# The tutor mentions a few points about diatoms.

In my Feb 4 post I introduced protists, which constitute a kingdom of eukaryotic, mainly single-celled organisms. Protists are divided into plantlike and animal-like ones.

Diatoms, from phylum Chrysophyta, are among the golden brown algae. They are plantlike protists, perhaps the most numerous of them. In the oceans, their abundance makes them a major food source at the base of the ecosystem. Furthermore, they are prominent producers of oxygen on Earth.

Diatoms have a two-valve structure, rather like the base and lid of a box. They are well known for having glass (silica) in their cell walls, which show striking patterns under a microscope.

Diatoms have been even more abundant in the past; today, those fossils constitute diatomaceous earth, which is mined for applications such as soundproofing, filtration, and scouring powders.

HTH:)

Source:

Mader, Sylvia S. Inquiry into Life, 9th ed. Toronto: McGraw-Hill, 2000.

Ritter, Bob et al. Biology. Scarborough: Nelson Canada, 1996.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows the easy concept, from graph theory, of elementary subdivision.

An elementary subdivision on a graph replaces one edge by two, with a new vertex installed between them.

Consider the following two graphs:

## Graph 2

Graph 2 is produced from Graph 1 by an elementary subdivision. First, edge DA is removed, then vertex P is added between D and A. Finally, edges DP and PA are drawn.

Elementary subdivisions are used to compare graphs for similarities.

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills: Addison-
Wesley, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor defines cut set with a couple of examples.

When one imagines a cut set, one imagines removing edges from a graph, but not the vertices they connect.

In this context, we are thinking of undirected graphs.

A cut set is the minimum number of edges whose removal will disconnect a given portion of the graph. Each cut set is specific to a portion of the graph to be disconnected.

Consider the graph below:

To disconnect just the vertex A from the graph, edges AB and AE need to be removed. The corresponding cut set is {AE, AB}. This may also be written {{A,E},{A,B}}, meaning “[the set containing] the edge from A to E or E to A, and the edge from A to B or B to A”.

{DP}, or {{D,P}}, is also a cut set: removing the edge DP separates the two cyclic portions of the graph.

When we disconnect one part of a graph from the rest, we say the number of components of the graph has increased by 1. A graph’s number of components is its number of isolated parts. Before any cut sets are applied, a connected graph (such as the one above) has 1 component: κ(graph)=1. A component, within itself, can be called a connected component.

HTH:)

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills: Addison-
Wesley, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor reviews the progress of thought about atoms through the ages.

Democritus, a Greek philosopher living around 400 BC, is credited as the first promoter of the concept of the atom. He proposed that all matter consists of tiny, indivisible particles, which he called atoms.

The next published refinement of Democritus’s idea was Dalton’s atomic theory of 1808. He agreed with Democritus, elaborating the idea of an element – a type of atom. Every element has its unique type of atom.

After further study came the Thomson model, in common use from the 1890s to 1911. It was also called the “plum pudding model”, and proposed that an atom consists of positively charged “dough” with negatively charged “raisins” scattered throughout. Thomson, who discovered the electron in 1897, realized that the electrons should be moving.

Rutherford, in 1911, carried out an experiment whose results led to the idea of the nucleus. The Rutherford model claims that the center of an atom is heavy and positive, while electrons orbit distantly around it. Most of the “volume” of an atom is empty space between the nucleus and the electrons.

The Bohr model (1913) refined the Rutherford model, suggesting that the electrons orbit at only certain distances (levels) from the nucleus, and that they can jump from one level to another. The levels are called shells.

Rutherford’s model is still used until high school; in grades 11 and 12, Bohr’s is studied.

In a future post I’ll explain Rutherford’s experiment and why it led to such a breakthrough.

Source:

Giancoli, Douglas C. Physics. New Jersey: Prentice Hall, 1998.

Bullard, Jean et al. Science Probe 10. Scarborough: Nelson Canada, 1996.

britannica.com

texasgateway.org

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor offers a solution to a textbook graph theory question.

On page 556 (Grimaldi) is the following question:

Let G be a loop-free connected undirected 3-regular graph (every vertex has degree 3), such that

|E| = 2|V| – 6

Find |V| and |E|.

Solution:

First, |V|= number of vertices; |E| = number of edges. The degree of a vertex means how many edges are incident with it.

We know, for any undirected graph, the sum of the degrees of the vertices is equal to twice |E|:

Σ(deg) = 2|E|

In our case, the graph is 3-regular, so Σ(deg) = 3|V|. Then

3|V| = 2|E|

3|V|/2 = |E|

Given originally in the question,

|E| = 2|V| – 6

which means

3|V|/2 = 2|V| – 6

Multipltying by 2 on both sides gives

3|V| = 4|V| – 12

Now, subtracting 4|V| from both sides yields

-|V| = -12

Finally, multiplying both sides by -1,

|V| = 12

Then

|E| = 2|V| – 6 = 2(12) -6 = 18

So, the graph has 12 vertices and 18 edges.

