Speak Maths · University
Shorter, not vaguer
Rewriting a paragraph in notation can lose nothing, or can lose everything that mattered. A good compact statement keeps the assumptions, the scope and the conditions, and keeps the quantifiers in the right order.
Draft: written by an AI helper and not yet checked by a teacher.
The alphabet
Read each sign aloud, the way it is written under it.
- Say: for every positive epsilonMeans: The condition “greater than zero” is part of the quantifier. Drop it and epsilon could be zero or negative, and the sentence changes.
- Say: there is a positive deltaMeans: Existence with its condition. The delta is chosen after the epsilon, and may be as small as needed, but never zero.
- Say: for every x there is a yMeans: The order of the quantifiers is part of the meaning. Swapped, the same letters say something much stronger: one y for all x.
- Say: if H then CMeans: Hypothesis and conclusion. A compact statement may shorten the wording but must keep every hypothesis: dropping one makes the claim false or different.
- Say: there is exactly one xMeans: Existence and uniqueness together. “Exactly one” is stronger than “at least one”, and dropping the word is a classic way of saying more or less than you meant.
- Say: A and BMeans: Both conditions at once. Each extra condition narrows the claim; never merge two conditions into one, or drop one, without saying so.
- Say: it is not true that for every x, P of xMeans: The denial of a whole statement. Pass the not through: every ∀ becomes ∃, every ∃ becomes ∀, and an implication becomes “and … fails”.
- Say: delta depends on epsilon and on aMeans: Shows what the chosen quantity may depend on. Everything in the brackets is allowed to change the delta; nothing else is.
The story: One paragraph, shortened with nothing lost
Tap a line to see its words.
- The paragraph: “when x is near a, f of x is near f of a”. Short, but how near?
- Say how near: inputs within delta, outputs within epsilon.
- A delta must exist, and it must be positive. Zero would say nothing.
- However small the epsilon asked for: epsilon first, then delta. The order is the heart.
- The delta may depend on epsilon and on the point a, because it is chosen after them.
- And x ranges over the domain D of f. Leave this out, and x could be a number where f does not exist.
- The whole paragraph in one line, with nothing lost: f is continuous at a.
- Its denial: every quantifier flips, and the arrow becomes “and the outputs are still far”.
- Check with a case: for f of x equal to two x, delta equal to epsilon over two works.
- And the limit of two x at one exists, is unique, and is two.
In words this story is 791 letters long; in symbols, 199.
Read it
What does it say?
Write it
Which notation says this?
However small a positive epsilon is, there is a positive delta.
One positive delta works for every positive epsilon.
If x is at least one, then x squared is at least x.
There is exactly one real x with x cubed equal to eight.
If both a and b are positive, then ab is positive.
Not every real x has x squared less than ten.
Delta depends on epsilon and on the point a.
There is one bound C that holds for all the terms a n.
For every positive epsilon, from some N on every term is within epsilon of L.
If x is strictly between zero and one, then x squared is smaller than x.
To deny it: some epsilon beats every delta.
For all positive a and b, the mean is at least the square root of the product.
x squared is four exactly when x is two or minus two.
Build it
Tap the tiles in order. One tile does not belong.
For every positive epsilon there is a positive delta.
If a and b are positive, their product is positive.
There is exactly one x with x plus two equal to five.
Not every x satisfies P.
Delta depends on epsilon and on a.
There is one bound C for all the terms.
x squared is four exactly when x is two or minus two.
Challenges
Creative puzzles that use every sign of this chapter. Take your time; there is no clock.
1 Smaller is safer
For f(x) = 2x, a delta works for an epsilon when delta is at most epsilon / 2. Rows are epsilons, columns are deltas. Judge each line.
| ✓ | ✓ | ||
| ✓ | ✓ | ||
| ✓ |
A tick W(ε, δ) means: this delta works for this epsilon.
The smallest delta serves all three epsilons here, but in the real definition the epsilons never stop getting smaller, so delta must follow epsilon: δ = ε / 2. In this small world both orders happen to be true.
2 A bound that is missing
The terms are a_n = n. B(n, C) means: the term n is at most the bound C. Judge each line; notice the condition n ≤ 3.
| ✓ | ✓ | ✓ | |
| ✓ | ✓ | ||
| ✓ | |||
A tick means: the row term is at most the column bound.
Drop “n ≤ 3” and the truth flips: the fourth term has no bound among 1, 2, 3. The condition is part of the statement.
