proper standard

Just a quick morning warm up doodle of my cleaning crew from that one fightclub prompt fic. C-382 that uniform is not up to proper practice standards. 

Sets and Subsets

Underlying the mathematics studied in algebra, geometry, combinatorics, probability, and almost every other area of mathematics is the notion of a set.

So far, intuition of a set has been relied on. For example, each time the word “collection” appeared in Chapter 1, or each time the word “universe” appeared in Chapter 2, the situation was dealing with a set.

Definition of a Set
The informal definition of a set is the “well-defined” collection of objects called elements, which are the members of the set. The word “well-defined” is used to imply that any element in a set can be considered and can be directly known whether this element is in the set or not. Therefore, sets that depend on opinion are avoided.

Generally, capital letters A, B, …, Z are used to denote a set and lower case letters a, b, …, z are used to denote elements in a set.

For a set A, if the element x is in the set A, or x is a member of set A, it is denoted as x ∈ A, and if the element y is not in the set A, or y is not a member of set A, then it is denoted as y ∉ A.

Set Notation
A set can be designated by listing its elements within set braces {}. For example, if A is the set that contains the first five positive integers, then A = {1,2,3,4,5}. Then 2 ∈  A, but 6 ∉  A.

Another standard notation used for designating the set A is A = {x | x is an integer and 1 ≤ x ≤ 5}. The vertical line | is read as, “such that,” and can be also written with a colon : instead. The phrase {x | …} is then read as, “the set of all x such that,” where the properties that determine the elements of the set are written after the colon.

Note that the notation A = {x | 1 ≤ x ≤ 5}  for A = {1,2,3,4,5} is only correct if it is stated, or demonstrated, that elements in A are agreed to be considered integers. If this agreement was not made, then the set {x | 1 ≤ x ≤ 5} could be a set that includes not only integers but the rest of the real numbers, including irrational numbers, between 1 and 5, inclusive.

When the type of elements representing the set are stated in advanced, then the universe, or the universe of discourse, denoted as 𝓤, is being specified. Once the universe is specified, only elements from the universe are selected to form a set. Therefore, if the universe consisted of only even integers, the set A = {x | 1 ≤ x ≤ 5} would only have the elements 2 and 4.

The following are equivalent statements to writing sets where the universe is of all real numbers:

For a,b ∈ ℝ, a < b, then:

[a,b] = {x ∈ ℝ | a ≤ x ≤ b}
[a,b) = {x ∈ ℝ | a ≤ x < b}
(a,b] = {x ∈ ℝ | a < x ≤ b}
(a,b) = {x ∈ ℝ | a < x < b}

The first statement is called a closed interval and the last statement is called an open interval. The second and third statements are called half-open intervals.

Finite and Infinite Sets
Given 𝓤 = {1,2,3,…} as the set of positive integers, consider the following sets:

A = {1,4,9,…,64,81} = {x² | x ∈ 𝓤, x² < 100} =  {x² | x ∈ 𝓤 ᴧ x² < 100}
B = {1,4,9,16} = {y² | y ∈ 𝓤 , y² < 20} = {y² | y ∈ 𝓤 , y² < 23}  = {y² | y ∈ 𝓤  ᴧ y² < 16}
C = {2,4,6,8,…} = {2k | k ∈ 𝓤}

Set A and set B are examples of finite sets, because there are a finite amount of elements in each set. Set C is an example of an infinite set, because there is an infinite amount of elements in the set.

Cardinality
For any finite set A, the cardinality of A, denoted as |A|, is the number of elements inside of A. The above example sets show that |A| = 9 and |B| = 4.

Note that the sets A and B are such that every element of B is also in the set A. This relationship occurs throughout set theory and its applications.

Definition of Subsets
If C,D are sets from a universe 𝓤, then C is a subset of D, denoted as C ⊆ D or D ⊇ C (D is a superset of C, D has more or equal elements than C), if the elements in set C are fewer or equal to the elements in set D. If the elements in set C are fewer than set D’s elements, where D has elements that are not in set C, then C is a proper subset of set D, denoted as C ⊂ D or D ⊃ C (D is a proper superset of C, D has more elements than C). The symbol ⊊ is sometimes used instead.

Subsets and Quantifiers
If C,D are sets from a universe 𝓤, and if C ⊆ D, then it can be written ∀x [x ∈ C => x ∈ D].

