$|a – b| < ε, \forall ε > 0 \iff a = b$ resurfaces on standardized tests to 17 year old (y.o.) students, who can memorize and regurgitate the proof to earn full marks.

But the glut of duplicates substantiates that most students can’t intuit it! How can I demystify this equivalence, to 17 y.o.? It's counterintuitive and paradoxical for $<$ and $=$ to become equivalent!

How does the $<$ uncannily transmute to an $=$?
Vice versa, how can $=$ eerily transmogrify into $<$?

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    $\begingroup$ What country are you in? Can you share a link to a standardized test that includes this? $\endgroup$
    – Sue VanHattum
    Commented Oct 16, 2023 at 3:24
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    $\begingroup$ Not an answer, but note that = isn't "transmogrifying" into < - rather, the = is showing an equivalence with a statement about < and - - it's the difference that is less than ... $\endgroup$ Commented Oct 16, 2023 at 9:06
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    $\begingroup$ As pointed out by @user7761803, the words become equivalent, transmute, transmogrify obfuscate the real meaning of the statement. Another (this time, quite trivial) example of stating equality in terms of inequalities could be: $a=b$ if and only if neither $a < b$ nor $a > b$. $\endgroup$ Commented Oct 16, 2023 at 14:58
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    $\begingroup$ To amplify other comments to this end: describing this mathematical statement as saying that $<$ and $=$ are becoming equivalent is like describing the algebraic identity $(a-b)\times(a+b) = a^2-b^2$ as saying that $\times$ and $-$ are becoming equivalent. The individual symbols are not being equated with each other; rather, the entire assertions in which those symbols appear are mathematically equivalent. $\endgroup$ Commented Oct 16, 2023 at 18:07
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    $\begingroup$ I understand the maths in question perfectly well, but if someone were to have described this as "= can become equivalent into an <" when I was learning it, that probably would've left me deeply confused and frustrated, because that isn't what's happening at all. $\endgroup$
    – NotThatGuy
    Commented Oct 17, 2023 at 7:30

13 Answers 13


I'd recommend to play a little game with them. You choose a number $a.$ They choose a number $b$ that they think will satisfy $|a-b| < \epsilon \,\,\, \forall \epsilon > 0.$

You try to state a counterexample, that is, an $\epsilon > 0$ such that $|a-b| \geq \epsilon.$ If you can do this, you win. Otherwise, if you can't, then they win.

Enough rounds of this, and they should see that the only way they can win is if they choose $b = a.$

Some additional notes:

  • To help students develop the right intuition, I'd recommend to refer to $|a-b|$ as the distance between $a$ and $b$ throughout the game. (And if they don't understand why it represents distance, then it would be worth pausing the game to review that, since it's such an important part of the intuition.)

  • [Noted by Michal Miśkiewicz] This game not only leads the students to intuitively grasp the equivalence, but also leads them into the proof: after enough rounds, they should also see that there's a simple rule for producing some $\epsilon$ for each $b.$

  • 11
    $\begingroup$ The two-player "game" of epsilon-delta is how I learned limits too. $\endgroup$
    – qwr
    Commented Oct 16, 2023 at 14:42
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    $\begingroup$ By the way, this leads not only to the discovery but also a proof (after enough rounds, they should also see that there's a simple rule for producing some $\epsilon$ for each $b$). $\endgroup$ Commented Oct 16, 2023 at 14:55
  • 3
    $\begingroup$ While such a game is great, it can also lead to false conceptions. You should intersperse the back-and-forth by sometimes not giving a counterexample, both to give the student time to think of one themselves and to make the point that for many problems nobody has been able to come up with a counterexample yet, and still one should not necessarily assume them to be true. $\endgroup$ Commented Oct 16, 2023 at 15:26
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    $\begingroup$ The stubborn student will confidently choose .999... and 1, claiming they are different numbers. This will lead to a game of "No, YOU'RE making a circular argument!" $\endgroup$
    – Dan Staley
    Commented Oct 17, 2023 at 23:54
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    $\begingroup$ @DanStaley I am having trouble thinking of the simplest way to present it, but if that is truly what they do, writing out .9999 as a series as 'a' and 1 as 'b' and flipping the game to challenge them to find epsilon would be another way to present this concept. Assuming they're willing to play in good faith. $\endgroup$
    – Chuu
    Commented Oct 18, 2023 at 22:13

If a student sees $=$ and $<$ somehow becoming equivalent here, they exhibit a common beginner mistake:

They don't actually read the statement they are considering.

