Maxima and Minima of Continuous Functions

For a given continuous function on a closed interval, if there is a point in this closed interval where the derivative of f(x) = 0, then this point is either the low point (minima) or high point (maxima) for the continuous function in this interval.

This very basic idea of calculus requires a few lemmas before we are able to prove it.

If you are not familiar with the concept of a function, continuous function, or a closed interval, start here.

A function is said to be bounded if there is a value L such that for all x ∈ [a,b], f(x) ≤ L.

Lemma 1: Nested Interval Property for the real numbers

Suppose that I₁, I₂, ..., I_n is a sequence of nested, closed intervals where:
(a) Each I_i+1 is contained with I_i
(b) Each I_i interval is of the form [a_i,b_i]
(c) The lim (i → inf) (b_i - a_i) = 0 (See here for review of lim notation and the concept of limit)
Then:
There exists 1 and only 1 point c such that {c} = I₁ ∩ I₂ ∩ ... ∩ I_n

Proof:

(1) We know that there can be at most 1 number since lim(i → inf) (b_i - a_i) = 0 since:

If there were more than 1 number, the lim(i → inf) (b_i - a_i) would be greater than 0.

The only way that it can be 0 is if b_i = c, a_i = c, and c-c=0.

(2) We know that there is at least 1 number that is common to all intervals since:

(a) a_i has a limit a_n that fits somewhere in the interval. For all intervals, it is clear that a_i ≤ a_n ≤ b_i

(b) Likewise, b_i has a limit b_n such that a_i ≤ b_n ≤ b_i

(c) So we see that a_n,b_n are both elements of all intervals.

(d) We further note that a_n = b_n since the lim(i → inf)(b_i - a_i)= 0.

(e) So, if we let c = a_n = b_n, then we are done.

QED

Lemma 2: If a function f is continuous on a closed interval [a,b], then f is bounded there.

(1) Assume that a function f is not bounded on [a,b]

(2) We can bisect the interval [a,b] into two halves which I will label I₁ and I₂.

(3) We can now pick an interval I which is unbounded. If both intervals are unbounded, then we can pick either one.

(4) We can repeat this process and create a sequence of nested, closed intervals which we can call I_i where each interval selects a subset which is unbounded.

(5) From Lemma 1, we know that there exists a point c which is common to all the intervals in #4.

(6) Because f is continuous, we know that there is a number ε such that f is bounded on the interval c - ε and c + ε [See here for definition of Continuous Functions]

(7) But one of the unbounded values in I_n must lie within (c - ε, c + ε)

(8) And this is a contradiction since from (#6), it must be bounded.

(9) Therefore, we reject our assumption.

QED

Lemma 3: Maximum value property of continuous functions

If a function f is continuous on the closed interval [a,b], then there exists a number c in [a,b] such that f(x) ≤ f(c) for all x in [a,b]

(1) Let I be the the closed interval [a,b]

(2) From Lemma 2, we know that I is bounded.

(3) Let λ be its least upper bound.

(4) We can divide I in half.

(5) At least one of these halves will have a least upper bound = λ (although it is possible that both have this least upper bound). Let I₁ be the division of I which contains λ as the least upper bound.

(6) We can keep dividing up I₁ in the same way until we have I_n which has λ as its upper bound and b_n - a_n = 0.

(7) From Lemma 1, we know that there exists a point c which is common to all these intervals.

(8) It follows from (6) that f(c)=λ since:

(a) There exists a positive value δ such that if x-c is in between -δ and +δ, then f(x)-f(c) is between -ε and ε. [From the definition of a continuous function, see here]

(b) From (a), we have that f(c) - ε is less than f(x) which is less than f(c) + ε

(c) Since ε can be arbitrarily small, we can have f(c) ≤ f(x) ≤ f(c) which means that f(x)=f(c) at some point.

(d) In this case, f(c) cannot be more than λ since λ is an upper bound. [See here for the definition of an upper bound]

(e) Likewise, f(c) cannot be less than λ since λ is the least upper bound. [See here for the definition of a least upper bound]

(f) Therefore, f(c) = λ

QED

Definition 1: Right Hand Limit: lim (x → a+) f(x)

lim (x → a+) f(x) = L if and only if:

if x is between a and a + δ, then f(x) - f(a) is between -ε and ε

Definition 2: Left Hand Limit: lim(x → a-) f(x)

lim(x → a-) f(x) = L if and only if:

if x is between a - δ and a, then f(x) - f(a) is between -ε and ε

Lemma 4: One-sided and two-sided limits

The limit lim (x → a) for f(x) exists and is equal to the number L if and only if the one-sided liimits lim (x → a+) f(x) and lim (x → a-) f(x) both exist and both are equal to the number L.

Proof:

(1) Assume lim(x → a) f(x) = L

(2) Then if x - a is between δ and - δ, then f(x) - f(a) is between -ε and +ε [ By the definition of continuous functions, see here]

(3) Now x - a is between δ and -δ implies that:
x is between a - δ and a + δ.

(4) Since a is greater than a - δ (since δ is a positive value), this implies that:
if x is between a and a + δ, then x is necessarily between a - δ and a + δ.

(5) (#4) combined with (#2) gives us:
lim (x → a+) f(x) = L. [See definition 1 above]

(6) Since a is less than a + δ, this implies that:
if x is between a - δ and a, then x is between a - δ and a + δ.

(7) (#6) combined with (#2) gives us:
lim (x → a-) f(x)= L. [See definition 2 above]

(8) So, that proves the first half of the above theorem.

(9) Assume that lim (x → a+) f(x) = L and lim(x → a-) f(x) = L

(10) From definition 1, we have if x is between a and a + δ, then f(x) - f(a) is between -ε and ε

(11) From definition 2, we have if x is between a-δ and a, then f(x) - f(a) is between -ε and ε

(12) Combining (10) and (11) gives us:
if x is between a-δ and a+δ then f(x) - f(a) is between -ε and ε [For either condition (10) applies or condition (11) applies]

(13) Then, applying the definition of limit (see here), we get:
lim(x → a) f(x) = L

QED

Clarification: Local Maxima and Minima

A maximum or minimum is local if it is true for a given interval. A maximum or minimum is absolute if it is true across a domain. A maximum is value that is greater than all other points in for the range in question. A minimum is a value that is smaller or equal to all other points in the range in question.

Theorem: Local Maxima

If a function f(x) is differentiable at c and is defined as an open interval containing c and if f(c) is a local maximum value of f(x), then f'(c)=0.

Proof:

(1) Assume that f(c) is a local maximum value for f(x) on the open interval (a,b).

