How to read a theorem, prove it, and check your own proof.
7 chapters written · 44 definitions · 51 theorems · start here
The story
Every later unit is written in one language: statements with "for every" and "there exists", and proofs that connect them. This unit teaches that language before it is needed, so that a proof in Unit 1 reads as a sequence of reasons instead of a wall of symbols.
It starts with reading (0.1: what a theorem says, and what its negation says), then doing (0.2: the five ways to prove something), then the objects every proof talks about (0.3 to 0.6: sets, functions, equivalence relations, the integers mod n, complex numbers and polynomials). It ends with writing (0.7): turning scratch work into a proof a grader accepts.
Nothing here depends on calculus. If a later proof feels like a jump, the missing skill is almost always in this unit.
Every theorem in the unit, in the order it is proved. Each one is used by the ones after it. The label says whether it is on the 375/376 exams (T), an Apostol extra (A), or beyond both (B).
Fact · BRules for negating Not(and) is or-of-nots; not(or) is and-of-nots; not(if P then Q) is P and not Q; not(for all) is exists-not; not(exists) is for-all-not.
Fact · BRules for not in Not in an intersection means missing from at least one; not in a union means missing from both; not in a complement means in the original set.
Corollary · BClosure properties of countable sets Subsets of countable sets are countable; the image of a countable set under any function is countable; a product of two countable sets is countable.
Theorem · BDivision algorithm For any integer a and positive integer b there are unique integers q and r with a=bq+r and 0≤r<b.
Proposition · BDivisibility facts Divisors pass through sums and multiples: if a divides b and c, it divides bx+cy for all integers x,y; divisibility chains; and a divisor of a non-zero number is at most as large in size.
Theorem · BEuclid's algorithm Repeat 'divide the larger by the smaller and keep the remainder' until the remainder is zero; the last non-zero remainder is the gcd.
Theorem · BBézout's identity The gcd of a and b is always an integer combination of them: gcd(a,b)=ax+by for some integers x,y. In fact it is the smallest positive number of that form.
Theorem · BThere are exactly n residue classes Every integer is congruent mod n to exactly one of 0,1,…,n−1, namely its remainder; so there are exactly n classes, and a≡b iff a and b have the same remainder.
Corollary · BDivision in Z_p If p is prime, every integer not divisible by p has an inverse mod p; and if gcd(c,n)=1, you may cancel c: ca≡cb implies a≡b (mod n).
Theorem · BLaws of complex arithmetic Addition and multiplication of complex numbers are commutative and associative, multiplication distributes over addition, 1=(1,0) is the identity, and every z=0 has an inverse.
Theorem · BThe n-th roots of unityzn=1 has exactly n complex solutions, ωk=e2πik/n for k=0,1,…,n−1, the corners of a regular n-gon on the unit circle; for n≥2 they add up to 0.
A mixed set of 12 quick checks drawn at random from the 73 in this unit's chapters. Each one says which chapter it comes from, so a miss tells you what to reread. Press "New set" for another mix.
Quick checkWhat is the negation of 'If n is odd, then n2 is odd'?
An if-then statement is false only when the 'if' part holds and the 'then' part fails. So its negation is an AND: some n is odd AND n2 is even. The first option is a different if-then statement, and it is a very common mistake (it was the one miss on CS 240 Exam 1).
Quick checkWhich of these is a statement in the sense just defined?
A statement must be true or false once and for all. The second one is true. The first is a predicate: it is true for x=3 and false for x=4, so it has no truth value until x is fixed or quantified. The third is a request, not a claim.
Your answerTake three statements P,Q,R. Their truth table has 8 rows. In how many rows is (P∨Q)⇒R false?
Fix R false. Then count the pairs (P, Q) that make 'P or Q' true.
An implication is false only when its hypothesis is true and its conclusion is false. So we need R false and P∨Q true. With R false, the pairs (P,Q) that make P∨Q true are TT, TF and FT. That gives 3 rows.
Quick check'You pass the course only if you score at least 40.' Which statement does this say?
'P only if Q' means P implies Q: passing forces a score of at least 40. The first option reverses it (that is 'P if Q', the converse). Scoring 40 may be necessary without being enough.
Your answerAmong the integers 1,2,…,20, for how many n does the converse of 'if n is divisible by 4 then n is even' fail? (It fails at n when n is even but not divisible by 4.)
Count the evens, then subtract the multiples of 4.
There are 10 even numbers in 1 to 20 and 5 multiples of 4 (4, 8, 12, 16, 20). So 10−5=5 even numbers are not multiples of 4: 2, 6, 10, 14, 18. The original statement has no failures at all, but its converse has 5.
