UBC Wiki - User contributions [en]

Course:MATH110/Archive/2010-2011/003/Teams/Thurgau

2011-02-04T07:05:21Z

MatthewJohnson:

{{
Infobox MATH110 Teams
| team name = Thurgau
| member 1 = Amir Mohamad Nazri
| member 2 = Ernest Tsui
| member 3 = Matt Johnson
| member 4 = Miguel Caruncho
}}
In workshop M.

[[Course:MATH110/003/Teams/Thurgau/Homework 11|Homework 11]] 
[[Course:MATH110/003/Teams/Thurgau/Homework 12|Homework 12]] 
[[Course:MATH110/003/Teams/Thurgau/Homework 13|Homework 13]]

Ernest - Incompetent 
Amir - Sleep 
Miguel - Hermit 
Matt - Heroically returned

Course:MATH110/Archive/2010-2011/003/Teams/Thurgau/Homework 13

2011-02-04T07:04:50Z

MatthewJohnson:

For this homework, we chose to investigate on the apparent magnitude of stars. For those unfamiliar with this term, the apparent magnitude of stars is basically the brightness of the star to an observer on earth, normalizing its value to what it should be if there was no atmosphere. Originally the way people measured the apparent brightness of stars was through the self-reported perception of observers on a magnitude scale of 1 to 6, with 6 being the faintest and 1 being the brightest. This was the earliest way that people measured apparent magnitude and it was said that each level of magnitude was twice as bright as the next grade, or in essence, a logarithmic scale.

A logarithmic scale is a scale that shows differences in magnitudes or scales through the logarithms of values, instead of actual values themselves. This way of presenting data is particularly useful for representing data that is over a very large range of values or for very uniformly patterned information (such as apparent magnitude, richter scale, etc.).

As the apparent magnitude increases, the stars get dimmer and dimmer. As for very bright stars their apparent magnitudes are less than 1, assigning them negative numbers. These negative values have been created to accommodate the new modern system that has gone beyond the 1-6 scale. For example, the start Sirius, the brightest star in the celestial sphere has a magnitude of -1.4 and the moon has one of -12.74. Out sun, Sol is at a magnitude of -26.74. The Hubble space telescope has even been able to find stars with the magnitude of 30. The difference between 1 magnitude to another is by a factor of 2.512. As a result, using the scale 1 - 6, as the values increase across the values, the brightness of the star decreases by a factor of 2.512. Hence, the difference between a star with a magnitude of 1 and a star with a magnitude of 6 is 100. Since a star with a magnitude of 6 is 5 steps larger on the magnitude scale, it is measured as (2.512)^(5), which is 100. Logarithms are essential towards the study of Astrophysics. The following is an example from a physics textbook of the apparent brightness of stars:
 (example)

{| class="wikitable"
!style="background: #f2e0ce; text-align: left; padding:3px;"| '''In the case of Absolute magnitudes, it is written as: '''
|-
|style="background: #fffaf5; padding:12px;"|
In the case of Absolute magnitudes, it is written as:
 ''' m-M=5log(d/10)''' where;
 '''M = Absolute Magnitude'''
 '''m = Apparent Magnitude'''
 '''d = distance (measured in parsecs)'''
 To calculate the Absolute Magnitude of a star (Brightness of a star seen from 10 parsecs), the equation can be rearranged to:
 '''M = -5log(d/10)+m '''
The following is how this equation is applied:
 
The star Thurgau is roughly 3.1 parsecs from Earth and an apparant magnitude of 3.6, so what would its Absolute Magnitude be?
 
