Download Week 7: School Mathematics (Before Calculus)

Week 7: School Mathematics (Before Calculus)
The standard curriculum in the U.S. is a death march to calculus… This is a
shame on a lot of levels.
Richard Rusczyk
Two Theorems from Elementary School
Typically, elementary school students learn many facts about numbers, few of which are
proved, at that time, in any way. Some arithmetic facts are not proved because they seem
intuitively obvious. Some facts are not proved because the proof would be too deep. These
two rationales for omitting proofs in the elementary school setting are not mutually
exclusive. We now consider a couple of facts that nearly all elementary school students
encounter in some form.
The first theorem is an extension of the fact that every integer is either even or odd – in
other words, any integer 𝑥 can be written as either 2𝑚 or 2𝑚 + 1 for some integer 𝑚. A
similar fact is that any integer 𝑥 can be written as either 3𝑚 or 3𝑚 + 1 or 3𝑚 + 2 for some
integer 𝑚. Likewise, any integer 𝑥 can be written as either 4𝑚 or 4𝑚 + 1 of 4𝑚 + 2 or
4𝑚 + 3 for some integer 𝑚. This pattern continues; here is the general fact.
Arithmetic Theorem 1: For any integer 𝑥 and natural number 𝑛, there is a unique integer
𝑚 and a unique natural number 𝑏, less than 𝑛, so that 𝑥 = 𝑛 ⋅ 𝑚 + 𝑏.
It might help to state this theorem in symbolic form:
(∀𝑛 ∈ ℕ)(∀𝑥 ∈ ℤ)(∃! 𝑚 ∈ ℤ)(∃! 𝑏 ∈ ℕ)(𝑏 < 𝑛 ∧ 𝑥 = 𝑛 ⋅ 𝑚 + 𝑏)
We will not prove this theorem immediately, but will do so in time.
The next theorem concerns the factoring of whole numbers. Given a whole number 𝑛, a
prime factorization of 𝑛 is a sequence of prime numbers, the product of which is equal to 𝑛.
For example, we say that a prime factorization of 504 is 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 7. We can also say
that 504 is 2 ⋅ 3 ⋅ 2 ⋅ 3 ⋅ 2 ⋅ 7, but this is considered to be just the same factorization (albeit
in a different order). If a number is prime, we say that the number is its own prime
factorization.
Arithmetic Theorem 2: Every natural number greater than 1 has a unique prime
factorization.
Proof of existence: Note that 2 is its own prime factorization. Likewise, 3 is its own prime
factorization. A prime factorization of 4 is 2 ⋅ 2. Next, 5 is its own prime factorization, and 6
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has a prime factorization 2 ⋅ 3. Suppose we know that all of the natural numbers, from 2 up
to some number 𝑛, have prime factorizations. Consider the next natural number, 𝑛 + 1. By
definition, this number is either prime or composite.
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
If 𝑛 + 1 is prime, then it is its own prime factorization.
If 𝑛 + 1 is composite, then by definition, 𝑛 + 1 is the product of two natural numbers
𝑘 and 𝑚, with 1 < 𝑘 < 𝑛 + 1 and 1 < 𝑚 < 𝑛 + 1. We already know that each of
these numbers can be written as the product of a sequence of primes, so 𝑛 + 1,
which is 𝑘 ⋅ 𝑚, can be written as the product of these two products, which gives us a
prime factorization of 𝑛.
By induction, every natural number greater than 1 has a prime factorization. ∎
The uniqueness of a natural number’s prime factorization can be proved with the help of
Arithmetic Theorem 1, but this proof is slightly too complicated to delve into its details in
this course.
Exercise 1: Without referring to Arithmetic Theorems 1 or 2, prove that any natural
number 𝑥 can be written as either 3𝑚 or 3𝑚 + 1 or 3𝑚 + 2 for some integer 𝑚.
Exercise 2: Using the previous exercise (but again without referring to Arithmetic
Theorems 1 or 2), prove that any integer 𝑥 can be written as either 3𝑚 or 3𝑚 + 1 or 3𝑚 +
2 for some integer 𝑚.
Exercise 3: Without referring to Arithmetic Theorems 1 or 2, prove that no integer can be
written in more than one of three forms from the previous two exercises (3𝑚 or 3𝑚 + 1 or
3𝑚 + 2).
Exercise 4: Give a rough outline of how we would prove that any natural number’s prime
factorization is unique.
Exercise 5: Find a counterexample to the following statement: For all natural numbers 𝑎, 𝑏,
and 𝑐, if 𝑏 ⋅ 𝑐 is a multiple of 𝑎, then either 𝑏 or 𝑐 must be a multiple of 𝑎.
Using the Theorems
Arithmetic Theorems 1 and 2 are fundamental building blocks for proving many arithmetic
facts. Those two theorems also provide guarantees for the effectiveness of certain
mathematical techniques learned in elementary school and beyond.
The proof of Arithmetic Theorem 1 will be dealt with later; for now, let’s take Arithmetic
Theorems 1 and 2 for granted and see how they can be used in proofs.
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Proposition: If 𝑛 is an integer that is not a multiple of 5, then 𝑛2 can be written as either
5𝑘 + 1 or 5𝑘 + 4 for some integer 𝑘.
Proof: Suppose 𝑛 is an integer that is not a multiple of 5. By Arithmetic Theorem 1, there
are four possible cases –