(2) Since c is differentiable, it means that right-hand and left-hand limits both exist and are equal to f'(c). [See Lemma 4 above]

lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)

lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)

(3) If Δx is greater than 0, then:

[f(c + Δx) - f(c)]/Δ x ≤ 0

This is true since f(c) ≥ f(c +Δx) for all small positive values of Δx since f(c) is a local maximum by assumption.

(4) If Δx is less than 0, then:

[f(c + Δx) - f(c)]/Δ x ≥ 0

This is true since in this case, we have a negative value f(c + Δx) - f(c) over another negative value Δx.

(5) So, we have from (#2):
f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≤ 0

(6) But we also have from (#2):
f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≥ 0

(7) Combining (#5) and (#6), we can conclude that f'(c) = 0 [since f'(c) ≥ 0 and f'(c) ≤ 0 ]

QED

Corollary: Local Minima

If a function f(x) is differentiable at c and is defined as an open interval containing c and if f(c) is a local minimum value of f(x), then f'(c)=0.

Proof:

(1) Assume that f(c) is a local minimum value for f(x) on the open interval (a,b).

(2) Since c is differentiable, it means that right-hand and left-hand limits both exist and are equal to f'(c). [See Lemma 4 above]

lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)

lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)

(3) If Δx is greater than 0, then:

[f(c + Δx) - f(c)]/Δ x ≥ 0

This is true since f(c) ≤ f(c +Δx) for all small positive values of Δx since f(c) is a local minimum by assumption.

(4) If Δx is less than 0, then:

[f(c + Δx) - f(c)]/Δ x ≤ 0

This is true since in this case, we have a positive value f(c + Δx) - f(c) over a negative value Δ x.

(5) So, we have from (#2):
f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≥ 0

(6) But we also have from (#2):
f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≤ 0

(7) Combining (#5) and (#6), we can conclude that f'(c) = 0 [since f'(c) ≥ 0 and f'(c) ≤ 0 ]

QED

References

• Edwards & Penny, Calculus and Analytic Geometry

Mathematical Limits

One of the most important ideas in calculus is the concept of the mathematical limit. Limits relate to continuous functions and the basic idea is that if as a value the argument x approaches the value a, the difference between the limit L and f(x) can be arbitrarily small.

Definition 1: Mathematical Limit:

A function f(x) has a limit of L at point a if given any number ε, there exists a positive number δ such that:
if x-a lies between -δ and +δ, then f(x) - L lies between -ε and ε

This definition is very similar to the definition of a continuous function (see here) and it is not surprising that the two concepts are very closely related.

In today's blog, I will need two definitions in order to prove the Squeeze Law relating to mathematical limits.

Definition 2: Open Interval :

x is an element of an open interval (α, β) if x is greater than α and x is less than β

Definition 3: Deleted Neighborhood

A deleted neighborhood is a set of points that result from deleting a single point in an open interval.

Lemma 1: Constant Law for Limits

if f(x) = C, then lim (x → a) f(x) = C

Proof:

(1) Let δ = 1

(2) if x - a lies between -δ and +δ, we know that f(x) = C.

(3) So, we know that f(x) - C = C - C = 0 which is less than any positive value ε

QED

Lemma 2: Product Law

if lim (x → a) f(x) = L and lim(x → a)g(x) = M, then
lim(x → a)[f(x)*g(x)] = L * M

Proof:

(1) Let ε be any nonzero value. We will prove that f(x)g(x) - LM lies between -ε and +ε

(2) Since the limit of f(x) = L, we know that there exists δ₁ such that:
if x - a is between -δ₁ and +δ₁, then f(x) - L is between -ε and +ε

Since by definition, if x - a is between -δ₁ and +δ₁, then f(x) is between -L and +L.

(3) We also know that there exists δ₂ such that:
if x - a is between -δ₂ and +δ₂, then f(x) - L is between -ε/(2M) and ε/(2M).

The definition for limits is that for any given positive value (ε), we can find a positive value (δ) to get the result (see above if review is needed).

(4) And there exists δ₃ such that:
if x - a is between -δ₃ and +δ₃, then g(x) - M is between -ε/(2L) and ε/(2L)

(5) Let δ = min(δ₁,δ₂,δ₃)

(6) Now, if x - a is between -δ and +δ, then:

(a) f(x) - L is between -ε/(2M) and +ε/(2M)

(b) g(x) - M is between - ε/(2L) and + ε/(2L)

(c) M[f(x) - L] is between (M)[-ε/(2M)] and (M)[+ε/(2M)] which is between -ε/2 and ε/2.

(d) f(x)[g(x) - M] is between (L)[-ε/(2L)] and (L)[+ε/(2L)] which is between -ε/2 and ε/2.

(e) If we add (c) + (d), we get:
f(x)M - LM + f(x)g(x) - f(x)M = f(x)g(x) - LM

(f) So, f(x)g(x) - LM is between (-ε/2 + -ε/2) and (+ε/2 + +ε/2) which means that it is between -ε and +ε

(7) So LM is the limit for f(x)g(x).

QED

Lemma 3: Squeeze Law

Suppose f(x), g(x), h(x) are functions such that
(a) f(x) ≤ g(x) ≤ h(x) for a deleted neighborhood (α, β) where point a is removed.
(b) lim (x→ a) f(x) = L = lim(x→a)h(x).

Then:
lim (x→ a) g(x) = L

Proof:

(1) Let ε be an arbitary number.

(2) Using the definition of limits, we know that there exists δ₁ and δ₂ such that:

if x-a lies between -δ₁ and +δ₁, then f(x)-L lies between -ε and +ε

if x-a lies between -δ₂ and +δ₂, then h(x)-L lies between -ε and +ε

(3) Let δ = min(δ₁,δ₂)

(4) We know that δ is greater than 0. [By the definition of mathematical limit]

(5) If x-a in between -δ and +δ, we know that f(x) and h(x) are both points of the open interval (L-ε, L+ε) [Again, from the definition of mathematical limit]

(6) So L-ε is less than f(x) ≤ g(x) ≤ h(x) which is less than L + ε

(7) Combining (#5) and (#6), this gives us that for any given ε, there exists a δ such that:
if x-a is in between -δ and +δ, then g(x)-L is between -ε and +ε

(8) From (#7), L is also a limit for g(x) as x approaches a.

QED

Lemma 4: Substitution Law

If lim (x → a) g(x) = L and lim (x → L) f(x) = f(L), then lim (x → a) f(g(x)) = f(L)

Proof:

(1) Let ε be any positive real value.