For all subsets C of sets D in 𝓤, then C ⊂ D => C ⊆ D.

However, C ⊆ D → C ⊂ D is not a tautology and therefore not a logical implication. This is because it may not always be true that if C is a subset of D, then C is also the proper subset of D.

The universal quantifier ∀x indicates that every element x in 𝓤 must be considered. For every element c in 𝓤, where c ∈ C is false, then the implication c ∈ C → c ∈ D is true. Therefore, every element c’ in 𝓤, where c’ ∈ C is true, only needs to be considered. If it is found that, for every c’ in 𝓤, c’ ∈ D is true, then C ⊆ D, or ∀x [x ∈ C => x ∈ D]. Otherwise, C is not a subset of D, denoted as C ⊈  D.

Logical Implication and Cardinality
If C,D are finite sets, then C ⊆ D => |C| ≤ |D| and C ⊂ D => |C| < |D|.

Example
For the universe 𝓤 = {1,2,3,4,5}, consider the set A = {1,2}. If B = {x | x² ∈ 𝓤}, then the members of B are 1 and 2. Therefore, set A and set B contain the same elements, and so set A and set B are equal.

Equal Sets
For a given universe 𝓤, the sets C and D from 𝓤 are equal sets, denoted as C = D, if C ⊆ D and D ⊆ C.

From this idea of equality, it can be concluded that order nor repetition of the elements in a general set do not matter. Therefore, {1,2,3} = {3,1,2} = {2,2,1,3} = {1,2,1,3,1}.

A set C is not a proper subset of set D if either C ⊈ D or C = D.

Subsets and Negation Laws
Given the set A and set B from a universe 𝓤, where A ⊆ B, then the negation of A ⊆ B, denoted as A ⊈ B, can be written in the form of a quantified statement.

A ⊆ B
<=> ∀x [x ∈ A => x ∈ B]

A ⊈ B
<=> ¬[∀x [x ∈ A => x ∈ B]]
<=> ∃x ¬[x ∈ A => x ∈ B]
<=> ∃x ¬[¬(x ∈ A) ᴠ x ∈ B]
<=> ∃x ¬[x ∉ A ᴠ x ∈ B]
<=> ∃x [¬(x ∉ A) ᴧ ¬(x ∈ B)]
<=> ∃x [x ∈ A ᴧ x ∉ B]

Therefore, set A is not a subset of set B, or A ⊈ B, if there exists an element x in the universe such that x is in set A but x is not in set B.

To show that a set A is not equal to a set B, but they might possibly be proper subsets of each other, their equality is negated.

A = B
<=> A ⊆ B ᴧ B ⊆ A

A ≠ B
<=> ¬(A ⊆ B ᴧ B ⊆ A)
<=> ¬(A ⊆ B) ᴠ ¬(B ⊆ A)
<=> A ⊈ B ᴠ B ⊈ A
<=> ∃x [x ∈ A ᴧ x ∉ B] ᴠ ∃x [x ∈ B ᴧ x ∉ A]

Therefore, two sets A,B are not equal if A is not a subset of B or B is not a subset of A. Either one can be true to show that A ≠ B is true. Therefore, either ∃x [x ∈ A ᴧ x ∉ B] or ∃x [x ∈ B ᴧ x ∉ A] can be used to define A ≠ B.

Additionally, showing two sets’ inequality can be written in quantified statements:

A = B
<=> A ⊆ B ᴧ B ⊆ A
<=> ∀x [(x ∈ A → x ∈ B) ᴧ (x ∈ B → x ∈ A)]

A ≠ B (Negating A = B)
<=> ¬[∀x [(x ∈ A → x ∈ B) ᴧ (x ∈ B → x ∈ A)]]
<=> ∃x ¬[(x ∈ A → x ∈ B) ᴧ (x ∈ B → x ∈ A)]        
<=> ∃x ¬[(¬(x ∈ A) ᴠ x ∈ B) ᴧ (¬(x ∈ B) ᴠ x ∈ A)]        
<=> ∃x [¬(¬(x ∈ A) ᴠ x ∈ B) ᴠ ¬(¬(x ∈ B) ᴠ x ∈ A)]  
<=> ∃x [(¬¬(x ∈ A) ᴧ ¬(x ∈ B)) ᴠ (¬¬(x ∈ B) ᴧ ¬(x ∈ A))]  
<=> ∃x [(x ∈ A ᴧ x ∉ B) ᴠ (x ∈ B ᴧ x ∉ A)]  
<=> ∃x [x ∈ A ᴧ x ∉ B] ᴠ ∃x [x ∈ B ᴧ x ∉ A]  

Again, either ∃x [x ∈ A ᴧ x ∉ B] or ∃x [x ∈ B ᴧ x ∉ A] can be used to define A ≠ B.