Quantifiers are a particularly common victim of this: They just get treated as decorations, rather than them (and their order) being as utterly crucial to the meaning of a statement as they are. Other common examples are not distinguishing between "element of" and "subset of"; or whether a property refers to a set or the elements thereof.

  • 8
    $\begingroup$ It's OP that's saying = and < become equivalent, which they're trying to teach to students (I'd agree that this is a mistake, though). $\endgroup$
    – NotThatGuy
    Commented Oct 17, 2023 at 7:35

You could start by using precise language. Numbers don't have arbitrary small differences. To get into "arbitrary small" domain, you need at least one constrained variable, like an $\epsilon>0$. If this $\epsilon$ will be greater than the difference between two numbers even though it can be chosen arbitrarily small within its domain, the two numbers under consideration have a difference of exactly zero. Not an "arbitrarily small" difference. What is arbitrarily small is not the difference of the numbers but a variable greater than the difference of the numbers.

If you start calling the difference between two numbers "arbitrarily small", you are teaching pupils to regurgitate meaningless words making the teacher happy.

  • 1
    $\begingroup$ The title of the question is a good example of the need for precise language, i.e. To 17-year olds how to explain that 2 numbers with arbitrarily small difference are equal?. They are not equal. They could be effectively equivalent for certain purposes only, if the difference between the two numbers is so small as to be irrelevant for the given context. E.G. buying 1 pound of butter is equivalent to buying 1.0000000000047 pounds of butter, because no reasonable baker would care about the difference. The two numbers are both unequal and equivalent in the implied context of baking. $\endgroup$
    – MikeOnline
    Commented Oct 18, 2023 at 21:10
  • $\begingroup$ This is a title I had to edit into the question, because the original one was even worse. If you can manage to come up with a better title without falling into arrant pedantry, please go ahead. $\endgroup$ Commented Oct 26, 2023 at 18:26

Here are a few possible leads.

  • Ask your students to prove the following equality, then discuss with them whether they find this intuitive or counterintuitive, and why: $$0.999999... = 1$$
  • Given two real numbers $a$ and $b$, consider the three quantities $d_☾(a,b)$, $d_☆(a, b)$ and $d_🪐(a, b)$ defined by: \begin{align*}d_☾(a, b) & = \min\left\{\varepsilon \geq 0 \,\mid\,\left|{a-b}\right|\leq\varepsilon \right\} \\ d_☆(a, b) & = \inf\left\{\varepsilon > 0 \,\mid\,\left|{a-b}\right|<\varepsilon \right\} \\ d_🪐(a, b) & = \left|{a-b}\right| \end{align*} then have your students prove that: $$d_☾ = d_☆ = d_🪐$$ then reword your previous theorem as:$$\forall a, b \in \mathbb{R},\quad d_☆(a,b) = 0 \iff a = b$$
  • Consider the definition of convergence for a sequence $(b_n)_{n \in \mathbb{N}}$ converging to a limit $a$: $$\forall \varepsilon > 0,\, \exists N \in \mathbb{N},\, \forall n \geq N,\, |a-b_n| < \varepsilon$$ then discuss the specific case of a stationary sequence, i.e the case where you have a number $b$ such that $\forall n \in \mathbb{N},\, b_n = b$.
  • Consider the following propositions for two converging sequences $(a_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$: \begin{align*} \forall n,\, a_n \leq b_n & \implies \lim_n a_n \leq \lim_n b_n \\ \forall n,\, a_n < b_n & \implies \lim_n a_n \leq \lim_n b_n \end{align*} And note how we are forced to accept the possibility of equality appearing in the limit.