(2) Because lim (y → L) f(y) = f(L), we also know that there exists a value δ₁ such that:

if (y - L) is between -δ₁ and +δ₁, then f(y) - f(L) is between -ε and +ε

(3) Because lim (x → a) g(x) = L, we know that there exists a value δ₂ such that:

if (x - a) is between - δ₂ and +δ₂, then g(x) - L is between -δ₁ and +δ₁

(4) But this means if y = g(x), then:

if (x -a ) is between -δ₂ and +δ₂, then y - L is between -δ₁ and +δ₁ and f(g(x)) - f(L) is between -ε and +ε

(5) This then proves that:

lim (x → a) f(g(x)) = f(L)

QED

Lemma 5: lim (x → a) (1/x) = 1/a if a ≠ 0

Proof:

(1) Let ε be any positive real number.

(2) Assume that a is greater than 0.

(3) abs(1/x - 1/a) = abs([a - x]/ax) = abs([x -a]/ax) = (1/a)abs(x-a)/abs(x)

(4) Let us assume that abs(a-x) is less than a/2.

We can do this since abs(x-a) approaches 0 as x moves toward a.

(5) Then x -a is between -a/2 and +a/2 which means that x is between a/2 and 3a/2.

(6) This gives us that abs(x) is greater than a/2 and 1/abs(x) is less than 2/a.

(7) So that abs(1/x - 1/a) = abs(x-a)*(1/a)*abs(1/x) which is less than abs(x-a)*(1/a)(2/a) = 2/a² * abs(x-a)

(8) Let δ be the minimum of a/2 and a²ε/2

(9) Then if x - a is between -δ and +δ, then:

abs(1/x - 1/a) is less than (2/a²)(a²ε/2) = ε

This then proves that lim (x → a) (1/x) = 1/a for when a is greater than 0.

(10) Assume that a is less than 0

(11) Then

abs(1/x - 1/a) = abs(x-a)/(-a)*1/abs(x)

(12) If we assume that abs(x-a) is less than -a/2, then:

x - a is between -a/2 and +a/2, then x is between -3a/2 and -a/2.

(13) So abs(x) is greater than -a/2.

(14) So 1/abs(x) is less than -2/a.

(15) In this case, then:

abs(1/x - 1/a) = abs(x-a)/(-a)*1/abs(x) which is less than 1/(-a)*(-2/a)*abs(x-a) = 2/a²*abs(x-a)

(16) Let δ be the minimum of -a/2 and a²ε/2

(17) Then if x - a is between -δ and +δ, then:

abs(1/x - 1/a) is less than (2/a²)(a²ε/2) = ε

This then proves that lim (x → a) (1/x) = 1/a for when a is less than 0.

QED

Lemma 6: Reciprocal Law

if lim(x → a) g(x) = L and L ≠ 0, then lim (x → a) 1/g(x) = 1/L

Proof:

(1) Let f(x) = 1/x

(2) lim (x → a) f(x) = lim (x → a) (1/x)

(3) Using Lemma 5 above, we have:

lim (x → a) f(x) = 1/L = f(L)

(4) Applying the Substitution Law (Lemma 4 above) gives us:

lim (x → a) 1/g(x) = lim (x → a) f(g(x)) = f(L) = 1/L

QED

Lemma 7: Quotient Law

if lim (x → a) f(x) = L and lim (x → a) g(x) = M ≠ 0, then:

lim (x → a) f(x)/g(x) = L/M

Proof:

(1) Using the Product Law above, we have:

lim (x → a) f(x)/g(x) = lim (x → a)f(x) * lim(x → a)1/g(x)

(2) Using the Reciprocal Law above:

lim (x → a) 1/g(x) = 1/M

(3) Combining step #1 and step #2 gives us:

lim (x → a) f(x)/g(x) = L*(1/M) = L/M

QED

Lemma 8: abs(a + b - c - d)) ≤ abs(a - c) + abs(b - d)

Proof:

(1) If (a-c),(b-d) are the same sign, then abs(a + b - c -d) = abs(a -c) + abs(b - d)

(2) If (a-c),(b-d) are not the same sign, then abs(a - c + b - d) is less than abs(a -c) + abs(b-d).

QED

Corollary 8.1: Addition Law

if lim (x → a) f(x) = L and lim(x → a)g(x) = M, then
lim(x → a)[f(x)+g(x)] = L + M

Proof:

(1) Let ε be any nonzero value.

(2) Since the limit of f(x) = L, we know that there exists δ₁ such that:
if x - a is between -δ₁ and +δ₁, then f(x) - L is between -ε/2 and +ε/2

Since by definition, if x - a is between -δ₁ and +δ₁, then f(x) is between -L and +L.

(3) Since the limit of g(x) = L, we know that there exists δ₂ such that:
if x - a is between -δ₂ and +δ₂, then g(x) - M is between -ε/2 and +ε/2

Since by definition, if x - a is between -δ₂ and +δ₂, then g(x) is between -M and +M.

(4) Let δ = min(δ₁,δ₂)

(5) Now, if x - a is between -δ and +δ, then:

(a) f(x) - L is between -ε/2 and +ε/2

(b) g(x) - M is between - ε/2 and + ε/2

(c) By Lemma 8 above, abs([f(x) + g(x)] - (L + M)) ≤ abs(f(x) - L) + abs(g(x) - M)

(d) abs(f(x) - L) + abs(g(x) - M) ≤ η/2 + η/2 = η

(7) So L+M is the limit for f(x)+g(x).

QED

References

• Mathematical Limit, Wikipedia
• Edwards & Penny, Calculus and Analytic Geometry
  

Continous Functions

In mathematics, a function is a mapping from a set of input to an output. Each input is called an argument and the set of all arguments that can be put into this function is called the domain. The set of all outputs is called the range.

The standard notation for a function is the form f(x,y,z) where x,y,z are the arguments. In most of the examples of this blog, I am only using one argument so this type of function is represented as f(x) where x is the argument.

An equation such as e=mc² is a function. In this case, c is a constant since it is the speed of light so e=mc² is a function with only one argument m. So, we could represent Einstein's equation as: f(m) = mc²

A function is said to be continuous if as a function moves through input moves through a continuum of values, the output also moves through a continuum of values. The important idea is that there are no gaps in a continuous function. If a function has gaps, then it is not continuous.

Definition 1: Continuous at a Point

A function is continuous at a point c if and only if:
for any arbitrary value ε, there exists another a positive value δ such that:
if x is a point that lies between c - δ and c + δ, then f(x) lies between f(c) - ε and f(c) + ε

The important idea here is that if a function is continuous at c, then we can choose an arbitary interval ε such that we know that f(x) lies between f(c)-ε and f(c) + ε

Definition 2: Continuous Function

A function is said to continuous if it is continuous at all points in its domain.

Definition 3: Closed Interval [α, β]

A closed interval can be thought of as a set of elements that make up a subset of the domain for a function. For an interval [α, β], this includes all values x such that α ≤ x ≤ β.