Using both the definitions of equality and subsets, a proper subset can then be defined using logical equivalence:

A ⊂ B <=> A ⊆ B ᴧ A ≠ B

Example
Let 𝓤 = {1,2,3,4,5,6,x,y,{1,2},{1,2,3},{1,2,3,4}}, where x,y are the 24th and 25th lowercase letters of the English alphabet and do not represent anything else. Then |𝓤| = 11.
If A = {1,2,3,4} ∈ 𝓤, then |A| = 4. So then:
Since A ≠ 𝓤, A ⊆ 𝓤 => A ⊂ 𝓤.
Since A is just one of the subsets of 𝓤, and {A} ≠ 𝓤, {A} ⊆ 𝓤 => {A} ⊂ 𝓤, however as it can be seen from the elements in the universe, {A} ∉ 𝓤.
If B = {5,6,x,y,A} = {5,6,x,y,{1,2,3,4}}, then |B| = 5. So then:
A ∈ B and since {A} ≠ B, {A} ⊆ B => {A} ⊂ B.
However, the set of the set of A, or {A}, is not in the set B, where {A} ∉ B.
Additionally, A ⊈ B, because A ≠ {A}, and so A ⊄ B either.

Subsets and Law of Syllogism
Let A,B,C be sets in a universe 𝓤.

Then, if A ⊆ B and B ⊆ C, then by the Law of Syllogism, A ⊆ C.

If A ⊆ B and B ⊂ C, then by the Law of Syllogism, A ⊂ C.

If A ⊂ B and B ⊆ C, then by the Law of Syllogism, A ⊂ C.

If A ⊂ B and B ⊂ C, then by the Law of Syllogism, A ⊂ C.

Example
To prove that A ⊆ C given that A ⊆ B and B ⊆ C, every x ∈ 𝓤 needs to be verified that if x ∈ A, then x ∈ C. Begin with an element x in A. Since A ⊆ B, then x ∈ A implies that x ∈ B. Since B ⊆ C, then x ∈ B implies that x ∈ C. Therefore, x ∈ A implies that x ∈ C. By the Law of Syllogism, A ⊆ C.

Example
To prove that A ⊂ C given that A ⊂ B and B ⊆ C, A ⊂ B implies that there exists an element b ∈ B such that b ∉ A. Since B ⊆ C, then b ∈ B implies that b ∈ C. Therefore, A ⊆ C and there exists an element b ∈ C and b ∉  A, so A ⊂ C.

Example
Let 𝓤 = {1,2,3,4,5}, A = {1,2,3}, B = {3,4}, and C = {1,2,3,4}. Then the following is true:
A ⊆ C: The elements in A are also elements of C.
A ⊂ C: The elements in A are also elements of C, yet C has elements that are not in A.
B ⊂ C: The elements in B are also elements of C, yet C has elements that are not in B.
A ⊆ A: The elements in A are also elements of A.
B ⊈ A: Not all elements in B are in the set A.
A ⊄ A: Since A = A, there are no elements that are in A that are not in A already.

Empty Sets
The empty set, or null set, is the unique set containing no elements, denoted as ∅ or { }.

Note that |∅| = 0, yet {0} ≠ ∅, because |{0}| = 1 ≠ |∅| = 0.

Also note that ∅ ≠ {∅}, because |{∅}| = 1 ≠ |∅| = 0.

For any universe 𝓤, let A ⊆ 𝓤. Then ∅ ⊆ A, and if A ≠ ∅, then ∅ ⊂ A.