3 Exactly one
Four objects, three properties. Judge each line; “exactly one” is stricter than “at least one”.
| ✓ | |||
| ✓ | ✓ | ||
| ✓ | |||
A tick means: the row object has the column property.
P_2 holds for two objects, so “exactly one” fails while “there is one” holds. The fourth object has no property at all, so “everything has one of them” is false.
4 Four orders
Pair each line with what it says.
Tap a line, then its meaning.
The same two letters and the same two quantifier signs in different orders give four different claims. This is the heart of every “for every ε there is δ” passage.
5 What happens when a condition is dropped
Pair each line with the verdict.
Tap a line, then its meaning.
The first line is a theorem; drop its hypothesis and the second line is false. The same numbers, a different claim.
6 Four denials
Pair each line with what it says.
Tap a line, then its meaning.
To deny an implication, keep the hypothesis and break the conclusion. To deny “and”, one failure is enough.
7 What may depend on what
Pair each line with what it says.
Tap a line, then its meaning.
Brackets after a letter list everything it may depend on; a quantifier order says the same thing in another way.
8 Each epsilon, some delta
This should say: for every epsilon there is a delta that works. One tile is wrong.
Tap the tile that is wrong.
With “for every delta” the line would demand that every delta works, which no function achieves.
9 If, not and
This should say: if x is at least one, then x squared is at least x. One tile is wrong.
Tap the tile that is wrong.
With “and” the line claims both facts about every x. With the arrow, the second follows from the first.
10 Exactly one
This should say: there is exactly one x with x plus two equal to five. One tile is wrong.
Tap the tile that is wrong.
“There is an x” does not say how many. The exclamation mark adds exactly one.
11 Pass the not through
This should say: not every x has P exactly when some x lacks P. One tile is wrong.
Tap the tile that is wrong.
When the not passes through a “for every”, the quantifier becomes “there is”. “For every x, not P” is much stronger.
12 The largest delta
What is the largest delta for which this holds for every x?
|2x − 6| < 0.5 means |x − 3| < 0.25. So delta = 0.25 is the largest; 0.5 or 1 let some x break it.
13 Deny it
Which line is the denial of the statement?
Flip every quantifier (for every ↔ there is) and deny what is inside, keeping the order.
14 The one and only
Find the only n in the set that makes the equation true.
n² − 3n − 4 = (n − 4)(n + 1) = 0, so n = 4 or n = −1; only 4 lies in the set.
15 Three conditions at once
How many whole numbers satisfy all three conditions?
Positive with square under 20: 1, 2, 3, 4. Of these, the even ones are 2 and 4: two numbers.
16 Break the arrow
Which x is a counterexample to the statement?
A counterexample makes the hypothesis true and the conclusion false: for x = −3, x² = 9 > 4 but −3 is not above 2.
17 The paragraph, shortened
f is continuous at a if, however small a positive epsilon is chosen, there is a positive delta so that every x in the domain D within delta of a has f of x within epsilon of f of a. Write it with nothing lost.
One way:
Check the list: both conditions > 0, the order ε then δ, the domain D, and the arrow from the hypothesis on x to the conclusion on f(x).
18 One bound for all
The sequence a is bounded: one positive number C is at least the size of every term. Write it.
One way:
Putting “there is C” first makes C independent of n. With n first, every list of reals would qualify.
19 One solution
The equation two x plus one equals seven has exactly one real solution. Write it.
One way:
“Exactly one” is the sign ∃!. (The solution is x = 3.)
20 Shorten, keep everything
Whenever a and b are positive and a is less than b, the square of a is less than the square of b. Write it compactly without dropping a condition.
One way:
Both positivity and a < b are hypotheses. Without positivity it is false: a = −3 and b = 2 give 9 < 4.
21 A delta that follows epsilon
For f of x equal to three x, delta can be epsilon over three, depending on epsilon only. Write it.
One way:
|3x − 3a| = 3|x − a| < ε exactly when |x − a| < ε/3. The delta needs no help from a.
22 Not every number
Not every natural number is even. Write it, then write it again without the leading not.
One way:
The second form is the denial passed through: some natural number is not divisible by two (for example 3).
Chapter finished. The symbols are yours.
Tell your own
Take a sentence from a paper you know, shorten it into notation, and then list what you kept: each assumption, each condition, the order of the quantifiers. If any is missing, put it back.
Kept on this device only.