Example: a closed interval on the real numbers

For example, we could set up a closed interval [-5,5] this would then include all numbers that are greater or equal to -5 and less than or equal to 5 including -5, -4.5, 0, and 4.

To show the importance of continous functions, let's consider the Weierstrass Intermediate Value Theorem.

Theorem: Weierstrass Intermediate Value Theorem

If f(x) is continuous at all parts of a closed interval [ α, β ] and if f(α) is less than 0 and if f(β) is greater than 0, then there exists a point γ ∈ [α, β] such that f(γ) = 0.

Proof:

(1) Since f(α) is less than 0, we know that there exists a value μ such that for all values x on [α, μ] f(x) is less than 0 and μ ≥ α.

(2) We can assume that if f(x) is less than 0, then it lies on the interval [ α, μ ] since:

(a) If there was a value ζ such that ζ is greater than μ and f(ζ) is less than 0, then it follows that for x greater than μ and less than ζ , f(x) ≥ 0. (Otherwise, we could assume that ζ ∈ [ α, μ ] )

(b) For the interval specified in (a) if f(x)=0, then we are done with the proof so to finish the proof, we can assume that this is not the case.

(c) If in the interval specified in (a) f(x) is greater than 0, then we can set β to a value greater than μ and less than ζ where f(x) is greater than 0.

(d) In this case, we have a smaller interval than the original but we can make the assumption that there is no value ζ greater than μ where f(ζ) is less than 0.

(e) If we prove that f(γ)=0 exists for this smaller interval, we are likewise proving that it exists in the bigger interval.

(3) Let S be the bounded set on [α, μ] (See the note here for the definition of a bounded set)

(4) Since we are dealing with real numbers, we know that there exists a least upper bound γ for S [See here for proof]

(5) Assume f(γ) = v which is greater than 0.

(6) Since we are dealing with a continuous function, by the definition 1 above, we know that there exists a value δ such that at point γ:
if x is between γ - δ and γ + δ, then f(x) lies between f(γ) - v/2 and f(γ) + v/2. [Where I chose ε = v/2 as the arbitrary value.]

(7) From (#6), we know that if x lies between γ - δ and γ + δ, then:
f(x) lies between v - v/2 and v + v/2 [Since f(γ) = v from #5]

(8) From (#7) we have thta:
f(x) lies between v/2 and 3v/2.

(9) Now μ is greater than γ - δ since:

(a) Assume μ ≤ γ - δ

(b) Then, γ - δ is an upper bound for [α,μ] since for all x ∈ [α,μ] x ≤ γ - δ

(c) And γ - δ is less than γ since δ is a positive value.

(d) But γ is the least upper bound from #4

(e) So we have a contradiction and we reject (a).

(10) And μ is less than γ + δ since μ ≤ γ since γ is the least upper bound.

(11) Since μ is in between γ - δ (#9) and γ + δ (#10), we can use (#8) to conclude that:
f(μ) is greater than v/2 which means by step #4 that f(μ) is greater than 0.

(12) But this is a contradiction since f(μ) is less than 0 by our original assumption in #1 so we can reject our assumption in #4 and conclude that f(γ) ≤ 0.

(13) Let's assume that f(γ)=v is less than 0.

(14) Applying #7, we can conclude that if x lies between γ - δ and γ + δ, then:
f(x) is less than 0 since f(x) is less than 3v/2 which ≤ 3(0)/2 ≤ 0 [From #13]

(15) We know that there exists a value ν such that ν is greater than γ and ν is less than γ + δ since:

(a) Let ν = (2*γ + δ)/2. [We can make this assumption be the definition of multiplication, addition, and division on real numbers, see here]

(b) We see ν is less than γ + δ since γ + δ = (2*γ + 2*δ)/2 and 2*δ is greater than δ since δ is a positive value.

(c) We see that ν is greater than γ since γ = 2*γ/2 and 2*γ is less than 2*γ + δ.

(16) Since ν lies between γ - δ and γ + δ, we can use step #13 to conclude that f(ν) is less than 0.

(17) But this also contradicts our assumption in #2 since it presupposes that there exists a value ν which is greater than μ where f(ν) is less than 0.

(18) So we can reject our assumption in #13.

(19) This then gives us that f(γ) =0 since it cannot be greater than 0 (from #12) and it cannot be less than 0 (from #18)

QED

References

• Continuous Function, Wikipedia
• Function, Wikipedia
  
• Weierstrass Intermediate Value Theorem, MathWorld
  

Dedekind Cut

The Dedekind Cut is mathematical construction created by Richard Dedekind to provide a definition for the real numbers.

The Dedekind Cut itself is defined in terms of the rational numbers.

Definition 1 - Dedekind Cut

A Dedekind cut α is defined as the subset of the rational integers Q (ratios of integers) which is less than α.

NOTE: Q is used to present the set of all rational numbers; R is used to represent the set of all real numbers, and Z is used to represent the set of all integers.

Example: π

We could create a Dedekind Cut around π. In this case, we could think of 3, 22/7, -4, etc. as elements of the Dedekind Cut. On the other hand, 32/10, 4, etc. would not be elements of the cut.

Definition 2 - Set of Real Numbers R

The set of real numbers is the set of Dedekind cuts α that have the following properties:

(a) α is not empty

(b) α contains no greatest element

For any element x ∈ α, there exists y ∈ α such that x is less than y.

(c) If x,y are rational integers where y is less than x, then x ∈ α → y ∈ α.

With these definitions, we have enough the construct the properties of the real numbers.

Definition 3: Additive Identity: 0

0 is defined as the set { x ∈ Q such that x is less than 0 }

Definition 4: Multiplicative Identity: 1

1 is defined as the set { x ∈ Q such that x is less than 1 }

Definition 5: Addition

α + β is defined as the set { x + y such that x ∈ α, y ∈ β }

Definition 6: Subtraction

α - β is defined as the set { x - y such that x ∈ α, y ∈ β }

Definition 7: Multiplication

α * β is defined as the set { x * y such that x ∈ α, y ∈ β }

Definition 8: Division

α / β is defined as the set { x/y such that β ≠ 0, x ∈ α, y ∈ β }

Definition 9: Irrational Numbers

A real number α is said to be irrational if α ∩ Q does not have a least element.

Now, I will present a proof for a fundamental property of reals using the Dedekind Cut.

Theorem 1: Every real number that is bounded above has a least upper bound.

(1) Let Α be a set of real numbers that is bounded by γ such that α ∈ Α → α ≤ γ

NOTE: Bounded above just means that there is an element γ that is greater or equal to all the elements that make up A.

(2) The union of all the sets that make up Α are themselves a real number because:

(a) Since each real number α is a Dedekind cut, we know that each α is the set of x ∈ Q such that x is less than some real number.