M = -5log(.31)+3.6
 
M = -5(-4.1086)
<br?
M = 20.5432

|}

Course:MATH110/Archive/2010-2011/003/Groups/Group 16

2010-10-20T04:44:44Z

MatthewJohnson: /* Homework 4 - due Wednesday October 20 */

{{Infobox MATH110 Groups
| group number = 16
| member 1 = [[User:ShamillaBirring|Shamilla Birring]]
| member 2 = Matt Johnson
| member 3 = [[User:ArabellaCynthiaOlomide|Arabella Olomide]]
| member 4 = [[User:GaiaSilvestri|Gaia Solvestri]]
| member 5 = [[User:ChristopherWong|Christopher Wong]]
| member 6 =
}}

===Contact Information===
* Shamilla Birring | shamilla_birring@hotmail.com
* Arabella Olomide | aco2@sfu.ca

-----

== Homework 4 - due Wednesday October 20 ==

'''Problem 1'''
Five persons named their pets after each other. From the following clues, can you decide which pet belongs to Suzan's mother?

-Tosh owns a cat,

-Bianca owns a frog that she loves,

-Jaela owns a parrot which keeps calling her "darling, darling",

-Jun owns a snake, don't mess with him,

-Suzan is the name of the frog,

-The cat is named Jun,

-The name by which they call the turtle is the name of the woman whose pet is Tosh,

-Finally, Suzan's mother's pet is Bianca.

First it is important to state all the information we know:

Pet owners:

-Tosh (male)

-Bianca (female)

-Jaella (female)

-Jun (male)

-Suzan (female)

Note: we are able to conclude the owners genders by looking for gender indicating words.
For example in the phrase "Bianca owns a frog that ''she'' loves", we can can conclude that Bianca is a female because it states that SHE loves her frog.

Pet owners and their pets: (Note: the pets names are in the parentheses)

Tosh -> cat (Jun)

Bianca -> frog (Suzan)

Jaella -> parrot

Jun -> snake

Suzan -> turtle

Note: The only pet not assigned to an owner is the turtle, therefore it is logical that the turtle is Suzan's pet

Note: Keep in mind that Suzan's mother's pets name is Bianca

We can make some further conclusions from the information provided in the question.
Conclusions:

1)

-The name of the turtle will be a female owner of the pet Tosh.
This is concluded from the phrase "The name by which they call the turtle is the name of the woman whose pet is Tosh"

-> because it is a woman, we recognize that the turtles name will be a females name

-> the turtle will have the name of the pet Tosh's owner

In effect of conclusion 1, we can recognize the name of Suzans turtle to be Janella. This can be done because all the friends names have been used as pets names except; Janella and Tosh. Therefore since the turtles name must be a females name we can conclude the turtles name is Janella.

Following we can conclude that Janella's pet parrot's name is Tosh since it states that "The name by which they call the turtle is the name of the woman ''whose pet is Tosh''".

Therefore adding these new discoveries to our previous knowledge we can update our information.

Pet owners and their pets: (Note: the pets names are in the parentheses)

Tosh -> cat (Jun)

Bianca -> frog (Suzan)

Jaella -> parrot (Tosh)

Jun -> snake

Suzan -> turtle (Janella)

Therefore, since the snake is the only pet that has not been named and the name Bianca has not been assigned, we can conclude that Jun's snakes name is Bianca.
Following, it is imperative to remember the Bianca is the snake,meaning the snake is the pet that belongs to Suzan's mother.

Hence, the snake named Bianca is Suzan's mother's pet.

'''Problem 2'''
Bohao, Stewart, Dylan, Tim and Chan are the five players of a basketball team. Two are left handed and three right handed, Two are over 2m tall and three are under 2m, Bohao and Dylan are of the same handedness, whereas Tim and Chan use different hands. Stewart and Chan are of the same height range, while Dylan and Tim are in different height ranges. If you know that the one playing centre is over 2m tall and is left handed, can you guess his name?