Note that I used a lot of formal notation in the second and third points, but the notion of "distance between two points" is meant to be rather intuitive and the fact that we have more than one manner to define it with formal symbols should spark a discussion. As for the definition of convergence of a sequence, you can illustrate it graphically, then ask your students what this graph becomes in the case of a constant sequence.

Also, if your students are not familiar with the definition of $\inf$, then great! Rewrite the definition of $d_☆$ using only $\min$ and discuss the special cases. This will force you to make things more explicit and deal with what the abstract $\forall \varepsilon > 0$ really hides.

There are several possible proofs for the equality $0.999... = 1$; one possible takeaway is "they are equal because you cannot squeeze a third number inbetween", ie there is no $x$ such that $0.999... < x < 1$. Or alternatively they are equal because the distance between them is 0. In fact it's a direct application of your $\forall \varepsilon$-proposition: since $0.9 < 0.999... < 1$, we have $\left| 1 - 0.999... \right| < 0.1$, and likewise $0.99 < 0.999... < 1$, and $0.999 < 0.999... < 1$, etc. with any finite number of 9 after the decimal point, and thus $\left| 1 - 0.999... \right| < 0.0001$ for any number of 0 before the 1, which can be rewritten as $\forall n \in \mathbb{N},\, \left| 1 - 0.999... \right| < 10^{-n}$, which is equivalent to $\forall \varepsilon > 0,\, \left| 1 - 0.999... \right| < \varepsilon$, which by your proposition is equivalent to $0.999... = 1$.

  • $\begingroup$ Thanks for your comment. But how's your first bullet point (on $0.999999...=1$ ) relevant? "the content of a proof of 0.9999⋯=1 is not that we must agree to define 0.9999… as 1. The content is to define 0.9999… as the sum above, then by deduction show this sum is equal to 1.". Kindly elaborate the relevance? $\endgroup$
    – user95017
    Commented Oct 17, 2023 at 0:22
  • 1
    $\begingroup$ @user95017 0.999... and 1 are two numbers that look like different numbers, but it turns out that they're equal. There are several different ways to prove that they are equal, but one takeaway is "they are equal because you cannot squeeze a third number inbetween", ie there is no x such that 0.999... < x < 1. Or alternatively they are equal because the distance between them is 0. $\endgroup$
    – Stef
    Commented Oct 17, 2023 at 8:44
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    $\begingroup$ @user95017 In fact it's a direct application of your proposition: 0.9 < 0.999... <= 1, thus |1 - 0.999...| < 0.1; likewise 0.99 < 0.999... <= 1, and 0.999 < 0.999... <= 1, and 0.9999 < 0.999... <= 1, thus |1 - 0.999...| < 0.01 and |1 - 0.999...| < 0.001 and |1 - 0.999...| < 0.0001, etc., thus 1 = 0.999... $\endgroup$
    – Stef
    Commented Oct 17, 2023 at 8:48
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    $\begingroup$ Thanks! Want to incorporate your comments right into your answer? $\endgroup$
    – user95017
    Commented Oct 18, 2023 at 9:29
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    $\begingroup$ I would also use the 0.99999... example and relate it to Zeno's paradox. The arrow does reach its target, therefore the sum of the infinite sequence must be equal to the distance to the target. It's much easier to understand things if you make them concrete, and we KNOW that the arrow reaches its target. $\endgroup$ Commented Oct 18, 2023 at 21:08

I admit that I cannot remember how I thought as a 17 year old, but I am pretty sure my intuitive approach would have been this:

From right to left:

  • Assume $a=b$
  • Hence, $|a-b|=0$
  • As given, we know that $\epsilon>0$.
  • $|a-b| < \epsilon$, qed.

I hope this part is so trivial that it is easy for them. If not, then you probably are fighting with cases of advanced boredom, severe uninterestedness and possibly hormonally induced unstableness in your pupils. </rant>

From left to right:

  • Preparing a proof by contradiction, assume $|a-b| > 0$, i.e. the opposite of the right-hand side.
  • Calculate $|a-b| / 2$, and define $\epsilon$ to be that. Make sure they recognize that this particular choice satisfies $\epsilon>0$, and thus conforms to the problem description.
  • Check whether $|𝑎–𝑏|<\epsilon$ holds (remember, per the task description, this should be true for any and all non-0 $\epsilon$) by plugging in the value of $\epsilon$ from the previous step: $|a-b| < |a-b|/2$. Clearly, this can not be true for any combination of $a$ and $b$.
  • This completes our proof by contradiction; hence $|a-b|=0$.