(b) Since each α is not empty, the union of all elements that make up α is not empty.

(c) We know that the union does not have a greatest element since if it did, this element would likewise be the greatest element for whichever real number that it is an element for but this is impossible since by definition, none of the real numbers have a greatest element.

(d) If x ∈ the Union and y is less than x, then y is necessarily an element of the Union, since y would necessarily be an element of the real number that x is an element of.

(3) The union of all sets that make up A is an upper bound for A since every element of A ⊆ Union of all sets of A.

(4) Now, γ is an arbitrary upper bound, so all we need to prove is that A ⊆ γ since this shows that A is necessarily less than or equal to any given upper bound.

(5) But this is easy to prove since x ∈ A → x ∈ γ since we defined each element of A as less than γ.

QED

Theorem 2: For any positive real number ε, there exists a natural number n such that:

0 is less than 1/n is less than ε

Proof:

(1) We can find a natural number n such that:

(n-1) ≤ (1/ε) ≤ n.

(2) Taking the reciprocal for each gives us:

1/(n-1) ≥ ε ≥ 1/n.

QED

Corollary 2.1: Between any two distinct real numbers, there exists a rational number.

Proof:

(1) Let x,y be real numbers such that y is greater than x.

(2) By Theorem 2 above, there exists an integer n such that: 1/n is less than y - x.

(3) Let m = floor(n*x) where floor(n*x) returns the highest integer that is less than n*x.

(4) From this, we know that:

(m/n) ≤ x and x is less than (m+1)/n.

(5) Further, we have:

(m + 1)/n = m/n + 1/n ≤ x + 1/n is less than x + y - x = y.

(6) So that, we have:

x is less than (m+1)/n is less than y.

QED

References

• Dedekind Cut, PlanetMath
• Dedekind Cut, Wikipedia
• Supremum, Wikipedia
  

Derivatives

A derivative in mathematics refers to the rate of change of a mathematical function. For purposes of this blog, I will only focus on functions of the form f(x), that is, a function in terms of a single variable which I will label x. For those who are not familiar with the concept of a mathematical function, see here.

It is assumed that the function f(x) is continuous. This is an important assumption since any function which is not continuous does not have a derivative. On the other hand, not all continuous functions have a derivative (see here).

For continuous functions, finding the rate of change (the derivative) comes down to finding the ratio of: (Δy)/(Δx) where x is a value as small as possible. In mathematical terms, we are finding this ratio as x approaches a limit of 0. We use the limit because of the problem of division by 0. Division by 0 is undefined but it is quite valid to take the limit that a value reaches at its divisor approaches 0.

For example, 1/x = infinity as x approaches 0

NOTE: This is different than saying 1/0 which is undefined. You can think of the concept of a limit to 0 as Newton's workaround for division by 0.

Sometimes, Δy is written as dy and Δx is written as dx. In both cases, the value is meant to be the smallest value whose limit is 0.

There are many conventions for representing the derivative of a function. In my blog, I will use three forms:

d/dx(x)
f'(x)
fⁿ(x)

Each one serves a different purpose. f'(x) represents the derivative of a given function f(x). fⁿ(x) implies the n-order derivative which is equivalent to f'(f'(f'(f'...(x))))).

d/dx makes it clearer which variable is being used in the derivative. For example:
f(x) = u + v
f'(x) = du/dx + dv/dx

We can now look at a more formal definition of a derivative in terms of a limit.

Definition 1: Derivative of f(x)

f'(x) = lim (Δx → 0) [f(x + Δx) - f(x)]/(Δx)

I will show how this equation is worked out in the lemmas below.

Definition 2: Differentiable

A function is said to be differentiable at a given point if a derivative exists for that point.

In today's blog, I will go over some very basic results using derivatives.

Lemma 1: Constant Rule: If a is a constant, (d/dx)(a) = 0

f(x + Δx) - f(x) = a - a = 0

QED

Lemma 2: Power Rule: (d/dx)(axⁿ) = nax^n-1

(1) f(x) = axⁿ

(2) f(x + Δx) - f(x) = a(x + Δx)ⁿ - axⁿ

(3) Applying the Binomial Theorem, gives us:


(4) So a(x + Δx)ⁿ - axⁿ, gives us:


(5) Then, dividing by Δx gives us:


(6) Finally, since Δx approaches 0, we can replace Δx with 0 to get:


(7) This then gives us:


QED

Example 1: (d/dx)(5x) = 5

f(x) = 5x

f'(x) = (d/dx)(5x) = d/dx(5x¹) = (1)5x⁰ = 5.

Example 2: (d/dx)(1/x) = -1/(x²)

f(x) = 1/x

f'(x) = (d/dx)(1/x) = (d/dx)(x^-1) = (-1)x^-2 = -1/(x²)

Example 3: (d/dx)(4x²) = 8x

f(x) = 4x²

f'(x) = (d/dx)(4x²) = (4)(2)x¹ = 8x

Lemma 3: Derivative of a Linear Combination

f(x) = a*g(x) + b*h(x) → f'(x) = a*g'(x) + b*h'(x) where a,b are constants.

(1) f(x) = a*g(x) + b*h(x)

(2) f'(x) = f(x + Δx) - f(x) = a*g(x + Δx) + b*h(x + Δx) - [a*g(x) + b*h(x)] =

= a*[g(x+Δx) - g(x) ] + b*[h(x + Δx) - h(x)] = a*g'(x) + b*h'(x)

QED

Example 1: (d/dx)(4x + 5) = 4

f(x) = 4x + 5

f'(x) = (d/dx)(4x) + (d/dx)(5) = 4 + 0 = 4

Example 2: (d/dx)(2x² + 3x) = 4x + 3

f(x) = 2x² + 3x

f'(x) = (d/dx)(2x²) + (d/dx)(3x) = 4x + 3

Lemma 4: Product Rule: f(x) = g(x)h(x) → f'(x) = g(x)h'(x) + g'(x)h(x)(1) Let f(x) = g(x)h(x)

(2) f'(x) = lim (Δx → 0) [ g(x+Δx)h(x + Δx) - g(x)h(x)]/Δx =

= lim(Δx → 0)[g(x + Δx)h(x+Δx)-g(x)h(x+Δx)+g(x)h(x+Δx)-g(x)h(x)]/Δx =

= lim(Δx→0)[g(x+Δx)h(x+Δx)-g(x)h(x+Δx)]/Δx +
lim(Δx→0)[g(x)h(x+Δx) - g(x)h(x)]/Δx =

= lim(Δx→0)[g(x+Δx)-g(x)]/Δx * lim(Δx→0)h(x+Δx) +
lim(Δx→0)g(x)*lim(Δx→0)[h(x+Δx)-h(x)]/Δx