Based on the clues for handedness, we know that one of tim or chan must be the same handedness as Bahao and Dylan, who must be right handed. This means that Stewart must be a Lefty.
Based on the clues for height, we know that Dylan or Tim must be the same height as Stewart and Chan, who must be under 2m tall. This means that Bahao must be over 2m tall

The only lefty we know is under 2m, and the only person we know over 2m is a righty. We know that dylan or tim must be over 2m, but Dylan is a righty. So the only way their is a lefty over 2m is if it is Tim

His name is Tim

'''Problem 3'''

This problem is quite difficult to map out so I will phrase it exactly how I did while finding solutions because that is what worked for me. In this exercise, the more we know about a particular person the better; however it could be possible that one person be suitable for more than one position and vice versa.

Adam: not a catcher, battery or infield, not shortstop

Bobo: not a catcher, not 3rd baseman, not shortstop, not 2nd baseman

Charles: not pitcher, could be outfielder, battery or infield

Ed: not 2nd baseman, not outfielder, not catcher, not 3rd baseman, bachelor, could be shortstop or 1st baseman

Hassan: not 3rd baseman, could be battery or infielder, not catcher, not 2nd baseman, could be either 1st baseman, shortstop, pitcher

Jason: not 3rd baseman, not catcher, bachelor. '''Jason is the 2nd baseman.'''

Matthieu: not centre field, shortest in height of the team

Pascal: not pitcher, could be outfielder, not shortstop, not catcher, not 2nd baseman, 3rd baseman, not right fielder, not centre fielder, bachelor. '''Pascal is the left outfielder.'''

Sung: married, not right fielder, not centre fielder, not left fielder, not 2nd baseman.

''Important notes'':

There are 5 bachelor, 4 married men, the catcher, the 3rd baseman and the pitcher are married.

'''Problem 4'''

'''Problem 5'''

Homer finally had a week off from his job at the nuclear power plant and intended to spend all nine days of his vacation (Saturday through the following Sunday) sleeping late. But his plans were foiled by some of the people who work in his neighbourhood.
On Saturday, his first morning off, Homer was wakened by the doorbell; it was a salesman of magazine subscriptions.
On Sunday, the barking of the neighbour's dog abruptly ended Homer's sleep.
On Monday, he was again wakened by the persistent salesman but was able to fall asleep again, only to be disturbed by the construction workers next door.
In fact, the salesman, the neighbour's dog and the construction workers combined to wake Homer at least once each day of his vacation, with only one exception.
The salesman woke him again on Wednesday; the construction workers on the second Saturday; the dog on Wednesday and on the final Sunday.
No one of the three noisemakers was quiet for three consecutive days; but yet, no pair of them made noise on more than one day during Homer's vacation. On which day of his holiday was Homer actually able to sleep late?

From the question, we are able to extract the information of which noisemakers appearing on which days as listed below:

Saturday: Salesman

Sunday: Dog

Monday: Salesman, Construction

Tuesday: ?

Wednesday: Salesman, Dog

Thursday: ?

Friday: ?

Saturday: Construction

Sunday: Dog

Two conditions are also told to help us solve the question:

'''No one of the three noisemakers was quiet for three consecutive days'''

'''No pair of them made noise on more than one day during Homer's vacation.
'''

Well looking at the preliminary information, we can see that there are three days that are unknown where one of them is when no noisemakers exists and also the day that Homer can sleep in, in which this is the answer to the question.

In order to solve the problem, we can first go back to the conditions of the situation, we are reminded that “no pair of them made noise on more than one day during Homer's vacation” therefore looking at the existing outcomes we are able to see that the only possible outcomes left are:
'''
“Salesman” and “Dog, Construction”'''

Next we can make use of the other outcome which is “no one of the three noisemakers was quiet for three consecutive days”. Looking at Wednesday we see that, that is the last time we are told that the Salesman appears. Yet if we are to meet the condition then the Salesman would have to appear at least once more either Thursday or Friday or else this would make him quiet for three consecutive days, seeing how both Saturday and Sunday are already occupied.

