To 17 year olds, how can I explain that two numbers with arbitrarily small difference are equal?

Question

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

Answer 1

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 $b = a.$

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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 $b.$

Answer 2

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.

Answer 3

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.

Answer 4

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 $0.999... = 1$.

Answer 5

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.

Answer 6

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 $\epsilon$.

Answer 7

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".

(EDIT)

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,∀ε>0

as

a < b, ∀ε>0

?

Answer 8

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.

Answer 9

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

2. $0 \leq y \leq 0 \iff y = 0$

finally

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

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).

Answer 10

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 $=$.

Answer 11

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.

Answer 12

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 $\forall \delta>0:x<\delta$.

Answer 13

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.