(3) Now, we note that:
lim(Δx→ 0)g(x) = g(x)

lim(Δx→0)h(x+Δx) = h(x)

(4) We also note that:
lim(Δx→0)[g(x+Δx) - g(x)]/Δx = g'(x)

lim(Δx→0)[h(x+Δx)-h(x)]/Δx = h'(x)

(5) Combining #2, #3, and #4 gives us:
f'(x) = g'(x)h(x) + g(x)h'(x)

QED

Lemma 5: Reciprocal Rule

If f(x) is differentiable at x and f(x) ≠ 0 and u(x)=1/f(x)

Then:

u'(x) = -f'(x)/[f(x)]²

Proof:

(1) From the definition of the derivative and the definition of u(x), we have:

u'(x) = lim (Δx → 0) (1/Δx)[1/f(x+Δx) - 1/f(x)] =

= lim (Δx → 0)[f(x) - f(x + Δx)]/[Δx*f(x + Δx)*f(x)]

(3) Since:

[f(x) - f(x + Δx)]/[Δx*f(x + Δx)*f(x)] =(-1)*(1/[f(x+Δx)f(x)])*([f(x+Δx)-f(x)]/Δx),

we can use the Product Rule for Limits (see Lemma 2, here) to get:

lim (Δx → 0)[f(x) - f(x + Δx)]/[Δx*f(x + Δx)*f(x)] =

= lim (Δx → 0)(-1) * lim (Δx → 0)(1/[f(x+Δx)f(x)]) * lim (Δx → 0) ([f(x+Δx)-f(x)]/Δx)


(5) Since:

lim (Δx → 0)(-1) = -1.

lim (Δx → 0)(1/[f(x+Δx)f(x)]) = 1/[f(x)]²

lim (Δx → 0) ([f(x+Δx)-f(x)]/Δx) = f'(x)

We have:

u'(x) = (-1)*( 1/[f(x)]²)* f'(x) = -f'(x)/[f(x)]²

QED

Lemma 6: The Quotient Rule

If f(x) and g(x) are differentiable at x and g(x) ≠ 0, and u(x)=f(x)/g(x) with u(x) differentiable at x, then:

u'(x) = [f'(x)g(x) - f(x)g'(x)]/[g(x)]²

Proof:

(1) Using the definition for u(x), we have:

u(x) = f(x)/g(x) = f(x)*1/g(x)

(2) Using the Product Rule (see Lemma 4 above), we have:

u'(x) = f(x)*D[1/g(x)] + f'(x)*[1/g(x)]

(3) Using the Reciprocal Rule (see Lemma 5 above), we know that:

D[1/g(x)] = -g'(x)/[g(x)]²

(4) Applying step #3 to step #2 gives us:

u'(x) = f(x)*[-g'(x)/[g(x)]²] + f'(x)/g(x) = [-f(x)g'(x) + f'(x)g(x)]/[g(x)]² = [f'(x)g(x) - f(x)g'(x)]/[g(x)]²

QED

Set of Integers

The set of integers can be constructed from the Natural Numbers.

We can define the set of integers as the set NxN where "x" is the Cartesian Product.

In set theory, the Cartesian Product is the combination of all the elements of one set with all the elements of the other set forming pair.

So, if N = { 1, 2, 3 ... }, then NxN = { (1,1), (1,2), (1,3), ..., (2,1),(2,2), ..., }

Here is the definition for integers:

Definition 1: Set of Integers: Z is the set NxN where (a,b) is the same number as (c,d) if and only if a + d = b + c.

NOTE: Z stands for Zahlen which is German for number.

From this perspective, each number corresponds to the set of all values (a,b) where a-b is equal to the number.

So 0 for example corresponds to { (1,1), (2,2), .... }

And -1 corresponds to { (1,2), (2,3), ... }

We can use the definition to show that (1,1) ~ (2,2) in that 1 + 2 = 1 + 2.

From this, we can now define addition, multiplication, and subtraction.

Definition 2: (a,b) + (c,d) = (a + c, b + d)

Example:

-1 + -3 = (2,3) + (5,8) = (2+5,3+8) = (7,11) = -4

Definition 3: (a,b) * (c,d) = (ac + bd, ad+bc)

Example:

-1 * -1 = (1,2)*(1,2) = (1*1 + 2*2,1*2 + 2*1) = (1 + 4,2 + 2) = (5,4) = 1

-1 * 3 = (1,2)*(4,1) = (1*4 + 2*1,1*1 + 2*4) = (4 + 2,1 + 8) = (6,9) = -3

Definition 4: (a,b) - (c,d) = (a + d, b + c)

Example:

4 - 3 = (5,1) - (4,1) = (5 + 1, 1 + 4) = (6, 5) = 1

(-3) - (-5) = (2,5) - (1,6) = (2 + 6, 5 + 1) = (8, 6) = 2

3 - 4 = (4,1) - (5,1) = (4 + 1, 1 + 5) = (5,6) = -1

Definition 5: (a,b) is less than (c,d) if and only if a + d is less than b + c

Example:

-3 is less than 4 since (2,5) is less than (5,1) since 2+1=3 is less than 5 + 5=10.

Lemma 1: Z is closed under addition.

(1) (a,b) + (c,d) = (a + c, b + d)

(2) a + c is a natural number and b + d is a natural number (since addition is closed for natural numbers, see here)

QED

Lemma 2: Z is closed under multiplication

(1) (a,b) * (c,d) = (ac + bd, ad + bc)

(2) ac, bd, ad, bc are all natural numbers since N is closed under multiplication (see here)

(3) ac + bd, ad + bc are natural numbers since N is closed under addition (see here)

QED

Lemma 3: Z is closed under subtraction

(1) (a,b) - (c,d) = (a + d, b + c)

(2) a + d, b + c are both natural numbers since N is closed under addition (see here)

QED

Lemma 4: Z has 0 as the identity element for addition

n + 0 = (n+1,1) + (1,1) = (n + 2,2) = n

QED

Lemma 5: Z has 1 as the identity element for multiplication

n * 1 = (n + 1, 1)*(2,1) = (2*(n+1) + 1*1, (n+1)*1 + 2) = (2n + 3,n + 3) = n

QED

Lemma 6: Each element of Z has an inverse element for addition that is determined by reversing the pairs (b,a) is the inverse element for (a,b)

n + (-n) = (n+1,1) + (1,n+1) = (n + 1 + 1, 1 + n + 1) = (n+2,n+2) = 0

QED

Lemma 7: Z is commutative on addition: a + b = b + a

a + b = (a+1,1) + (b+1,1) = (a + b + 2,2)
b + a = (b+1,1) + (a+1,1) = (b + a + 2,2) = (a + b + 2,2) [Since natural numbers are commutative on addition, see here]