Example
To prove that for any universe 𝓤, A ⊆ 𝓤, then ∅ ⊆ A, or ∀x {x ∈ 𝓤 | x ∈ ∅ => x ∈ A}. Assume that ∅ ⊆ A is false, where ∅ ⊈ A, or ∃x {x ∈ 𝓤 | x ∈ ∅ ᴧ x ∉ A}. Then there exists an element x in 𝓤 such that x ∈ ∅ but x ∉ A. However, x ∈ ∅ is impossible, so the assumption that ∅ ⊈ A is rejected and find that ∅ ⊆ A. Additionally, if A ≠ ∅, or ∃x,y [(x ∈ ∅ ᴧ x ∉ A) ᴠ (y ∈ A ᴧ y ∉ ∅)], then it is true that there is an element y ∈ A and y ∉ ∅ for A ≠ ∅ to be true, and so ∅ ⊂ A.

Power Sets
If A is a set from a universe 𝓤, the power set of A, denoted as ℘(A) or 2ᴬ, is the set of all possible subsets of A.

Power Sets and Cardinality
For any finite set A, where |A| = n ≥ 0, then it is found that A has 2ⁿ possible subsets. Therefore, ℘(A) = 2ⁿ = 2ˡᴬˡ.

The following example uses the method of set theory to solve instead of combinations.

Example
The following is the xy-plane that shows two possible paths from (2,1) to (7,4), where each path is made up of individual steps going one unit to the right (R ) or one unit upward (U).

Notice there is always a total of three units upwards and a total of five units right.
For figure (a), it lists R, U, R, R, U, R, R, U. For figure (b), it lists U, R, R, R, U, U, R, R.
In total, there are 8 possible steps to get from (2,1) to (7,4), which forms the set {1,2,3,…,8}. Figure (a) shows that the path has its upward moves located in positions 2, 5, and 8 in the list. This forms the set {2,5,8}. For figure (b), the path has its upward moves located in positions 1, 5, and 6 in the list, which forms the set {1,5,6}.
If there was a set {1,3,7} of {1,2,3,…,8}, then the corresponding list for this set would be U, R, U, R, R, R, U, R.
As a result, the number of possible paths from (2,1) to (7,4) is the number of 3-element subsets of {1,2,3,…,8}, because every subset has a cardinality of 3. Then there are C(8,3) = 8!/(3!5!) possible subsets.
If instead of considering the subsets of the upward positions, but consider the rightward positions, the subset would then have a cardinality of 5. Again, there are C(8,5) = 8!/(5!3!) possible subsets.

Example
Consider the following combinatoric identity for integers n,r where n ≥ r ≥ 1:

Instead of proving this identity using algebra or hypothetical situations, it will be proven using set theory and combinatorics.
Let A = {x,a₁,a₂,…,an}. The cardinality of A is then n + 1 (because there is an x). Consider all the subsets of A that only contain r elements, where |A| = r. Then there are C(n + 1,r) possible subsets.
Each of these subsets fall into two cases, which is the r-element subsets that contain the element x and the r-element subsets that do not contain the element x.
To obtain an r-element subset C that contains the element x, where x ∈ C and |C| = r, then put the element x in C and then select the remaining r – 1 elements. One is subtracted from r, because the cardinality of C is not r + 1 (r-elements + x element). The cardinality of C is just r. Since x is already selected, there are n options left (a₁,a₂,…,an) to form C. This can be done in C(n,r – 1) ways.
To obtain an r-element subset B that does not contain the element x, where x ∉ B and |B| = r, then the r number of elements of B are selected from {a₁,a₂,…,an}, which has a cardinality of n. Therefore, there are C(n,r) ways to form the set B.
By the rule of sum, it shows that C(n + 1,r) = C(n,r – 1) + C(n,r).

Pascal’s Triangle and Pascal’s Identity
The following is Pascal’s Triangle:

The two triangles represent the pattern where the top two binomial coefficients add up to a sum of the bottom binomial coefficient, where C(3,0) + C(3,1) = C(4,1). This can be easily seen when each binomial coefficient is replaced with its numerical value:

It is clear now that C(3,0) + C(3,1) = 1 + 3 = 4 = C(4,1).