To sum it up; I used the following concepts, and your 17 year olds would be able to figure this out if they are familiar with them:

  • To prove a $"<=>"$, you can split it in the two "directions" (i.e. $"=>"$ and $"<="$). Divide & conquer.
  • Many proofs use the principle of "proof by contradiction", i.e. assuming the opposite of what you wish to prove, and proving that from that follows something untrue. Obviously, this is already a slightly higher-level concept, if they have difficulty with that, then it's time to draw the good old boolean diagram for "=>" on the board... (And N.B.: an additional pointer would be that proofs of the form "for all ..." often are amenable to proof by contradiction since the opposite is "there is no ...", and then you just assume there is one "..." and use that "..." in the proof. In other words, it's often mechanically straight-forward to do.)
  • "for all $\epsilon$" may seem daunting, but the actual proof in this case doesn't require any kind of set-theoretic construct at all; it just requires to pick instances for the numbers which satisfy the conditions from the problem description. It should not be difficult to see that if you can pick any $\epsilon$ at random, then it holds for all $\epsilon$. If this does provide difficulties, again, you can have them try to prove the opposite, which is "there exists at least one $\epsilon$ for which the opposite holds", and this will quickly lead them to the impossibility of that in a more intuitive manner.

I am glossing over the Axiom of Choice here when asking them to "pick any $\epsilon$" and would reserve that topic for a latter date.

  • 1
    $\begingroup$ This would not work if your student was intuitionist. $\endgroup$ Commented Oct 22, 2023 at 3:27
  • $\begingroup$ @MichaelBächtold I agree. This answer is WAY too formalistic and theoretical. I seek intuition, NOT proofs. $\endgroup$
    – user95017
    Commented Nov 22, 2023 at 4:09
  • $\begingroup$ @user95017, I notice that you added that comment to almost every answer - I appreciate that you give a reason for accepting a different one, but it's not really necessary; There Can Only Be One, anyways, so it's fine. :) That said, the way I interpret the word "intuitive" it would expand to an intuition about how to structure a (possibly formalistic) proof. I had not interpreted the question as to "how can they just 'see' the statement and automatically 'know' (out of thin air) the solution". If that is the normal meaning of "intuitistic" in teaching circles, then I stand corrected. ;) $\endgroup$
    – AnoE
    Commented Nov 22, 2023 at 10:24

Draw a number line and label a point $a$. Then ask them to draw which points $b$ satisfy the condition $|a-b| < \epsilon$ for all $\epsilon$.

  • $\begingroup$ Isn't this the same as Justin Skycak's answer? Should this be a comment? $\endgroup$
    – user95017
    Commented Oct 18, 2023 at 9:22
  • $\begingroup$ @user95017 The answer you link suggests playing an oral game; this answer suggests drawing a number line! The two might seem equivalent to you because you can easily conjure a number line in your head, but visualising mathematics is an issue for most students. $\endgroup$
    – Stef
    Commented Oct 18, 2023 at 10:55
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    $\begingroup$ @Stef Can responder elaborate this too snippety answer? $\endgroup$
    – user95017
    Commented Nov 22, 2023 at 4:11

Can your audience understand that p -> q is equivalent to non q -> non p ?

And that is easy to prove: suppose non q does not imply non p, therefore non q can coexist with p; but p -> q, therefore non q can coexist with q, which is a contradiction.