QED

Lemma 8: Z is commutative on multiplication: a*b=b*a

a*b = (a+1,1)*(b+1,1) = ((a+1)(b+1) + 1*1,(a+1)*1 + 1*(b+1)) =
= (ab + a + b + 1 + 1, a + 1 + b + 1) =
= (ab + a + b + 2,a + b + 2) = ab

b*a = (b+1,1)*(a+1,1) = ((b+1)(a+1) + 1*1,(b+1)*1 + 1*(a+1)) =
= (ba + a + b + 2, a + b + 2) =
= (ab + a + b + 2, a + b + 2) = ab [Since natural numbers are commutative on multiplication, see here]

QED

Lemma 9: Z is associative on addition

(a+b) + c = [(a+1,1) + (b+1,1)] + (c+1,1) = (a+b+2,2) + (c+1,1) = (a+b+c+3,3)=
= a+b+c

a + (b+c) = (a+1,1) + [(b+1,1) + (c+1,1)] = (a+1,1) + (b+c+2,2) = (a+b+c+3,3) =
= a+b+c

QED

Lemma 10: Z is associative on multiplication

(a*b)*c = [(a+1,1)*(b+1,1)]*(c+1,1)=
= [(a+1)(b+1) + (1)(1),(a+1)(1) + 1(b+1)](c+1,1) =
= (ab + a + b + 2, a+1+b+1)(c+1,1) = (ab+1,1)(c+1,1) =
= ((ab+1)(c+1) + (1)(1), (ab+1)(1)+(1)(c+1)) =
= (abc + ab + c + 1 + 1, ab + 1 + c + 1) =
= (abc + ab + c + 2, ab + c + 2) = abc

a*(b*c) = (a+1,1)[(b+1,1)(c+1,1)] =
= (a+1,1)((b+1)(c+1) + (1)(1),(b+1)(1)+(1)(c+1)) =
= (a+1,1)(bc + b + c + 1 + 1,b+1+c+1) =
= (a+1,1)(bc+b+c+2,b+c+2)=(a+1,1)(bc+1,1)=
= ((a+1)(bc+1) + (1)(1),(a+1)(1) + (1)(bc+1)) =
= (abc + a + bc + 1 + 1,a + 1 + bc + 1) =
= (abc + 1, 1) = abc

QED

Lemma 11: Z is distributive: (a+b)c =ac + bc.

(a+b)c = [(a+1,1)+(b+1,1)](c+1,1) = [(a + b + 2,1+1)](c+1,1) = (a+b+1,1)(c+1,1) =
= ((a+b+1)(c+1) + (1)(1),(a+b+1)(1) + (1)(c+1)) =
= (ac + bc +c + a + b + 1 + 1,a + b + 1 + c + 1) =
= (ac + bc + 1, 1) = ac + bc.

QED

References

• Integers, Wikipedia
• Integer, PlanetMath
• Integer, Dr. Noach Dana Picard's Page

Natural Numbers

The natural numbers are the numbers that are most familiar to us; they are the counting numbers that usually begin at 1 (some mathematicians hold that the natural numbers include 0). This also matches with the history of mathematics where the use of 0 was a major innovation.

The mathematician Kronecker once said:
"God made the natural numbers. Everything else is the work of man." (Quoted from here)

At the same time, this leads us to an interesting question. Is it possible to define the natural numbers in terms of a set of axioms similiar to what Euclid did for geometry?

The successful definition of natural numbers was done by two men: Richard Dedekind in his book What is a Number and the logician Guiseppe Peano. Today, these axioms are known as Peano's Axioms.

Here are a set of postulates based on Peano's Axioms:

I. The natural numbers is a set of numbers starting with 1 where each number has a unique successor which we will characterize by S(a) where a is a natural number.

II. All numbers but 1 have are themselves successors to a unique number.

III. a = b if and only if S(a) = S(b)

IV. If a property is possessed by 1 and possession by a value a implies it is also true of S(a), then we say it is true of all numbers. (Axiom of Induction)

Definition 1: m + n

m + 1 = S(m)
m + S(n) = S(m + n)

Example:

1 + 2 = S(1 + 1) = S(S(1)) = S(2) = 3

Definition 2: m * n

m * 1 = m
m * S(n) = m + (m * n)

Example:

2 * 2 = 2 + (2*1) = 2 + 2 = 4

Definition 3: m is less than n

m is less than n if there exists a natural number d such that m + d = n.

Definition 4: m - n

S(m) - 1 = m
S(m) - S(n) = m - n

If m is less than or equal to n, then it is undefined.

Lemma 1: Natural numbers are closed under addition.

(1) m+1 = S(m)
(2) Assume that there exists a value n such that m + n is a natural number.
(3) m + S(n) = S(m+n)
(4) Which must also be a natural number by Axiom I.

QED

Lemma 2: Natural numbers are closed under multiplication

(1) m*1 = m
(2) Assume that there exists a value n such that m * n is a natural number
(3) m*S(n) = m + (m*n)
(4) Which is a natural number by Lemma 1 above.

QED

Lemma 3: Natural numbers are not closed under subtraction.

m - n is undefined for the case where m ≤ n.

QED

Lemma 4: 1 + m = S(m)

(1) Case n=1:

1 + 1 = S(1)

(2) Assume that it is true up to n.

(3) 1 + (n+1) = 1 + S(n) = S(n+1)

(4) So, it is true by Axion IV.

QED

Lemma 5: Natural numbers are commutative under addition: m + n = n + m

(1) For case n=1:

m + 1 = S(m)

(2) From Lemma 4, 1 + m = S(m)

(3) Assume it is true up to n so that m+n=n+m

(4) m + (n+1) = m + S(n) = S(m+n)

(5) (n+1) + m = S(n) + m = S(n+m) = S(m+n)

(6) So, we apply Axiom IV and we are done.

QED

Lemma 6: 1*m = m*1

(1) 1*1 = 1*1

(2) S(m)*1 = S(m) [By Definition 2 above]

(3) 1*S(m) = 1 + m = m + 1 = S(m)

(4) We are done since all numbers are either 1 or a successor to another number. (by Axion II)

QED

Lemma 7: Distributive Law: (a + b)m = am + bm

(1) Case m = 1:

(a + b)1 = a + b [By Definition 2]

a(1) + b(1) = a + b [By Definition 2]

(2) So we assume that it is true up to n so that (a+b)n = an + bn.