If 𝑛 = 5𝑚 + 1 for some integer 𝑚, then 𝑛2 = 25𝑚2 + 10𝑚 + 1 = 5(5𝑚2 + 2𝑚) + 1.

If 𝑛 = 5𝑚 + 2 for some integer 𝑚, then 𝑛2 = 25𝑚2 + 20𝑚 + 4 = 5(5𝑚2 + 4𝑚) + 4.

If 𝑛 = 5𝑚 + 3 for some integer 𝑚, then
𝑛2 = 25𝑚2 + 30𝑚 + 9 = 5(5𝑚2 + 6𝑚 + 1) + 4.

If 𝑛 = 5𝑚 + 4 for some integer 𝑚, then
𝑛2 = 25𝑚2 + 40𝑚 + 16 = 5(5𝑚2 + 8𝑚 + 3) + 1.
In each case, 𝑛2 is either 5𝑘 + 1 or 5𝑘 + 4 for some integer 𝑘. ∎
The next proposition is a somewhat surprising fact about prime numbers, which mostly do
not follow any convenient patterns.
Proposition: When we square any prime number greater than 3, the result is always a
multiple of 12, plus 1.
Proof: Let 𝑥 be a prime number greater than 3. By Arithmetic Theorem 1, there is an
integer 𝑚 such that 𝑥 = 6𝑚 + 𝑏, and 𝑏 is either 0, 1, 2, 3, 4, or 5. Now,

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we know that 𝑥 cannot be 6𝑚, because 𝑥 is prime.
if 𝑥 = 6𝑚 + 1, then 𝑥 2 = 36𝑚2 + 12𝑚 + 1 = 12(3𝑚2 + 𝑚) + 1, which is a multiple
of 12, plus 1.
𝑥 cannot be 6𝑚 + 2, because this is 2(3𝑚 + 1), which is not prime.
𝑥 cannot be 6𝑚 + 3, because this is 3(2𝑚 + 1), which is not prime.
𝑥 cannot be 6𝑚 + 4, because this is 2(3𝑚 + 2), which is not prime.
if 𝑥 = 6𝑚 + 5, then 𝑥 2 = 36𝑚2 + 60𝑚 + 25 = 12(3𝑚2 + 5𝑚 + 2) + 1, which is a
multiple of 12, plus 1.
In the only possible cases, it turns out that 𝑥 2 is always a multiple of 12, plus 1. ∎
In trying to prove the proposition above, because the statement is related to multiples of
12, it would be a natural first instinct to write a proof with 12 cases: 𝑥 = 12𝑚 + 𝑏, with 𝑏 as
either 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, or 11. In fact, the proof can be written this way
successfully, and it would not be a bad idea to try writing that version of the proof yourself.
It requires a little extra insight to see that a shorter proof can be written successfully with
only 6 cases.
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A proof of this form, involving several cases, is often called a “proof by exhaustion.”
Proposition: There are infinitely many prime numbers.
Proof: Suppose there are a finite number of primes. Let 𝑁 be the product of all these prime
numbers. Consider the natural number 𝑁 + 1, which is clearly larger than 1. For any prime
number 𝑝, this number 𝑁 + 1 can be written as 𝑝 ⋅ (product of the other primes) + 1, which
means that 𝑁 + 1 cannot be a multiple of 𝑝, because of the uniqueness part of Theorem 1.
But if 𝑁 + 1 is not a multiple of any of the prime numbers, it contradicts Theorem 2. So
there are infinitely many prime numbers. ∎
Proposition: If the product of two natural numbers is a multiple of a prime, then one of
those two natural numbers must be a multiple of that prime.
Proof: Let 𝑎 be prime and 𝑏 and 𝑐 be natural numbers, and suppose 𝑏 ⋅ 𝑐 is a multiple of 𝑎.
This means that 𝑏 ⋅ 𝑐 = 𝑚 ⋅ 𝑎 for some natural number 𝑚. We consider three cases: 𝑚 = 0,
𝑚 = 1, and 𝑚 > 1.