This pattern is known as Pascal’s Identity, which was proven in the previous example:

Standard Sets
The following are standard sets that are commonly used as universes in mathematics:

ℝ = the set of all real numbers = {x ∈ ℝ}
ℝ* = the set of real, nonzero numbers = {x ∈ ℝ ᴧ x ≠ 0}
ℂ = the set of complex numbers = {x + yi | x,y ∈ ℝ ᴧ i² = -1}
ℤ = the set of all integers = {0,1,-1,2,-2,3,-3,…}
ℤ⁺ = the set of all positive integers = {1,2,3,4,…} = {x ∈ ℤ | x > 0}
ℚ = the set of all rational numbers = {x/y | x,y ∈ ℤ ᴧ y ≠ 0}
ℚ⁺ = the set of all positive rational numbers = {x/y | x,y ∈ ℤ⁺}
ℚ* = the set of all nonzero rational numbers = {x/y | x,y ∈ ℤ ᴧ x,y ≠ 0}

PDF reference: 155

Page 134. Question 1.

All of the sets are equal.

Page 134. Question 3.

a) True, because 1 is a member of the set A.
b) False, the set of 1 is not a member of the set A.
c) True, the set of 1 is a subset of the set A.
d) False, the set of the set of 1 is not a subset of A, because {1} ∉ A.
e) True, the set of 2 is a member of the set A.
f) True, the set of 2 is a subset of the set A.
g) True, the set of the set of 2 is a subset of A, because {2} ∈ A.
h) True, the set of the set of 2 is a proper subset of A, because no other elements in A besides {2} are members of {{2}}.

Page 134. Question 7.

a)

A ⊂ B
<=> A ⊆ ᴧ A ≠ B
<=> ∀x [x ∈ A → x ∈ B] ᴧ ∃x [x ∈ B ᴧ x ∉ A]

Note that only one of the statements from the disjunction is needed to define A ≠ B.

b)

A ⊄ B
<=> ¬[∀x [x ∈ A → x ∈ B] ᴧ ∃x [x ∈ B ᴧ x ∉ A]]
<=> ∃x ¬[¬(x ∈ A) ᴠ x ∈ B] ᴠ ∀x ¬[x ∈ B ᴧ x ∉ A]
<=> ∃x [¬¬(x ∈ A) ᴧ ¬(x ∈ B)] ᴠ ∀x [¬(x ∈ B) ᴠ ¬(x ∉ A)]
<=> ∃x [x ∈ A ᴧ x ∉ B] ᴠ ∀x [x ∉ B ᴠ x ∈ A]

Page 134. Question 9.

a) All of the subsets of a set A are proper subsets of A, except one subset C, where C = A, C ⊆ A.

Therefore, this one subset that isn’t a proper subset of A is removed. Then:

|℘(A)| – |{C}| = 2ˡᴬˡ - 1 = 63
2ˡᴬˡ = 64
|A| = 6

b) Since there’s an equal amount of even and odd cardinality subsets, 2|P(B)| = 128 = 2ⁿ => n = 7 = |A|.

Page 134. Question 11.

Let the set S = {penny, nickel, dime, quarter, half-dollar}. Then |S| = 5.

a) The total number of ways she can leave a tip is the total number of possible subsets of S, which is |P(S)| = 2⁵. However, this includes not giving any of her coins, which is not allowed. Therefore, the set of the empty set ∅ is subtracted, where there are a total of |P(S)| - |{∅}| = 2⁵ – 1 = 31 ways to tip at least one coin. The set of the empty set symbolizes that no coins were given for tips.

Using combinations, there a total of five cases:

Case 1: She tips just 1 coin. C(5,1) ways.
Case 2: She tips 2 coins. C(5,2) ways.
Case 3: She tips 3 coins. C(5,3) ways.
Case 4: She tips 4 coins. C(5,4) ways.
Case 5: She tips all 5 coins. C(5,5) ways.
The total sum of these combinations is 31.

**b) This time, not only is the restriction such that she must give at least one coin, the new restriction is she doesn’t give all of the coins away.  The total number of ways she can leave a tip is the total number of possible subsets of S, which is |P(S)| = 2⁵. However, since she must keep at least one coin, the possibility of having no coins is avoided, and so the set of the empty set does not need to be subtracted.

*not completed*

c) Let C = {penny, nickel} ⊆ S. Since all of the possible subsets of C will never add up to 10 cents, it is subtracted from the total amount of possible tip combinations.

|P(S)| - |P(C )| = 2⁵ – 2² = 28 ways to tip at least 10 cents.