The situation is the same for Dog and Construction. The days that Dog must appear in order to meet the condition is Thursday and Friday while for Construction it is Tuesday and Thursday. Already we have established that the possible outcomes left are Salesman and Dog/Constructions, from this we can be sure that the outcome Dog/Construction appears on Thursday as this is the only day they have in common. Therefore this makes the Salesman appear on Friday. The information can now be organized as shown below,

Saturday: Salesman

Sunday: Dog

Monday: Salesman, Construction

Tuesday: '''No Noisemaker'''

Wednesday: Salesman, Dog

Thursday: '''Dog, Construction'''

Friday: '''Salesman'''

Saturday: Construction

Sunday: Dog

From this, we are able to see that the only day that no noisemakers appeared was on Tuesday. Therefore, the only day that Homer was actually able to sleep in was on Tuesday.

== Homework 3 - due Wednesday October 13 ==

Thanks Shamilla,

That sounds like a great idea.

Arabella.

Hey guys,
since we did not discuss dividing up the 25 math problems assigned in class, I was thinking everyone could do 5 problems each, going in order by how the group members names have been listed. For instance, my name is first, therefore I would be assigned to complete the first 5 problems. Hopefully you guys agree, if not just let me know. Also I think it would be a great idea if we added our email addresses next to our names above.

[[User:ShamillaBirring|ShamillaBirring]]

NOTE: I plan on adding images... as soon as i figure out how to do so. If someone could explain how to do so, that would be helpful 
[[User:ShamillaBirring|ShamillaBirring]]

All you ever wanted to know about adding images can be found [http://wiki.ubc.ca/Help:Adding_Media#Inserting_an_Image_onto_a_UBC_Wiki_Page here]. 
--[[User:DavidKohler|DavidKohler]]

'''25 PROBLEMS'''

'''QUESTION 1''' -
A bus traveled from the terminal to the airport at an average speed of 30 mi/hr and the trip took an hour and 20 min. The bus then traveled from the airport back to the terminal and again averaged 30 mi/hr. However, the return trip required 80 min. Explain

BUS 1: Terminal Airport to Airport

BUS 2: Airport to Terminal Airport

Bus 1: Speed = 30mi/hr ; Time = 1hour 20mins

BUS 2: Speed = 30mi/hr ; Time = 80mins

By comparing the speed of both buses it is evident that Bus 1 and Bus 2 are travelling at the same speeds, however it is not so clear whether the time taken for each bus to reach their destination is the same. In order to make this more clear we must put both times in the same units.

Bus 1: Time = 1hour 20mins

1hour = 60 mins

Therefore 1hour 20mins = (60mins + 20mins) = 80mins

Hence, Bus 1 took 80mins to reach the airport (Bus 1 time = 80mins)

Now that both the bus times are in the same units we can compare the times

Bus 1: Time = 80mins

Bus 2: Time = 80mins

Therefore it is now evident that both buses took the same amount of time to travel the same distance at the same speed.

Hence, the return trip required 80 mins, because the bus continued to travel at the ''same speed'' of 30mi/hr and ''same distance'' (between the terminal airport and the airport) as the initial trip which required 80 mins.

'''QUESTION 2''' -
A lady did not have her driver's license with her when she failed to stop at a stop sign and then went three blocks down a one-way street the wrong way. A policeman saw her, but he did not stop her. Explain.

Information that can be extracted from the question:

1. Lady does not have her driver's licence with her

2. Lady did not stop at a stop sign

3. Lady went three blocks down a one-way street the wrong way

4. Policeman saw the lady, but did not stop her

After analyzing the question and the conditions, it is clear that it does not state in the problem that the lady was in a vehicle, which if the case she would have to have her driver's licence with her, stop at a stop sign, be prohibited from driving down a one-way street the wrong way and the policeman would be obliged to stop the her. Therefore, it can be understood that the lady is a '''pedestrian''' and therefore she does not need her drivers licence to walk, she is not obligated to stop at a stop sign, she can walk down a one-way street the wrong way and under these conditions the policeman has no reason to stop the lady.