Quick checkA theorem says 'If A is invertible, then detA=0.' Which statement is guaranteed to be true as well?
That is the contrapositive: not-conclusion implies not-hypothesis. The first option is the converse and the third is the inverse; neither follows from the theorem alone, since each is equivalent to the other. (Both happen to be true too, but that needs its own proof, in 2.7.)
A 'for every' claim fails as soon as one element fails, and 02=0 is not greater than 0. The third option is wrong because the quantifier ∀x fixes x by covering all values, so the sentence is a statement (a false one).
Your answerLet x and y range over {1,2,3,4,5,6}. For how many values of x is the statement '∀y:xy≤12' true?
A 'for every y' is only as strong as its worst y. Which y is the worst?
The statement must hold for every y, so it must hold for the worst case y=6. We need 6x≤12, so x≤2. The values x=1 and x=2 work, and x=3 fails because 3⋅6=18.
Your answerFor how many integers n from 1 to 10 is the statement 'if n is prime, then n is odd' false?
Look for a prime that is even.
The implication is false only when n is prime and n is not odd. The primes up to 10 are 2, 3, 5, 7, and the only even one is 2. So n=2 is the single counterexample.
Quick checkWhich is the negation of 'Some student passed every exam'?
The statement is 'there is a student s such that for every exam e, s passed e'. Negating: for every student s, there is an exam e that s did not pass. That is 'every student failed at least one exam'. The first option is far stronger (every student failed everything).
Quick checkWhich 'proof' is a valid proof that n2 is even for every even integer n?
The statement is about every even integer, so the proof must start from an arbitrary even n=2k. Checking cases, however many, never covers all of them. Examples can only disprove a 'for every', or prove a 'there exists'.
Quick checkYou want to prove a statement about every integer n by cases. Which choice of cases is valid?
Cases must cover every integer. The first pair misses n=0. The second misses n=7. The third uses the remainder on division by 3, which is always 0, 1 or 2. Cases may overlap, but they may not leave a gap.
Your answerWhat is the smallest positive integer n for which n!>2n?
Compute n! and 2^n side by side for n = 1, 2, 3, 4.
1!=1<2, 2!=2<4, 3!=6<8 and 4!=24>16. This is why the base case in the example above is n=4. After that the step shows the inequality never fails again.
Your answerWhat is the largest amount of rupees that CANNOT be paid with 3-rupee and 5-rupee coins?
List which of 1, 2, ..., 7 can be made as 3a + 5b.
Try 1 to 7: 1, 2, 4 and 7 cannot be made (3 and 5 and 6 can). From 8 on, everything can, by the example above. So the answer is 7. The general formula for coins a,b with no common factor is ab−a−b, here 15−3−5=7.
Quick checkWhich of these is the best fit for STRONG induction rather than ordinary induction?
Ordinary induction suffices when case k+1 is built from case k alone: the first and third statements both are (the third was done above with 7k+1−1=7(7k−1)+6). Factoring k+1=ab needs the cases for a and b, which can be anywhere below k+1.
Your answerTake the first six primes 2,3,5,7,11,13 and form N=2⋅3⋅5⋅7⋅11⋅13+1=30031. What is the smallest prime that divides N?
You do not need all of N's factors. Divide 30031 by primes after 13: 17, 19, 23, 29, ...
N leaves remainder 1 when divided by each of 2, 3, 5, 7, 11, 13, so none of them divides it. Trial division by 17, 19, 23, ... first succeeds at 59, and 30031=59⋅509. So N itself is NOT prime, but its prime factors 59 and 509 are new primes outside the list. That is exactly what the proof needs: a prime factor not on the list.
Quick checkWhich statement is equivalent to x∈/A∖B?
x∈A∖B means x∈A and x∈/B. Negating an 'and' gives an 'or', and negating x∈/B gives x∈B. So x∈/A∖B means x∈/A or x∈B (rule 4). The first option keeps the 'and', and the third forgets to flip x∈/B.
Quick checkTo prove that A⊆B (that A is NOT a subset of B), what must you do?
A⊆B says 'for every x, if x∈A then x∈B'. Its negation is 'there is an x with x∈A and x∈/B', so one such x is enough. The first option is far too strong (it says A and B are disjoint), and the third is about B⊆A.
Quick checkWhich is the complement of A∩B∩C (all inside U)?
x∈/A∩B∩C means the 'and' of three membership statements fails, so at least one fails: x∈/A or x∈/B or x∈/C. Complementing swaps intersection and union. The first option is the complement of the union.
If 2x=2y then x=y, so f is injective. But no odd integer is ever an output, since 2n is even. So 1 is not hit, and f is not onto Z. As a map Z→2Z (the even integers) it would be a bijection.