(this can be easily visualized with Venn diagrams: p->q means that everything that is p is also q, so the circle representing p is included in the circle representing q; non p and non q are everything outside of these circles; so if circle p is included in circle q, the exterior of circle q is included in the exterior of circle p)

Going back to your example:

statement p : "two numbers a and b are different"

implies statement q: "there exists ε>0 such that |a–b| >= ε" (Since a and b are different, therefore their difference differs from 0, the absolute value of that difference is a strictly positive number, so pick epsilon anywhere in the interval (0, |a-b|) )

the negation of statement q : "there is no ε > 0 such that |a-b| >= ε"

which is equivalent to your statement "for every ε > 0, |a-b| < ε" (if a certain property does not hold for ANY epsilon, then the opposite(complementary) property must hold for EVERY epsilon)

non q implies non p : "numbers a and b are not different "

An intuitive explanation of the above: I would use the visualization of the numbers on the real axis, while making an analogy with maps and zoom levels:

statement p: "points A and B are distinct on the map"

implies statement q : "there exists a certain zoom level that would clearly show points A and B apart (separated by a specific distance)" . Of course, not all zoom levels would show this, and there is no need to; also, there is not one single zoom level, but once we found one, any other zoom level more detailed than that would also be ok.

non q: "for any given zoom level, (no matter how detailed), points A and B still appear to overlap" (with the reals numbers axis, we can zoom indefinitely, unlike real world maps)

than students must understand that non p is true: "points A and B are not distinct"

Note : with regard to the fact that the sign "<" magically transformed into an "=": show your students a sequence of strictly negative numbers converging to zero. Here, "strictly less than zero" also turns, in the end, into an "equal to zero".


One more point :

It's counterintuitive and paradoxical for < and = to become equivalent

No. It would have been paradoxical for "a < b" to turn, at some point, into "a = b". But this is not the case. The sign "<" is not applied between a and b, but between |a-b|, on one side, and epsilon, on the other side. Your students should pay attention to this aspect. Comparing the difference of a and b with some third term really tells nothing about the relation between a and b.

Please check this: is there any chance that your students read



a < b, ∀ε>0


  • $\begingroup$ There are two important points of logic here: "p -> q" <=> "not q -> not p" is the first, and "not forall e, p(e)" <=> "exists e, not p(e)" is the second. $\endgroup$
    – Stef
    Commented Oct 17, 2023 at 12:20
  • $\begingroup$ I would be leaving my PC soon, but I was curious about what you tried to highlight. With regard to the formulation of your second point, at first sight, this seems rather incorrect. But perhaps that was exactly what you tried to point out: is there any flow in my proof? If so, could you please indicate where exactly? I have labeled my statements with p, q, not p and not q . Also, I am pretty confident that no where in my proof is there any terms such as "not forall"; I am only dealing with terms such as "there exists", "there is no", "for every" $\endgroup$ Commented Oct 17, 2023 at 12:54
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    $\begingroup$ I did not see a flaw in your proof, no. It's just that your post starts with the question "Does your audience understand that p->q is equivalent to not q -> not p?" and I think it's a great question, but then you mention "statement q: there exists ε>0 such that |a–b| >= ε" as if it was obvious that this was what statement q should be. My point is, it's maybe not obvious to the students, that the negation of "for all ε>0, |a–b| < ε" is "there exists ε>0 such that |a–b| >= ε". $\endgroup$
    – Stef
    Commented Oct 17, 2023 at 16:26
  • $\begingroup$ Thanks, but this is too formalistic and academical. I seek intuition and motivation. Can you please remove your gray blockquotes? Too difficult to read black text on gray background! $\endgroup$
    – user95017
    Commented Oct 18, 2023 at 9:21
  • $\begingroup$ I suggest using italics (text between asterix) instead of the grey background (text between backquotes) $\endgroup$
    – Stef
    Commented Oct 18, 2023 at 10:25

Ultimately I think these things often come down to a matter of phrasing it in a clear manner, more than needing a new pedagogical approach. Especially since they can regurgitate the correct proof (as you say). I might explain it like this.

Suppose $a$ and $b$ are numbers. Then $|a-b|$ is a number. If $a$ and $b$ are the same, then $|a-b|$ is obviously zero. If $a$ and $b$ are different, then $|a-b|$ is, in fact, a positive number. It's not, like, a magically small number, smaller than all other positive numbers, because that doesn't make sense and that doesn't exist. It's just a regular number, and it's positive.