Therefore, the lady is a pedestrian.

'''QUESTION 3''' -
One of three boxes contains apples, another box contains oranges, and another box contains a mixture of apples and oranges. The boxes are labeled APPLES, ORANGES and APPLES AND ORANGES, but each label is incorrect. Can you select one fruit from only one box and determine the correct labels? Explain.

It is understood from the problem that there are three boxes filled with either/or apples and oranges and the boxes have been incorrectly labeled APPLES, ORANGES, and APPLES AND ORANGES.

Therefore, say the boxes are labelled as described below:

Box 1: APPLES

Box 2: ORANGES

Box 3: APPLES AND ORANGES

Because the boxes have been labelled incorrectly this means that Box 1 does not contain only apples and can therefore only contain either "oranges" or "apples and oranges", this is the same situation for each box.

Hence,

Box 1: Labelled APPLES

Fruit options inside the box: Oranges ; Apples and Oranges

Box 2: Labelled ORANGES

Fruit options inside the box: Apples ; Apples and Oranges

Box 3: Labelled APPLES AND ORANGES

Fruit options inside the box: Apples ; Oranges

Now that we understand the conditions and situation of the questions we must focus on the question; Can you select one fruit from only one box and determine the correct labels?

Consider that you have box 3 which is labelled incorrectly as APPLES AND ORANGES. The fruit options for box 3 are "apples" or "oranges", if an apple were picked out of that box, this would mean that this box contains apples only. Therefore, box 2 which is labelled incorrectly to contain ORANGES and has the fruit options "apples" or "apples and oranges", by the process of elimination one knows that box 2 cannot contain apples since we have already discovered that box 3 contains apples, meaning that the only fruit option left for box 2 is "apples and oranges". Therefore, so far box 3 contains apples and box 2 contains apples and oranges, leaving box 1 with only oranges.

Therefore, with the example above we were able to pick out one fruit from a box and in effect conclude which box contained either apples, oranges, or both apples and oranges.

Hence, yes it is possible to select one fruit from only on box and determine the correct labels.

'''QUESTION 4''' -
I am the brother of the blind fiddler, but brothers I have none. How can this be?

Therefore, this question states that "this person" is a brother to their sibling (whom is not gender defined) and that "this person" does not have any brothers.
Hence, because this person has a sibling and has no brothers, one can assume that the sibling is a female. This assumption is accurate because "this person" is a brother to his sister whom is the blind fiddler, and can still have no brothers of his own.

Therefore, this statement "I am the brother of the blind fiddler, but brothers I have none" is true if the blind fiddler is a female and therefore the boy's sister.

'''QUESTION 5''' -
Two quarters rest next to each other on a table. One coin is held fixed while the second coin is rolled around the edge of the first coin with no slipping. When the moving coin returns to its original position, how many times has it revolved?

The conditions of the question:

1. two quarters

2. one quarter is fixed while the second coin is rolled around the edge of the first coin

3. there is no slipping

It is important to recognize that it is two quarters being used, meaning that they are the same size (have the same diameter, perimeter and area). Hence, for the second quarter to go around the first quarter it will only revolve once. A more simpler manner to picture this question is to imagine two identical squares (as opposed to the quarters) under the same conditions (have same width and length and therefore the same perimeter and area). Each sides of the second square will get to touch a side of the first square, while only revolving once.

Therefore, the second quarter will only revolve once in order to return to it's original position.

"QUESTION 10" -
Suppose one-half of all people are chocolate eaters and one-half of all people are women. (i) Does it follow that one-fourth of all people are women chocolate eaters? (ii) Does it follow that one-half of all men are chocolate eaters? Explain.

Being a man or a woman is independent of preference for chocolate.

i) No, it does not follow.
ii) No, it does not follow.

"QUESTION 11" -
A woman, her older brother, her son, and her daughter are chess players. The worst player’s twin, who is one of the four players, and the best player are of opposite sex. The worst player and the best player have the same age. If this is possible, who is the worst player?