Quick checkf:A→B and g:B→C are functions, and g∘f is injective. What must be true?
If f(x)=f(y) then g(f(x))=g(f(y)), so x=y because g∘f is injective. That makes f injective. But g need not be: let A={1}, f send 1 to 1 in B={1,2}, and g send both 1 and 2 to the same c. Then g∘f is injective and g is not.
Your answerThere are 12 months in a year. What is the smallest number of people that guarantees two of them have a birthday in the same month?
Find the largest group in which everyone can still have a different month, then add one.
With 12 people, one born in each month, nobody shares a month, so 12 do not guarantee it. With 13 people, a function from people to months cannot be injective by the pigeonhole principle (13>12), so two people share a month.
Quick checkA student 'lists' Z as 0,1,2,3,… and then continues −1,−2,−3,… after all of those. What is wrong?
A listing must give every element a definite counting-number position. The integer −1 would come after infinitely many entries, so there is no n with f(n)=−1 and the list is not onto. The fix is to alternate, as in the next theorem: 0,1,−1,2,−2,… so that each integer has a finite position.
Quick checkYou build a list a1,a2,a3,… of a non-empty set A. Every element of A appears at least once, and some appear several times. What can you conclude?
Part (c) of the theorem: a surjection N→A is enough, because the proof of (c) implies (b) keeps only the first position of each element and so removes the repeats for you. What you can never do is leave an element out. If some element has no position, the list is not onto.
Quick checkBetween any two rational numbers there is another rational (for example their average). So the rationals are 'dense', and there is no next one. Does that make Q uncountable?
A listing only needs every element to get a finite position. It does not need to respect the order of size. The rationals are listed in the order of the grid above, not on the number line. And Z is countable, so the third option has a false premise.
Quick checkIn the diagonal argument, why can we not build y by 'adding 1 to the diagonal digit'?
Two issues. A 9 plus one is not a digit, and, more seriously, if y may contain long runs of 9s or 0s, then y could equal some xn as a number while differing from it in the digit we looked at. Using only the digits 4 and 5 avoids both.
Quick checkOn R, define x∼y iff xy≥0. Which property fails?
xx=x2≥0, so it is reflexive, and xy=yx makes it symmetric. But 1∼0 (since 1⋅0=0≥0) and 0∼−1, while 1⋅(−1)=−1<0, so 1∼−1. The element 0 is related to everything and links the positives to the negatives.
Quick checkOn A={1,2,3} let R={(1,1),(2,2),(1,2),(2,1)}. Which property fails?
Reflexive needs (3,3)∈R and it is missing, so 3∼3. Symmetric holds because (1,2) and (2,1) both appear, and transitive holds because the only chains stay inside {1,2}. Elements that are related to nothing are the usual way reflexivity fails.
Quick checkFor ∼5 on Z (differ by a multiple of 5), which statement is false?
7−2=5 and −3−2=−5 are multiples of 5, so [7] and [−3] are other names for [2]. But 3−2=1 is not a multiple of 5, so [3] is a different class. Five classes in all: [0],[1],[2],[3],[4].
In the first, 2 lies in two blocks (they overlap). In the third, the empty set is not allowed as a block (and 3 is missing). The second covers every element exactly once.
Your answerOn {1,2,…,10}, how many ordered pairs (a,b) are related by 'same remainder on division by 5'?
Find the classes first, then count k2 for a class of size k.
Each remainder 0,1,2,3,4 is shared by exactly two numbers ({5,10}, {1,6}, and so on), so there are 5 classes of size 2. Each contributes 22=4 ordered pairs, and 5⋅4=20.
Your answerHow many equivalence relations on {1,2,3,4} have exactly two classes?
List the possible block sizes for two blocks, then count the ways to choose the blocks for each.
Two blocks means block sizes 3 + 1 or 2 + 2. For 3 + 1, choose the single element: 4 ways. For 2 + 2, split into two pairs: {12∣34}, {13∣24}, {14∣23}, 3 ways. Total 4+3=7.
Quick checkDefine g:Q→Z by g(ba)=a+b. Is g well defined?
A rule on classes must give the same answer for every representative. 21 and 42 are the same rational number, but the rule sends one to 3 and the other to 6. Being an integer is not the issue; consistency across representatives is.
Quick checkWhat is the remainder when −7 is divided by 3?
In this chapter remainders are always between 0 and n−1. We write −7=3⋅(−3)+2, so the quotient is −3 and the remainder is 2. The answer −1 comes from −7=3⋅(−2)−1, which has a negative remainder and is not allowed.