Note that the set of the empty set is not subtracted, because she will have at least a penny or nickel with her.

Page 134. Question 13.

a) If |A| = 5, and |S| = 30, there is a total of C(30,5) ways to form the set A.

b) If |A| = 5, and 5 is the smallest element already chosen, assuming no repetition of 5, there is a total of C(25,4) ways to form the set A. The subset {1,2,3,4,5} was subtracted from S, and there are four elements left to be selected.

c) Assuming no repetition of the smallest element:

Case 1: The smallest element is {1}: C(29,4) (29 elements greater than 1, 4 elements left to select)
Case 2: The smallest element is {2}: C(28,4) (28 elements greater than 2, 4 elements left to select)
Case 3: The smallest element is {3}: C(27,4) (27 elements greater than 3, 4 elements left to select)
Case 4: The smallest element is {4}: C(26,4) (26 elements greater than 4, 4 elements left to select)

Therefore, there is a total of C(29,4) + C(28,4) + C(27,4) + C(26,4) ways to create the set A.

Page 135. Question 21.

Let S = {1,2,3,4,…n}. Then |S| = n, and applying Pascal’s Identity, the total number of ways to form a subset with five elements is then C(n,5). Consider all of the 5-element subsets that contain 7. To obtain a subset A such that it contains the element 7, assume 7 is already chosen in the subset, leaving n – 1 possible selections to choose from, with 4 remaining elements to chose, making a total of C(n – 1,4) ways to create a subset of S such that 7 is in the subset.

Since the question states that one quarter (¼) of these subsets contain the element 7, then the following is true:

Lab Example *not completed

Let A = {1,2,…,7}. Find a) the number of subsets of A, b) the number of nonempty subsets of A, c) the number of subsets of A with even elements, d) the number of subsets of A with odd elements, e) the number of subsets of A containing 1 but not containing 2.

For a), |℘(A)| = 2⁷

For b), since there is only one empty set for A, the number of nonempty subsets of A is 2⁷ - |{∅}| = 2 ⁷ – 1.

For c) and d), since there is the same amount of odd and even subsets in A, the number of even subsets of A is 2⁷/2 and the number of odd subsets of A is 2⁷/2.

For e), consider the set which does not contain either 1 or 2: B = {3,4,5,6,7}. Then |B| = 2⁵.

Then |B| + |{1,3,4,5,6,7}| = 2⁵ + 1 is the number of subsets of A that contain 1 but does not contain 2.

one of the most ridiculous arguments people make in favor of uncritically using proper grammar 100% of the time is that it helps make communication clearer.

like sure, sometimes, but only when it’s in the appropriate context?? all language use is context-specific. there are contexts where using strictly standard proper English not only isn’t the clearest form of communication, but actively HINDERS clear communication.

like when people on tumblr, or any other internet media really, completely refuse to use “internet english”… not only does it seem context inappropriate, but there’s also just things that can’t be said. how do you translate the different meanings of “omg” and “OH MY GOD” into proper english? what about bb? idgaf? the difference between a single word sentence that ends in a period and one that doesn’t?? how can you even begin to express all the shades of meaning and functionality that :/ has into proper english??? proper english just doesn’t have the creativity and multi-modality that internet english has, so you CAN’T express yourself as clearly, at least within this context.

obviously i have many, many beefs with the whole concept of “proper” english, but it’s just so ridiculous to me that people can look at the way internet speak works and complain that it’s a degenerate form of “correct” speech that’s “ruining” the way we talk, like wtf

anonymous asked:

Have you noticed any interesting correlations between demographics and the two political questions in your survey? The "catgirl effect" or anything else?

So, the first thing to note is that my political questions were set up as basically a political compass. They were social and economic freedom axes, and they were on a scale of 1 to 5, with 1 being furthest left and 5 furthest right (because English speakers read ordered lists left to right).

The first question was social freedom and had an answer of 1 being “The government should keep out of one’s private affairs” and 5 being “The government should regulate proper social and moral standards”. The second question was economic freedom and had an answer of 1 being “The government should have strong control over the economy” and 5 being “The government should avoid interfering with the economy”. Answers from 2 to 4 were taken as intermediate.