(3) (a+b)S(n) = (a + b) + (a+b)n = a + b + an + bn = a + an + b + bn

(4) a(S(n)) + b(S(n)) = a + an + b + bn

(5) So by the Axiom of Induction we are done (see Axiom IV)

QED

Lemma 8: Natural numbers are commutative under multiplication: m * n = n * m

(1) Case n=1:

m * 1 = 1 * m [From Lemma 6 above]

(2) Assume that it is true up to n so that we have m*n=n*m

(3) m*(n+1) = m + mn [By definition 2]

(4) (n+1)*m = nm + m = m + nm = m + mn

(5) So we are done by the Axion of Induction (Axiom IV above)

QED

Lemma 9: Natural numbers are associative under addition:
(a + b) + n = a + (b + n)

(1) Case n = 1

(a + b) + 1 = S(a + b)

a + (b + 1) = a + S(b) = S(a + b)

(2) (a + b) + S(n) = S(a + b + n)

(3) a + (b + S(n)) = a + S(b + n) = S(a + b + n)

(4) It is true now in all cases since all numbers are either 1 or are a successor (by Axiom II above)

QED

Lemma 10: Natural numbers are associative under multiplication:
(a*b)*n = a*(b*n)

(1) Case n = 1

(a*b)*1 = a*b

a*(b*1) = a*b

(2) (ab)*S(n) = S(n)(ab) = (n+1)(ab) = abn + ab

(4) a*(b*S(n)) = a*(b*(n+1)) = a*((n+1)*b) = a*(bn+b) = (bn+b)*a = abn+ab

QED

References

• Peano Axioms, Wikipedia
• Peano Arithmetic, Planet Math
• Natural Numbers, Dr. Noach Dana Picard's Page
  

Continued Fractions; Loose Ends

In today' s blog, I will provide details that are used in the general proof for Continued Fractions.

Lemma 1: If x is a positive integer, then x² - 4 is not a square.

(1) First, I will show that for all positive integers greater than 2, the difference between one square and the next highest square is greater than 4 therefore no square x² - 4 can exist.

(2) For n=2, the smallest difference is 9 - 4 = 5 which is greater than 4.

(3) We assume that this is true up to n ≥ 2.

(4) So (n+1)² - n² = n² + 2n + 1 - n² = 2n + 1 ≥ 2(2) + 1 = 5.

(5) In other words, the minimal difference between any two successive squares is at least 5.

(6) Now, I will show that all differences 2 or less don't work either.

3² - 2² = 9 - 4 = 5
2¹ - 1² = 1
1² - 0² = 1

The only possible differences then are 1, 2, or a number ≥ 5.

QED

Quadratic Equation Solved

One of the most useful formulas of all time historically is the solution of the equation:
ax² + bx + c = 0.

The equation above is known as the quadratic equation.

Theorem: (-b ± √b² - 4ac)/2a is the solution to the quadratic equation.

(1) First, we multiply both sides by 4a and get:

4a²x² + 4abx + 4ac = 0

(2) Next, we add b² - b² to the equation:

4a²x² + 4abx + b² + 4ac - b² = 0

(3) Now, we add b² - 4ac to both sides which gives us:

4a²x² + 4abx + b² = b² - 4ac

(4) Further, we know that:

(2ax + b)² = 4a²x² + 4axb + b²

(5) Combining #4 and #3, gives us:

(2ax + b)² = b² - 4ac

(6) Now, taking the square root of both sides gives us:

2ax + b = ±√b² - 4ac

(7) Now, using basic algebra, we get to:

x = (-b ±√b² - 4ac)/2a.

Irrational Numbers

Irrational numbers are numbers that cannot be formed by the ratio of two integers. In Ancient Greece, the mathematician and philosopher Pythagoras created a secret society whose sole goal was to study the universe in terms of numbers. Pythagorus had discovered that musical harmonies result from perfect ratios between string sizes and became convinced that all the universe could be studied in terms of these perfect ratios. For example, the most famous discovery of his group was the Pythagorean Theorem which states that the hypotenuse (the diagonal) of a right triangle is equal to the sum of the squares of its sides. In other words: if c is the length of the hypotenus, a,b are the length of the other sides, then c² = a² + b².

It turns out that the Pythagorean Theorem is itself a proof for the existence of irrational numbers. For example, if a=1, b = 1, then c = √2. Here in lies one of the most famous irrational numbers.

Theorem: if p is a prime then √p is irrational.

(1) Assume that there exists two values a,b such that: a/b = √p and such that a,b are the lowest positive fraction such that gcd(a,b)=1.

NOTE: We know that in fractions, if a,b had any common factors, then we could divide them off and still maintain the same ratio.

(2) Then a²/b² = p and therefore a² = pb².

(3) So, we see that p divides a. So there exists a value a' such that a=pa'. [This is true by Euclid's Lemma since p is a prime and it divides either a or a]

(4) So, we get (pa')² = pb² which means that p²a'² = pb².

(5) Dividing both sides by p, we get: pa'² = b².

(6) But now we see that p divides b (again by Euclid's Lemma). This is a contradiction since a,b do not have any common divisors so we can reject our initial assumption.

QED

Lemma 1: if α is irrational, a,b are rational, and b ≠ 0, then a + bα is irrational

(1) Let x = a + bα

(2) bα = x - a

(3) α = (x - a)/b

(4) Now, this proves that x is not rational. Since if x is rational, then (x-a)/b would be rational, but this is impossible since (x-a)/b = α which is irrational.

QED

Lemma 2: if α is irrational, then 1/α is irrational

(1) Let y = 1/α

(2) Then, α = 1/y.

(3) This proves that y is not rational. If it were, then 1/y would be rational but it isn't since 1/y = α which is irrational.

QED

Lemma 3: if a,b are rational and α is irrational, and a + bα is rational, then a=b=0.

(1) By Lemma 1 above, if a + bα is rational, then b = 0.

(2) Since a + bα = 0 we know that a = 0 - bα = 0 - 0 = 0

QED

Lemma 4: For any positive real number ε, there exists a positive irrational number that is less than ε

Proof:

(1) Let p be a prime such that √p is greater than 1/ε [We can make this assumption based on Euclid's Theorem about infinite primes, see here]

(2) By the Theorem above, we know that √p is an irrational number.

(3) Now, if based on step #1, we know that:

1/√p is less than ε

(4) We also know that 1/√p is an irrational number from Lemma 2 above.

QED

Corollary 4.1: For any two distinct rational numbers, there exists an irrational number that is in between.

Proof:

(1) Let x,y be two distinct rational numbers where x is greater than y.

(2) We can see that if ε is any positive number less than x - y, then, y + ε lies in between x and y.

(3) By Lemma 4 above, we know that there exists an irrational number α that is less than ε.

(4) Further, we know that y + α is also irrational [by Lemma 1 above] and we know that y + α lies in between x and y.

QED