Quick checkWhich of these integers can be written as 12x+18y with integer x,y?
gcd(12,18)=6. Every combination 12x+18y is a multiple of 6, because 6 divides both terms. Conversely every multiple of 6 is a combination: 6=12⋅(−1)+18⋅1, and scaling gives 30=12⋅(−5)+18⋅5. Neither 5 nor 7 is a multiple of 6.
Quick check6∣4⋅9=36, but 6∤4 and 6∤9. Why does this not contradict Euclid's lemma?
The lemma is about a PRIME divisor. 6=2⋅3 can split its prime factors between the two numbers (2∣4 and 3∣9), so it divides the product without dividing either factor. A prime cannot be split.
Your answerHow many of the classes [0],[1],…,[35] in Z36 have an inverse mod 36?
Use the theorem: invertible iff coprime to 36=22⋅32.
[k] has an inverse iff gcd(k,36)=1, that is iff k is divisible by neither 2 nor 3. Among 0,…,35, half are odd (18), and of those a third are multiples of 3 (6 of them), leaving 18−6=12: the numbers 1,5,7,11,13,17,19,23,25,29,31,35.
Quick checkIn Z6 you are told 2x≡2y(mod6). What can you conclude?
gcd(2,6)=2=1, so 2 has no inverse mod 6 and cannot be cancelled freely. Writing 2x−2y=6k gives x−y=3k, so x≡y(mod3) and that is all you get. For example 2⋅1≡2⋅4(mod6) but 1≡4(mod6).
Your answerTake the RSA primes p=7 and q=11 with public exponent e=7. What is the private exponent d, as the smallest positive number with ed≡1(mod(p−1)(q−1))?
First compute (p−1)(q−1), then find the inverse of 7 modulo that number with the extended Euclid.
(p−1)(q−1)=6⋅10=60. We need 7d≡1(mod60). Euclid: 60=8⋅7+4, 7=1⋅4+3, 4=1⋅3+1, so 1=4−3=4−(7−4)=2⋅4−7=2(60−8⋅7)−7=2⋅60−17⋅7. So d≡−17≡43. Check: 7⋅43=301=5⋅60+1.
Quick checkIn the pair definition of C, why is i2=−1?
Nothing about i is assumed. The product was chosen first, and i2=−1 falls out of it. Defining i as −1 would be circular, since we have not yet said that −1 has a square root.
By the theorem, ∣1+i∣=2 and ∣2−2i∣=22, and the moduli multiply: 2⋅22=4. Check: (1+i)(2−2i)=2−2i+2i−2i2=4. The option 32 adds the moduli instead of multiplying them.
Your answerGive the argument of −1−i in radians, in the range from −π to π.
Decide the quadrant first.
The point (−1,−1) is in the third quadrant, with cosθ=sinθ=−21. The angle in (−π,π] is −43π (equivalently 45π). The calculator value tan−1(1)=4π is the first-quadrant mistake.
Quick checkHow many different complex roots can a polynomial of degree 3 have?
The theorem gives an upper bound only. (x−1)3 has degree 3 and only one distinct root. The fundamental theorem (below) says 3 roots exist when you count repeats.
Quick checkWhich is the best first sentence of a proof that 2n≥n+1 for all n≥1?
Naming the technique in the first line tells the reader what structure to expect (a base case, then a step). The first option uses 'obviously' in place of a check. The third assumes the conclusion, which is the circular mistake from 0.2.
Quick checkA proof says: 'Let ε>0. Set δ=ε/2.' Which letter is arbitrary and which is constructed?
'Let ε>0' hands the choice to the reader, so the proof must work for every positive ε. 'Set δ=ε/2' is the writer's own construction, allowed to depend on ε because it comes later.
Quick checkIn a 2.1-style proof that a set of matrices is a vector space, which may you use without proof?
Facts about real numbers come from earlier 'bricks'. The second is a theorem about vector spaces you must cite (and it holds once you know it is a vector space). The third is exactly what you are being asked to check.
Your answerScratch work on the claim 'if f and g are increasing then fg is increasing': take f(x)=g(x)=x and compare the points −2 and −1. Compute (fg)(−2)−(fg)(−1).
(fg)(x) = x times x.
(fg)(x)=x2, so (fg)(−2)=4 and (fg)(−1)=1. The difference is 3>0, so fg is larger at the smaller input −2, and fg is not increasing. One computed example is enough to disprove a 'for every' claim.
Quick checkWhich word should never appear as a reason in a proof?
'Since' and 'hence' connect steps and point to reasons. 'Obviously' replaces a reason. If a step is obvious, a short reason costs nothing; if it is not, the word hides a gap.