So, I understand that for something to be rigorous it is inclined to meet strict and occasionally extreme criteria. Within the scope of mathematics, this therefore implies that anything that is truly rigorous may not defy any of the rules of mathematics, and as it should with everything in mathematics, but my question is: do the abstract rules align with the geometric rules when it comes to proving statements in question?

For example, I will take the proof that $$\lim_{x\to0}\frac{\sin(x)}{x} = 1$$

the most renowned proof of this statement entails the following geometric shape:enter image description here1

based on this shape, the following inequality can be seen: $$\triangle ACB\leq ACB \leq \triangle ADB$$ which can be rewritten as $$\frac{1}{2}\cdot1\cdot1\cdot\sin(x)\leq\frac{1}{2}1^2x\leq\frac{1}{2}\cdot1\cdot\tan(x)$$ $$\frac{1}{2}\sin(x)\leq\frac{1}{2}x\leq\frac{1}{2}\tan(x)$$ then, dividing through by $\frac{1}{2}$, $$\sin(x)\leq x\leq \tan(x)$$ and then by $\sin(x)$, $$1\leq\frac{x}{\sin(x)}\leq\frac{\tan(x)}{\sin(x)}$$ and since $\tan(x)=\frac{\sin(x)}{\cos(x)}, \frac{\tan(x)}{\sin(x)}=\frac{1}{\cos(x)}$ $$1\geq\frac{\sin(x)}{x}\geq\cos(x)$$ from here, the limit can be taken: $$\lim_{x\to0}1\geq\lim_{x\to0}\frac{\sin(x)}{x}\geq\lim_{x\to0}\cos(x)$$ $$1\geq\lim_{x\to0}\frac{\sin(x)}{x}\geq1$$ hence, by the squeeze theorem $$\lim_{x\to0}\frac{\sin(x)}{x}=1$$

the reason for my question is this: why, despite the fact that we identify the inequality by means of observation, and that hence it can clearly be observed that the three identified areas are each one bigger than the other, do we still use the $\leq$ symbol as oppose to the $<$ symbol? Isn't it true that these areas can never be exactly equal to each other? Hence, does the use of the symbol truly act in accordance with the "extreme conditions" of abstract mathematics? (Note: in this context, I am referring to abstract mathematics as anything that is not represented visually)

I understand that explaining the scenario for one geometric proof does not mean it applies to all geometric proofs, but the question is merely a curious one, and anyone with more experience in mathematics than myself is more than welcome to explain any flaws in my logic. Thank you.

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    $\begingroup$ It is quite common to use $\leq$ even when one knows that $<$ holds. I don't know why. Perhaps it's because $\leq$ is better behaved under limits, so we want to preserve the clarity of keeping the same symbol after taking limits? Maybe it's because the squeeze theorem uses $\leq$, and we want it to be as clear as possible that it actually applies in our situation? $\endgroup$ – Arthur Sep 22 '17 at 14:20
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    $\begingroup$ @RossMillikan No, he inverted all the fractions. The mistake is that all the $\leq$ should've been $\geq$ from that point on. $\endgroup$ – Arthur Sep 22 '17 at 14:24
  • $\begingroup$ That's a reasonable explanation, but if the majority of mathematical problems stem from reality; all mathematical equations must have a set of rules they must meet in order to be truly representative of reality, and the question is therefore whether or not $\leq$ is truly representative of the reality to which this specific geometric proof applies. If that makes sense. @Arthur $\endgroup$ – joshuaheckroodt Sep 22 '17 at 14:25
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    $\begingroup$ To address your question about $\le$ versus $<$, think about logical disjunctions, i.e. statements of the form "P or Q". In general, if you know "Q" is true, you may conclude that "P or Q" is true. As a special case, if you know "$a<b$" is true, then you may conclude that the disjunction "$a=b$ or $a<b$" is true, which is precisely the definition of "$a \le b$". So, there is no mathematical objection to be made here. $\endgroup$ – Lee Mosher Sep 22 '17 at 14:27
  • $\begingroup$ The inequalities are written as $\leq$ instead of expressing the extra information contained in the short-hand symbol $<$ because that extra information will be lost anyway after taking limits: $a_n\to L_1$, $b_n\to L_2$ and $a_n<b_n$ implies only that $L_1\leq L_2$. One can't derive $L_1<L_2$ in general from those hypotheses. $\endgroup$ – Hellen Sep 22 '17 at 14:28

It is perfectly valid to go through and replace $\leq$ by $<$ for most of this proof, since for any given $x<\pi/2$ the areas are all different. (However, even if we get to $1<\frac{\sin x}x<\cos x$, when we take the limit we have to put $\leq$ back in. $f(x)<g(x)$ only implies $\lim f(x)\leq \lim g(x)$.)

But you really don't have to use $<$. Saying $f(x)\leq g(x)$ does not imply that sometimes $f(x)<g(x)$ and sometimes $f(x)=g(x)$; all it means is that for every $x$ one or the other is true, and it is perfectly fine if it is the same one every time.


There's nothing fundamentally different about a geometric proof compared to other proofs. It refers to a style of proof, where the steps are designed to be pictured, operating on some kind of prototypical example. When people don't follow along with a picture, the proofs tend to be more dense and less intelligble. Such proofs don't always need to be accompanied by a diagram, but there's always an assumption that the reader is keeping some kind of picture in their head or on paper.

While it is common in geometric proofs to be less rigourous, it is not essential! The choice to be less rigourous is usually chosen because rigour is often accompanied by mind-numbing detail, which spoils the evocative nature of geometric proofs. The details can still be filled in, and there should be rigorous rules that underpin them.

For example, one can show both rigorously and geometrically that any equilateral triangle has angles $\pi/3$, like so:

Suppose $\Delta ABC$ is equilateral. Then $AB = BC = AC$. Since $AB = BC$, using a theorem about isoceles triangles, we obtain $\angle BAC = \angle BCA$. Similarly, from $BC = AC$, we obtain $\angle BCA = \angle ABC$. Using the angle sum theorem, we see that $\angle BCA + \angle ABC + \angle BAC = \pi$, and since they're equal, $\angle ABC = \pi /3$ (and similarly for the other angles).

Everything in the above proof could be verified by a computer, from Euclid's axioms (or maybe just inner product space axioms). It's perfectly rigorous, but it's difficult to read through without having some picture in your head. That's what makes it a geometric proof.


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