The situation is not possible. The woman's daughter and son must be either the best or worst players. They are of opposite sex and they have the same age, since the woman is older than both of them (she is their mother) . The woman's older brother is the oldest player.

"QUESTION 12" -
A Manhattan fellow had a girlfriend in the Bronx and a girlfriend in Brooklyn. He decided which girlfriend to visit by arriving randomly at the train station and taking the first of the Bronx or Brooklyn trains that arrived. The trains to Brooklyn and the Bronx each arrived regularly every 10 minutes. Not long after he began his scheme the man's Bronx girlfriend left him because he rarely visited. Give a (logical) explanation.

For example, the Bronx train arrives at 12.00, 12.10. 12.20...and the Brooklyn train arrives at 12.09, 12.19, 12.29...

"QUESTION 13" -
If a clock takes 5 seconds to strike 5:00 (with 5 equally spaced chimes), how long does it take to strike 10:00 (with 10 equally spaced chimes)?

Considering the time between chimes: it takes 45/4 = 11.25 seconds to strike 10.00.

"QUESTION 14" -
One day in the maternity ward, the name tags for four girl babies became mixed up. (i) In how many different ways could two of the babies be tagged correctly and two of the babies be tagged incorrectly? (ii) In how many different ways could three of the babies be tagged correctly and one baby be tagged incorrectly?

i) There are six ways that two of the four babies can be correctly tagged.
ii) There are no ways in which three of the four babies correctly tagged.

"QUESTION 15" -
Alex says to you, “I'll bet you any amount of money that if I shuffle this deck of cards, there will always be as many red cards in the first half of the deck as there are black cards in the second half of the deck.” Should you accept his bet?

Alex is right, I shouldn't accept his bet. The deck of cards is made up of an equal number of black and red cards (respectively 1/2 of the total number of cards).

'''21. Sven placed exactly in the middle among all runners in a race. Dan was slower than Sven, in 10th place, and Lars was in 16th place. How many runners were in the race?'''

Looking at the question, we are the given positions of the runners as followed:

Sven: Middle place of all runners

Dan: 10th place

Lars: 16th place

From the details of the questions, we can first of all try to determine a range for which place Sven came in as we only know he is in the middle amongst all runners. First we know that Dan, who was slower than Sven came in 10th place thus we can assume that Sven came in places ranging from 1st to 9th. Taking into account that Sven is in the middle we can say that the maximum number of runners is 18 as the Sven can only be placed up to the 9th at max. Next, as we are told that Sven is placed exactly in the middle therefore we can deduce that the number of runners is an odd number, as an even number cannot produce a runner exactly in the middle. Lastly we take into account the placement of Lars who came in 16th, as mentioned earlier we deduced that the maximum number of runners is 18 therefore shortening the range to 16-18.

In conclusion, the only possible number that fits with all the criteria is 17; therefore there were 17 runners in the race.

'''22. During a vacation, it rained on 13 days, but when it rained in the morning, the afternoon was sunny, and every rainy afternoon was preceded by a sunny morning. There were 11 sunny mornings and 12 sunny afternoons. How long was the vacation?'''

'''23. Suppose you overhear the following conversation: Paul: How old are your three children? Paula: The product of their ages is 36 and the sum of their ages is the same as today's date. Paul: That is not enough information. Paula: The oldest child also has red hair. If you were Paul could you determine the ages of Paula's children? Explain.
'''
We are told that Paula has three children and that the product of their ages is 36, therefore our first step would be to find the factors of 36 in sets of 3. This gives us the following sets:

• 1,1,36

• 1,2,18

• 1,3,12

• 1,4,9

• 1,6,6

• 2,2,9

• 2,3,6

• 3, 3,4

Next we are told that the sum of their ages is the same as a date, therefore we know the sum cannot exceed 31 as that is the maximum number of days in a month. By calculating the sums of each set of factors we are given the following:

• 1+ 1+36 = 38

• 1+2+18= 21

• 1+3+12 = 16

• 1+4+9 = 14

• 1+6+6 = 13

• 2+2+9= 13

• 2+3+4 = 11

• 3+3+4= 10

From this we see that we can eliminate the set 1,1,36 as its sum is greater than 31.
Next we are told that the information is still not enough, the only reason for this may be that there are sums of the same number thus making Paul unable to determine which the correct set is. Looking at the sums we can see that the number 13 is duplicated making 1,6,6 and 2,2,9 possible answers.