If 𝑚 = 0, then 𝑏 ⋅ 𝑐 = 0, so either 𝑏 or 𝑐 must be 0 and therefore a multiple of 𝑎.
If 𝑚 = 1, then 𝑏 ⋅ 𝑐 = 𝑎, so since 𝑎 is prime, either 𝑏 or 𝑐 must be equal to 1 and the
other must be equal to 𝑎.
If 𝑚 > 1, then by Theorem 2, 𝑚 can be expressed as a product of prime factors.
Therefore 𝑎 ⋅ 𝑚 is the product of 𝑎 and these factors. So since 𝑏 ⋅ 𝑐 = 𝑎 ⋅ 𝑚, the
prime number 𝑎 is one of the unique prime factors of 𝑏 ⋅ 𝑐. Now, 𝑏 and 𝑐 can each be
written as products of prime factors, so 𝑏 ⋅ 𝑐 is the product of every prime factor of
either 𝑏 or 𝑐. Since the prime factorization of 𝑏 ⋅ 𝑐 is unique, 𝑎 must be one of these
prime factors of either 𝑏 or 𝑐.
In each case, either 𝑏 or 𝑐 is a multiple of 𝑎. ∎
Exercise 6: Prove that for any integer 𝑥, if 𝑥 2 is a multiple of 3, then 𝑥 is a multiple of 3.
Exercise 7: Use the previous exercise to prove that √3 is irrational.
Exercise 8: Prove that if 𝑥 is an integer greater than 5 and 𝑥 − 1 and 𝑥 + 1 are both prime,
then 𝑥 is a multiple of 6.
Exercise 9: Prove that no square number ends with the digit 2.
Exercise 10: Use Arithmetic Theorem 2 to prove that if 𝑝 is prime, then √𝑝 is irrational.
Exercise 11: Prove Arithmetic Theorem 1. (Hint: the proof will be like a combination of
Exercises 1–3, but more general.)
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Some Properties of Functions
From basic algebra through the most modern ideas of mathematics, functions play a
steadily increasing role. Up to at least the early years of college-level mathematics,
essentially all functions are concerned only with real numbers – that is, a typical function
takes real numbers as “input” and gives real numbers as “output.” But in general, a function
can have any kind of input and any kind of output.
The domain of a function is the set of all objects that are allowed to be “plugged into” the
function. A codomain of a function is a set containing all objects that might be “output” by
the function.
Example: The function given by the formula 𝑟(𝑥) = 5 − √2 + 𝑥 , according to the usual
understanding of an algebra student, has domain [−2, ∞). The interval (−∞, 5] is a
codomain of this function.
Note that any function with codomain ℝ is known as a real-valued function. These are the
functions dealt with throughout grade school and most of college.
There is a special notation, used extremely rarely in early mathematics courses but
increasingly useful for more modern areas, that is used to designate a function along with
its domain and codomain.
Definition: We say that 𝑓 is a function from set 𝐴 to set 𝐵 iff 𝐴 is the domain of 𝑓 and 𝐵 is a
codomain of 𝑓. This is indicated by the notation 𝑓: 𝐴 → 𝐵.
As an example, for the function 𝑟 mentioned above, we can say that 𝑟 ∶ [−2, ∞) → (−∞, 5].
Definition: A function 𝑓: 𝐴 → 𝐵 is one-to-one iff
(∀𝑎 ∈ 𝐴)(∀𝑏 ∈ 𝐴)(𝑎 ≠ 𝑏 → 𝑓(𝑎) ≠ 𝑓(𝑏)).
Definition: A function 𝑓: 𝐴 → 𝐵 is onto 𝐵 iff
(∀𝑏 ∈ 𝐵)(∃𝑎 ∈ 𝐴)(𝑓(𝑎) = 𝑏).
Definition: A function 𝑓: 𝐴 → ℝ, with 𝐴 ⊆ ℝ, is increasing iff
(∀𝑎 ∈ 𝐴)(∀𝑏 ∈ 𝐴)(𝑎 < 𝑏 → 𝑓(𝑎) < 𝑓(𝑏)).
Definition: A function 𝑓: 𝐴 → ℝ, with 𝐴 ⊆ ℝ, is decreasing iff
(∀𝑎 ∈ 𝐴)(∀𝑏 ∈ 𝐴)(𝑎 < 𝑏 → 𝑓(𝑎) > 𝑓(𝑏)).
Exercise 12: Prove that the function 𝑟, given before, is one-to-one.
Exercise 13: Prove that the function 𝑟, given before, is onto (−∞, 5].
Exercise 14: Prove that the function 𝑟, given before, is decreasing.
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Exercise 15: Prove that the typical algebra function given by the formula 𝑓(𝑥) = 𝑥 2 + 5
(a) … is not one-to-one.
(b) … is not onto ℝ.
(c) … is not increasing.
(d) … is not decreasing.
Exercise 16: Prove that if 𝑔 ∶ ℝ → ℝ is increasing and ℎ ∶ 𝐴 → ℝ is decreasing, then the
function given by 𝑔(ℎ(𝑥)) is decreasing and one-to-one.