Then obviously it's greater than some positive number, because it's a positive number, and it's bigger than some other positive numbers (actually as the problem is stated, $\varepsilon$ can be $|a-b|$, but that's sort of a distraction pedagogically).

Note that this uses the contrapositive of the troublesome direction; that is, instead of $\forall \varepsilon > 0 |a-b|<\varepsilon \Rightarrow a=b$, I used $a\ne b \Rightarrow \exists \varepsilon>0 |a-b| \geq \varepsilon$, which is certainly easier to think about. If your students also don't understand the contrapositive then maybe this is a good time to spend some time there, because it's really helpful.

I do think this plays into some confusion about the $0.999...=1$ theorem, which is often confusing to students who (a) don't understand and won't accept that decimal expansions are infinite series, (b) don't understand or like series, and (c) end up imagining the difference is something somehow between 1 and every number less than 1.

Concern (c) is really the point of both examples, but in that example you have to deal with confusions about (a) and (b); luckily in your example, you don't.

  • $\begingroup$ This answer is too formalistic and academical. I seek intuition, NOT proofs. $\endgroup$
    – user95017
    Commented Nov 22, 2023 at 4:13
  • $\begingroup$ I suggest adding a comma , and maybe some whitespace between $\forall \varepsilon > 0$ and $|a-b| < \varepsilon$; right now it reads a bit like a single inequality "$\varepsilon > 0|a-b|$" $\endgroup$
    – Stef
    Commented Jan 4 at 14:04

We need to know a couple of properties of the absolute value: that $|x| \geq 0$ and $|x|=0$ if and only if $|x|=0$.

Then the easy part goes like this:

$a=b \implies |a-b|=|0|=0 \implies \forall ε>0, ε>|a–b|$

We see in this case the $>$ in $ε>|a–b|$ actually comes from the $>$ in $ε>0$.

Now going the other way, the chain of implications is as follows:

$\forall ε>0, ε>|a–b| \implies |a-b| \leq 0$

$\implies |a-b|<0 \wedge |a-b|=0 \implies |a-b|=0 \implies a-b=0 \implies a=b$

The first implication is the hard one, and the only part that involves any Real analysis, arising from the Dedekind cut definition of the Reals. For beginners, I'd divide the real number line into two disjoint sets $x \leq 0$ and $x > 0$, and observe that all the the numbers in the first are safe, but all the numbers in the latter eventually get chopped off by some even tinier $ε$.

The second implication follows from the definition of $\leq$, the third from $|x| \geq 0$, the fourth from $|x|=0 \iff |x|=0$, and the fifth by adding $b$ to both sides. This is enough to see that the $>$ is turned into $\leq$ by negation, and then the $<$ part dropped, leaving an $=$.

  • $\begingroup$ This answer is too formalistic and academical. I seek intuition, NOT proofs. $\endgroup$
    – user95017
    Commented Nov 22, 2023 at 4:13

There's two things I have found which help with the intuition of this. The first is to break $|a-b|<\epsilon$ into parts: $a-b<\epsilon$ and $a-b>-\epsilon$. I think the fact that you can "transmute $<$ into $=$" is less astonishing when you have both a $<$ and a $>$ written out explicitly.

The second is a way of thinking about numbers that I found helpful for the famous $0.9999\ldots=1$ argument. In this approach, we ask what $1$ or $0.9$ or $0.99$ means, and start to ask what $0.9999\ldots$ means. The approach I have found successful is one I borrow from how we handle time: say "1", "0.9" or "0.999..." is a "name" for a number, not the number itself. (This is similar to the Peano arithmetic argument that "2" is the name for the number $S(S(0))$, where S is the successor function)

From there, we can discuss two names of real numbers, $a$ and $b$. What must be true about them? What can we say about two named numbers $a$ and $b$ that are equal, knowing the above oddness of $0.999\ldots=1$ above. Its easy to see how this is just a special case of the general version. And after going through those steps, it starts to be more intuitive to see that the original $|a-b|<\epsilon$ handles these odd cases well.