Lastly, we are told that the oldest child has red hair, the clue here is that there is a eldest sibling, ruling out the set 6,6,1 as there is not one single elder sibling. Therefore, this we can determine the ages of Paula’s children to be 9,2 and 2.

'''24. Two candles of equal length were lit at the same time. One candle took 6 hr to burn out and the other candle took 3 hr to burn out. After how much time was one candle exactly twice as long as the other candle?
'''
From the details of the question we are given the following conditions:

1.) Two candles are of equal length

2.) One candle took 6hr to burn out (we can call this Candle A)

3.) One candle took 3hr to burn out (we can call this Canlde B)

Firstly to make this question more clearly, we can give the candles an actual length giving that both candles are of equal length. Let’s assume that the candle length is 60cm.Taking into account the time it takes each candle to burn out we can come up rate of how fast each candle burns by dividing the length by the time, thus giving us the following information.

Candle A: 60cm/6hr = 10cm/hr

Candle B: 60cm/3hr = 20cm/hr

Second, we can shorten our time to under 3hrs as we know that by 3hrs, Candle B would’ve already burned out leaving only one candle left. Now we can use a trial and error method, by trying the times of each hour. After the first hour, the length of Candle A would be 50cm while the length of Candle B would be 40cm; one is not exactly twice as long as the other. Next after the second hour, the length of Candle A is 40cm while the length of Candle B is 20cm; one candle is exactly twice as long as the other.

Therefore after 2 hours, one candle will be exactly twice as long as the other candle.

'''25. Two candles of length L and L + 1 were lit at 6:00 and 4:30, respectively. At 8:30 they had the same length. The longer candle died at 10:30 and the shorter candle died at 10:00. Find L.'''

User:MatthewJohnson

2010-09-20T07:54:15Z

MatthewJohnson:

My name is Matthew Johnson, I am a first year studying commerce from Seattle and England. I live in Place Vanier on campus and I love to play the drums and piano. In general, I am just a big fan of music, and hopefully one day will be able to combine music and business. I also really enjoy sports and the more extreme things in life. Hit me up if you ever want to go skydiving!

Pythagorean Theorem

Everybody knows that A^2 + B^2 = C^2. It was one of the first things we learned in geometry way back in high school. It seems so simple but the actual discovery and of course the proving of this theory were much more difficult. Pythagoras, the alleged creator of the theory, was supposedly born on the Greek island of Samos. After settling in Italy, he created a sect of scholars called the Pythagoreans, who believed in the power of numbers. While the Pythagoreans hold the name of the theory, it is not for sure that they invented it, only that they proved it. They did so by thinking of a right triangle as three seperate squares. One side of each square made the three sides of the triangle. If we can agree the one side of a square times itself makes the area of the sqaure, then we can say that A^2 is the area of square A. The thing that they noticed was that the areas of squares A and B were equal to square C. Downsizing from squares they could then prove what came to be known as the Pythagorean Theorem.

Sites Used:

http://www.cut-the-knot.org/pythagoras/
http://www.notablebiographies.com/Pu-Ro/Pythagoras.html

User:MatthewJohnson

2010-09-19T20:13:50Z

MatthewJohnson: Created page with 'My name is Matthew Johnson, I am a first year studying commerce from Seattle and England. I live in Place Vanier on campus and I love to play the drums and piano. In general, I a…'