An interesting fact about the data as a whole: From just a glance at the bar graph for the social freedom question, it’s obvious that my readership leans socially liberal really hard. The most common answer was 1, and 1 and 2 together account for 89% of my responses. Which is… wow. Like, I’m not surprised, but I’m also definitely not appealing to conservatives.

Meanwhile, the data on economic freedom was way more balanced. Since it was only a 5 point scale, I can’t really be sure, but it certainly looks normally distributed about a point only slightly right of centre. The modal and median answers are both ‘3′, while the mean is 3.33. My readership is only a little bit more conservative economically than one might expect by pure chance. Apparently, I have very broad appeal on this side.


So, what correlations to perform? Obviously, the first one is: Does support for social freedom correlate with support for economic freedom? The answer, of course, is no. Left/Right split continues to exist, as evidenced by the correlation coefficient being -0.251. However, the fact that it’s not much higher does prove that there are some general-libertarians or general-authoritarians about. Let’s go find them!

Firstly, we want to define what these things mean. For the purpose of this test, I defined someone as being on the Left if they answered between 1 and 2 for both social and economic freedom, as these were the two left-leaning answers. I defined someone as Right if they answered between 4 and 5 for both. I defined someone as Libertarian if they answered between 1 and 2 for social freedom and between 4 and 5 for economic freedom. I defined someone as Authoritarian if they answered between 4 and 5 for social freedom and between 1 and 2 for economic freedom. I defined everyone who answered 3 for one question and 2-4 for another as a Dirty Centrist. All else were uncounted.

When I actually ran the numbers, I found 3 Authoritarians and 0 actual Conservatives. Pretty much everyone is either a liberal, libertarian, or centrist - as could be expected based on how everyone is super into social freedom. After looking at that, I decided that I would simply run correlations on support for economic freedom, as that’s the only thing with notable variance.


Now, for the demographics! Here I had a little trouble finding groups large enough to find out meaningful information. However, I was able to get useful information from the following demographics:

Age: Basically what it says on the tin, but with the actual calculations being done using ages estimated as the expected middle of the age range selected on the survey.

How long they’ve followed me: Similarly to age, this one was given as a range, but I’m picking an arbitrary midpoint for calculatory reasons.

Log of time they’ve followed me: Natural log of the above, in case any effects are logarithmic.

How much they like the blog: Simple 1-5 scale. Easiest correlation.

Gender: Broken into Cisfemale, Cismale, Transfemale, Transmale, AFAB Other, and AMAB Other.

Nationality: Americans vs literally everyone else. (”Americans” is still bigger)

Religion 1: Atheists-and-agnostics vs Theists

Religion 2: Jews (religious and non-), ex- and current-Christians, and Others.

As only age, length of following, and blog-enjoyment are scalar, only those three will be reported as a correlation. Everyone else will just have their average value reported.

This is being done with a sample of 250-260 (based on what people did and didn’t skip) in Google Sheets. Using this, I can calculate Pearson’s r, but I can’t figure out the p values. I’m going to talk about results over 0.1 as if they matter and under 0.1 as if they don’t, but this is so not how that should be done. If anyone wants to help me do Real Statistics to this stuff, hit me up!


Finally, the results:

Firstly, there is no correlation among my readers between age and economic liberty. The Pearson’s r between the two is 0.038, which is literally nothing. (Social liberty had a moderate negative correlation (in keeping with stereotypes) of -0.197, but everyone was super into social liberty, so it basically doesn’t count.)

The correlation between people liking my blog and supporting economic liberty is also very weak (0.090). I wouldn’t read much into it. The weakness means that I suppose people who disagree with me really like hearing what I have to say anyway?

The amount of time people have spent following my blog has a more significant effect. The r here is 0.162 for linear time and 0.177 for log time, so I suppose the effect is relatively linear? Of course, what that effect is is uncertain. It could be that long term exposure to me helps turn people libertarian, or that libertarians are more likely to stick around long term. Whatever it is, it’s minor enough that I doubt it’s worth worrying about.

Now, the thing this ask focused on: Gender! For gender, we have the following values for average level of economic freedom support: Cisfemale = 3.0, Cismale = 3.5, Transfemale =  3.8, Transmale = 3.1,   AFAB Other = 3.0, and AMAB Other = 3.3.