Another approach is to turn it around. Because $|a – b| < ε, \forall ε > 0 \iff a = b$, we can also write the opposite: $\exists ε > 0|a – b| \ge ε \iff a \ne b$. It becomes the definition of the minimum requirements for two named numbers to be different.

  • $\begingroup$ I suggest adding a comma between "0" and "|a-b|" in the last paragraph. Also this kind of "opposite" is sometimes called "contrapositive" $\endgroup$
    – Stef
    Commented Oct 18, 2023 at 10:58

I think there are lots of aspects of the original equivalence that distract from the key point.

  • It's presented as an equivalence; let's just focus on the more conceptually different direction $(\forall \varepsilon>0:|a-b|<\varepsilon) \Longrightarrow a=b$.
  • It's all written in symbols: let's use more natural language: "If $|a-b|<\varepsilon$ for every $\varepsilon>0$, then $a=b$".
  • There are really three steps here: concluding that $|a-b|\leq 0$, then $|a-b|=0$, and finally $a=b$. Again, the first bit is (hopefully) the main difficulty. So let's get rid of this norm business: "If $x<\varepsilon$ for every $\varepsilon>0$, then $x\leq 0$."
  • Even the $<$ is unnecessary: all you really need is "If $x\neq\varepsilon$ for every $\varepsilon>0$, then $x\leq 0$."
  • Now it's fairly straightforward: "I'm thinking of a (real) number. I know it's not equal to any positive number. So it must be either negative or zero."

Once the last line is clear, you can start working upwards. And then you can think about why $\forall \varepsilon>0:x\leq \varepsilon$ would imply $\forall \delta>0:x<\delta$.

  • $\begingroup$ Nice breakdown. I agree with everything, except I'd put a bit of a caveat for "Even the < is unnecessary". With statements that begin with $\forall \varepsilon > 0$ we usually only care about small values of $\varepsilon$. For instance, the apparently-weaker statement $\forall 0 < \varepsilon < 0.01,\, |a-b| < \varepsilon$ is actually equivalent to $\forall \varepsilon > 0,\, |a-b| < \varepsilon$. However, if you want to replace $<$ with $\neq$, then you need to consider all values of $\varepsilon$, not just small values, and this is a big conceptual shift about the role of $\varepsilon$. $\endgroup$
    – Stef
    Commented Oct 18, 2023 at 10:40
  • $\begingroup$ Likewise if the apparently-weaker statement $\forall n \in \mathbb{N},\, |a-b| < 10^{-n}$, is actually equivalent to $\forall \varepsilon > 0,\, |a-b| < \varepsilon$; and of course that doesn't work if you only have $\forall n \in \mathbb{N},\, |a-b| \neq 10^{-n}$. $\endgroup$
    – Stef
    Commented Oct 18, 2023 at 10:42

If for every non-zero $\epsilon$, $a - b$ is not $\epsilon$, then by deduction $a - b = 0$. On the other hand, when $a - b = 0$ it's clearly true that for every non-zero $\epsilon$, $a - b$ is not $\epsilon$, since $a - b$ is $0$ which is not $\epsilon$.

Thus, $\forall \epsilon \neq 0,\, a - b \neq \epsilon \iff a = b$. This is the basic principle.

Take the absolute value and this becomes $\forall \epsilon > 0,\, |a - b| \neq \epsilon \iff a = b$.

$\forall \epsilon > 0,\, |a - b| \neq \epsilon$ and $\forall \epsilon > 0,\, |a - b| < \epsilon$ are equivalent, so the result follows.

