hypotenuse. So $$c^2 = r^2 + h^2$$.

Taking all we have $$a^2 = d^2 + h^2$$

And $$b^2 = d^2 +$$ two rectangles, $$r\times d$$ and square $$r\times r$$.

ANd $$c^2 = h^2 +$$ the square $$r\times r$$.

So $$a^2 = b^2 + c^2 -$$ two rectangles $$r\times d$$ and two squares $$r\times r$$.

Now we can "glue" an $$r\times d$$ rectangle to an $$r\times r$$ square to get an $$r\times (d+r) = r\times b$$ rectangle.

So $$a^2 =b^2 +c^2 - 2(r*b)$$. But we need to express that $$r$$ variables in terms of $$a,b,c$$ and angle $$A$$.

Well $$r, h,$$ and $$c$$ are the sides of a right triangle so $$r = c*\cos A$$ and .... that's it.

......

$$a^2 = b^2 + c^2 - 2bc\cos A$$[1].

So, was that mechanical, or explanitory? intuitive or being led by the nose?

Well, I don't know. IMO a good proof should be explanatory. But proofs also rely on an idealized reader with the assumption that every step will be absolutely ingested with utter comprehension.

But we are human. We stumble and sometimes see things and sometime get blinded.

.....

[1] This assumed $$m\angle A < 90$$. If $$m\angle A = 90$$ we have a right angle and $$a^2 = b^2 + c^2 - 0$$ and $$\cos 90 = 0$$..... And if $$90 < m\angle A < 180$$ we can do something very similar but with $$d = b+r$$.

I don't know if this helps but I offer a proof from a paper I am writing in hopes that it shows that such can be presented intuitively. Some authors seem to think the reader has the the same background knowledge from [sometimes] years of research (and insights gained) that when into developing a proof. My proof could have been done by merely presenting the equations and showing how they relate. I hope mine shows consideration for the reader and let's you see that there are ways to prove-with-insight. Let me know, good or bad, because it will help me too.

• The proof seems to be laid out clearly, but it is probably beyond me (I only know maths up to A-level – 6th form in UK). I really appreciate the idea behind it though. How would you advise me in my approach to proofs that are rigorous but not particularly insightful? – Joe Aug 28 at 17:49
• It often helps to draw a pictures like the triangle(s) in your question. Draw a group of circles if the question is about arranging coins, drawing a graph using Wolfram Alpha, or draw squares to prove a tiling problem. I had to enter millions of formulas in spreadsheets, sort, gather and examine the results to find the set pattern and other topics in my paper. Perhaps you could try some of these things on paper and learn a programming language or use spreadsheet. – poetasis Aug 28 at 19:39
• What I'm saying is, try to find examples of what the guy is talking about which is the reverse of what I did. Suppose you were just given my equations above. They work but where did they come from? If you tried values of (n,k) from (1,1) to (4,4), you easily see set pattern and how the components could fit together to make the equations. Some of the insights will come when you learn more math. Example: The coefficients of a given power of x are equal on both sides of an equation. If $ax^2 +bx+c=(y-1)x^2+(z-2)x+3$ then $a=y-1, b=z-2, c=3$, no matter how complicated the expressions. – poetasis Aug 28 at 19:49
• One user Blue has been very helpful to me. He tries and succeeds in making math intuitive. – poetasis Aug 28 at 20:01
• Your theorem makes no sense, though. You quantify some $n,k\in\mathbb N$ yet they don't appear anywhere in the statement ($A^2+B^2=C^2$) that follows. – Hirshy Aug 30 at 11:30