So, yes! The Libertarian Transgirl stereotype is supported by the data! Being a transwoman is a +0.8 to economic freedom preference over being a ciswoman. Another intriguing result is that, while degree of economic freedom support varies about half a point by gender among AMABs, AFABs of all genders are pretty solidly centrist. All of these groups were within one point of each other, though, because the overall distribution is clustered near the centre.

Americans have an economic freedom preference of 3.3, while non-Americans have one of 3.4, which fits with the average of 3.33. I feel tempted to make a jab about America not being the land of the free anymore but, honestly, the numbers are so close that this is probably just noise. It does mean that my non-American followers aren’t significantly more left-wing, though - despite what you might expect for a bunch of Europeans and Australians :p

Theistic followers have an economic freedom preference of 3.2, while non-Theistic followers have on of 3.4. Again, pretty close to each other. Because religion has historically been a battle ground for social issues, I checked for differences in opinion there. However, among my followers, there seems to be complete agreement on social issues: Both theists and non- are at 1.7 (1.3 points to the left) on average.

People of a Christian background are at 3.3 for economic freedom; people of a Jewish background are at 3.3; and people of all other backgrounds are also at 3.3. Similarly, on social issues, everyone looks exactly the same.


Anyway, what the results show is that basically nothing matters. There is a moderate correlation between following my blog for a long time and being libertarian, and between being a transwoman and being a libertarian. But, otherwise, nothing of much importance is going on. My followers cluster hard near the centre for economic freedom and at the left for social freedom, and variance from that point is minimal and has little relationship with group-membership.

(If you want to take this survey and contribute to all the ~cool science~, you can do so here.)

Schizoglossia

Schizoglossia is a linguistic term used to refer linguistic insecurity or language complex about one’s mother language. It is common in societies where there are two language varieties and one is seen as incorrect and the other as proper. For example: Standard French versus Haitian French (often seen as a “creole”) or Standard American English versus AfroAmerican English. In these cases, one variety is seen as “bad” and its speaker might want to correct some uses that are representative of it for some of the prestigious one. Those negative attitudes usually make the speakers be ashamed of the language that doesn’t have prestige, either openely or indirectly (by using linguistic characteristics, such as pronunciation, of the other language).

The term was coined by Einar Haugen in 1962.

Reference: Einar Haugen. “Schizoglossia and the Linguistic Norm” Monograph Series on Languages and Linguistics. Number 15 - 1962

wsj.com
There Is No ‘Proper English’

Once we dispense with the idea that Standard English is “correct,” there are social and linguistic gains. Teaching the conventions of the language is vital but has long been bound up with two deeply mistaken beliefs: first that non-Standard dialects are “improper” English; and second, that literacy depends on bogus rules like the supposed ban on split infinitives (“to boldly go”).
People should not be stigmatized for the way they speak, and they certainly should not have stupid, made-up linguistic superstitions drilled into their heads.

There seems to be a growing trend in people being critical of those who criticize or question a strong lifter’s competition lift. “How can you even critique a 900+ deadlifter? DYEL BRAH?”

Yes. You can. And you should.

Make no mistake, a 900 DL with a soft lockout, a 1000+ high squat, or a 700+ TNG bench are all impressive feats. That is seriously strong, even if they aren’t done to proper standards.

So, why does powerlifting somehow find itself immune to criticism?

If an NFL WR makes an insane catch in the end zone but only plants one foot it isn’t a TD. It doesn’t matter that they are an amazing athlete doing what the rest of us only wish we could do. The rules say it isn’t a TD. No lift.

If Michael Phelps swam the fastest 100yd free relay split of all time but left the blocks 0.001 second too soon the whole performance is DQed. If the officials let it slide because “it was awesome and you can’t swim that fast” then that would be a disgrace to the Olympic Games or FINA. No lift.

So yeah, if you’re strong as balls but don’t lock out your deads in competition then you shouldn’t be exempt from people saying that your lift was not good.

“More important than petition is adoration, for in it truth will come to us — the truth of life. Everyday cares will find their proper place and our standards will become rightly adjusted. This truth will comfort us; it will put in order what the entanglements and illusions of life have thrown into confusion. It will heal us spiritually so that we may begin anew.”
(Fr. Romano Guardini; The Art of Praying)

If we are to know the Lord, we must go to Him. Listen to Him in silence before the tabernacle and approach Him in the sacraments.
(Pope Francis)