  • $\begingroup$ Sorry, but this is too academical and formalistic. I seek intuition and motivation...NOT proofs or arguments. $\endgroup$
    – user95017
    Commented Oct 18, 2023 at 9:20
  • $\begingroup$ Well the intuition here is that you have a dressed up version of the first result. So the idea is to break up the problem into two parts like that. $\endgroup$
    – user23141
    Commented Oct 18, 2023 at 9:27
  • $\begingroup$ I don't think it's too academical; you can rewrite the formulation if you find it too formal. For instance: "We know that |a-b| is a nonnegative real number. We have assumed that there is no positive real number $\varepsilon$ such that $|a-b|$ is $\varepsilon$. Thus |a-b| must be 0 as it's the only choice left for a nonnegative number." $\endgroup$
    – Stef
    Commented Oct 18, 2023 at 10:27
  • $\begingroup$ Fun thing here is that statements that begin with $\forall \varepsilon > 0$ usually only really care about small $\varepsilon$, ie you could write $\forall 0 < \varepsilon < 0.01$ instead of the mighty $\forall \varepsilon > 0$ and it would be an equivalent statement. But with this $|a-b| \neq \varepsilon$ reasoning, you're only using the $\neq$ part of the statement, not the full $<$, and thus you really do need all possible positive values of $\varepsilon$. I said "fun thing" but maybe that's something that might confuse the students even more. $\endgroup$
    – Stef
    Commented Oct 18, 2023 at 10:31

One way to understand something is to think about what it means for it to be false. I'd also argue there is three concepts being connected.

$|a – b| < ε, \forall ε > 0 \iff a = b$

The first bit is

  1. $y < ε, \forall ε > 0 \iff y \leq 0$

and the second is

  1. $0 \leq y \leq 0 \iff y = 0$


  1. $|a - b| = 0 \iff a = b$

and we can break those three down in turn.

For 1, this is a property of the reals. It isn't a given. If ε was a real number, but $y$ was from a space that had a value greater than 0 but less than every real number, then it wouldn't hold. There are extensions of the real numbers that act exactly like that, with a variety of infinitesimals.

In short, it isn't a given that 1 is true. So you have to talk about it being true in the context of the real numbers before you can expect someone to believe it is true. You don't even have to talk about infinitesimals, but you do have to make this fact intuitive to the student.

Once you have that, the second connects $\leq$ and $\geq$ to $=$.

Finally, the 3rd is a property of $||$ - that it is 0 if and only if the elements are the same. This isn't a given, there are spaces where distance doesn't work like that. But it holds in the real numbers.

In effect, we convert "how do I go from $<$ to $=$" into "how do I go from $<$ to $\leq$, and how do I go from $\leq$ and $\geq$ to $=$, and how do I go from distance to elements", which is much more natural.

The hard one here, I think, is step 1. But by removing the need to fold in the next two steps into the intuition for step 1, you can focus on the real problem. And the conversion from $<$ to $\leq$ is far easier to grasp; in effect, the interesting part is that we lose the power of $<$ and get a weaker $\leq$.

I'd start with $y \leq 0$ and show $y < ε, \forall ε > 0$ because that is easy.

For the next part, I'd look at this negation trick:

$!( \exists ε > 0, y \geq ε ) \iff ( y < ε, \forall ε > 0 )$

This exists makes your situation concrete, and implies an adversarial game. Your y has the property that someone cannot find an ε greater than zero that is equal to, or smaller than y.

So we need to show

$!( \exists ε > 0, y \geq ε ) \implies y \leq 0$

And, once you actually describe what this means, it should be really obvious. This is the adversarial game -- y has the property that you cannot find an epsilon greater than zero that is equal or smaller than y.

If they have some maturity, you can also use $(!x \implies !y) \iff (y \implies x)$, but that might be stretching it:

$y > 0 \implies \exists ε > 0, y \geq ε$

which is getting to be a rather trivial claim (set ε equal to y).

  • $\begingroup$ This answer is too formalistic and academical. I seek intuition, NOT proofs. $\endgroup$
    – user95017
    Commented Nov 22, 2023 at 4:12
  • $\begingroup$ @user95017 I mean, I use formulas, but I am not providing a proof. Rather, I am breaking down why it is actually true and thinking about why it might not be true. If we are going to teach intuition we should teach intuition that doesn't mislead! The underlying reason the original problem works relies on those 3 hidden quirks of the question: that there is no real bigger than zero but smaller than every other real bigger than zero, that |a-b|=0 only when a=b, and that |•| is positive. Get those 3 facts intuitively, and the overall fact becomes obvious. $\endgroup$
    – Yakk
    Commented Nov 22, 2023 at 14:06

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