HTH:)

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills: Addison-
Wesley, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor gives an example with friction force.

I’ve never been curling. Even so, I can imagine the following question resonates with many curlers and spectators:

A curling stone is released at 1.2m/s. If the coefficient of friction between the ice and the stone is 0.010, how far will it travel before coming to rest?

Solution:

From my previous post, the force of friction is calculated by

Ff = μFN

In this case, the stone is on level ice, so

Ff = μFN = μmg

The force of friction is the only unbalanced force acting on the stone, so its acceleration is due only to friction. From Newton’s Second Law,

a = F/m = μmg/m = μg = 0.010(9.8) = 0.098m/s2

Since the acceleration opposes the velocity, its sign is negative:

a = -0.098m/s2

Now, we can calculate the distance the stone travels using kinematics:

where v1= initial velocity = 1.2m/s, v2 = 0m/s

Therefore,

02=1.22 + 2(-0.098)d

0 = 1.44 – 0.196d

We add 0.196d to both sides:

0.196d = 1.44

We divide both sides by 0.196:

d = 1.44/0.196 = 7.3469…m (7.3m in sig figs)

Apparently the stone will glide 7.3m before stopping.

Source:

Heath, Robert W et al. Fundamentals of Physics. D.C. Heath Canada Ltd., 1981.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shows a basic example of calculating friction.

The force of friction, Ff, on a flat surface is given by

Ff = μFN

where

μ = coefficient of friction (often looked up from a table, or given)

FN = normal force: pushing force from surface to object at 90° to surface

Nearly always, FN = mg, where

m = mass of object

g = acceleration due to gravity (typically 9.8m/s2 on Earth)

Example 1: Calculate the force of friction between the tires and the road, dry conditions (μ = 0.40), for a 1400 kg car.

Solution:

Ff = μFN = μmg = 0.40(1400)(9.8)= 5488N or 5500N in sig figs

I’ll be covering more about friction in coming posts:)

Source:

Heath, Robert W. et al. Fundamentals of Physics. D.C. Heath Canada Ltd., 1981.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor attempts to strip the pigeonhole principle to bare wires with a simple problem.

I discussed the pigeonhole principle in my May 23/14 and Oct 29/14 posts. The concept is that, with n+1 pigeons going to n pigeonholes, one hole must host at least two pigeons. Apparently simple, the concept can be used to solve some surprising problems.

Example:

Imagine selecting a set of seven numbers from the Naturals {1,2,…,10}. Prove that, among those chosen, one of the numbers must exceed another by exactly 3.

Proof:

Consider the numbers chosen:

1 ≤ x1 < x2 < x3 < …< x7 ≤ 10

Next, add three to each of the numbers chosen:

x1 +3 < x2+3 < x3+3 <….< x7+3 ≤ 13

Both are sets of distinct numbers. If the second set and the first are disjoint (have no common members), then the sets together total 14 distinct numbers. These are the pigeons.

However, neither set of numbers can have values outside {1,2,…,13}. These possible values are the pigeonholes.

With 14 numbers, but only 13 possible values, one of the numbers from the second set must share the same value as one in the first. Therefore, xi+3=xm for some i, m. By the pigeonhole principle, one of the original set’s members, with three added, equals another of its members.

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills: Addison-
Wesley, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor defines, in the context of symbolic logic, contradiction, with a couple of examples.

For those new to symbolic logic, my previous post leads back to others that will fill the gaps.

A contradiction is a compound statement that is always false. The fundamental example is p ∧ ¬ p, which must be a contradiction, since both p and “not p” can’t be true simultaneously.

(p ⊻ q) ∧ (p ∧ q)

p ⊻ q is only true when one of p,q is true, but not both. However, both p,q must be true for p ∧ q to be. Therefore, the bracketed statements can’t be true simultaneously, meaning that the central “and” will always be false.

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills: Addison-
Wesley, 1994.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor follows up about tautology with another example.

For grounding about the symbols, etc, readers may want to refer to my Feb 12 post.

In my Feb 13 post I defined tautology with a simple example. Today, I’ll give a more involved one.

A tautology is a compound statement which is always true. Consider the following example:

(¬p ∨ q) ⊻ (p ∧ ¬q)

The central sign ⊻ is exclusive or; it’s true if one of the inputs is, but not both.

Let’s imagine both p, q false (0). Then the first bracket is true, the second false, so the entire statement is true.

If p is true, q false, then first bracket is false, but the second bracket is true. Once again, the full statement is true.

If q is true, then the first bracket is true, the second false: the full statement is true.

Source:

Grimaldi, Ralph P. Discrete and Combinatorial Mathematics. Don Mills: Addison-
Wesley, 1994

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.