Friday, December 19, 2008

Proof by Contradiction

Hi everyone! Although this is the last day of school for this year, we still had lessons in class that we talked about for the whole period...

Before we went into groups and answer group questions, Mr. K. recalled the sentence that we talked about the other day:

This sentence is false.

This sentence is a Paradox which means it can't be true and false at the same time, but it should be true and false at the same time. A Paradox is one kind of Contradiction that goes round and round.

When we went into our groups, we were given logic questions that we needed to answer. Here's the first question.

On of Isle of View, a knight always tell the truth and a knave always lie. A person who just came met Mr. A and Mr. B. Mr. A said "one of us is a knave". Which one is the knave, and which one is the night?

We need to answer this question by Proof by Contradiction, which means we have to assume the opposite of the statement and prove that it's right. This was how we tackled this question.

First, we choose the Direct Reasoning of the statement, which in this case was
1) Assume Mr. A is a knight
. We can then tell that
2) His statement is true,
therefore our conclusion is that
3) Mr. B is a knave and Mr. A is a knight.

Using the Indirect Reasoning, we take the opposite statement and check if it leads to contradiction...

1) Assume Mr. A is a knave
2) His statement is false
3) They are both knights

This part of the reasoning leads to a contradiction, which was what we were looking for. Therefore, Mr. A is a knight.

We looked at two more questions that deals with logic, obviously :) It was pretty much straight forward... We do have homework, however that we need to do. It's Exercise 50 all questions.

Next scribe is Charizze

Pre-test on Monday - January 5, 2009



HAPPY HOLIDAYS!

see you all next year...

Today's Slides: December 19

Here they are ...



Getting Ready For The Logic Test

A few links you may find useful in preparing for your test on Thursday:



And here are a few quizzes (refresh the page for more) ...



And finally, here are some logic puzzles to practice with ...



Study hard and do your best work; work you're proud of!

Thursday, December 18, 2008

Logic

Today in class we learn about Conditional Statement.

Conditional Statement = is a statement having a "if" and "then".
= it consist of an Hypothesis "if" and a Conclusion "then".

Example: "If two lines are parallel then they do not intersect".

There are 3 Related Conditionals:

1. Converse = is formed by interchanging the hypothesis and conclusion of the original statement.
Ex. "If two lines do not intersect then the lines are parallel.

2. Inverse = is formed by negating the hypothesis and conclusion of the original statement.
= the word "not" is added to both parts of the sentence.
Ex. "If two lines are not parallel then they do intersect.

*note: In algebra we also use Inverse, remember when we subtract positive we get negative and when we add both negative we get positive, well its the same thing in Logic. It is easy to remember Inverse if we know this.

3. Contrapositive = is formed by negating the hypothesis and the conclusion and interchanging the resulting negative.
Ex. "If two lines do intersect then the lines are not parallel.

In class we also talk about biconditional.

Biconditional =
it is a logical operator connecting two statements if and only if, where hypothesis and conclusion. Its like they are both going on the same direction.

-more information-

Logical Relationships Between Conditional Statement

Watch Video on Conditional Statements

Conditional Statements - Converse and Biconditional


and that`s it for today, don`t forget to do the homework Exercise 49 and finishing the homework from the blog.

next scribe is Jeamille.




Today's Slides: December 18

Here they are ...



Using Venn Diagrams in Word Problems

In class today, we went over some questions on Venn diagram sets. You can see the answers to in the slides that Mr. K posted.

After doing a bit of that, we then learned how to apply our knowledge of Venn diagram sets and rules to problems that you would see in real life.

Let's see how we would solve problems that look like this with Venn diagrams:

"In a group of students 12 are taking chemistry, 10 are taking physics, 3 are taking both and 5 are taking neither. How many students are in the group?"

First we would draw our 'universe' and the Venn diagram that goes in it.:
('U' being universe, 'C' being chemistry, and 'P' being physics)

After you have everything set up, it's time to add in some numbers. Let's start with the easiest possible ones first.

Seeing how there are 5 students who aren't taking either chemistry or physics, we can tell right away that they're going to be placed somewhere in the 'universe', outside of the Venn diagrams.

The 3 students who are taking both courses are th
e next obvious choice. Since they are taking chemistry AND physics, they belong in the intersection, where the 'C' and the 'P' circle meet.

Now we'll figure out how many students are taking each course.

Automatically your brain might think that the number of students taking chemistry is 12 and the number of students taking physics is 10, but neither statement is true.


The actual number of students taking chemistry is 9, and the number of students taking physics is 7. Why? Because we exclude the 3 students that are taking both courses.

We subtract those 3 students from the pool of students taking chemistry (12 - 3 = 9) and physics (10 - 3 = 7).

We do this because in circles 'C' and 'P', we only want the number of students who are ONLY taking either chemistry or physics, excluding the students that are taking both.

Using the information we've gathered, we can determine the total number of students by adding up all the numbers. You should then finish off with this:
And that's how you solve a word problem with a Venn diagram.

We did a couple more examples, which all use pretty much the same procedure.

We did however, do one that involved 3 circles in the Venn diagram instead of 2. It seemed a little confusing, but solving it involved the same procedures as the previous examples. You just had to be very careful with your numbers.

That's about it for my scribe post, thanks for reading. The next scribe will be..camiLLe

Wednesday, December 17, 2008

Today's Slides: December 17

Here they are ...



Sets and Venn Diagrams

Below are some definitions we learn't in class today.

SET: a collection of objects, may be things, people, numbers, etc.
SUBSET: a "smaller" collection of objects taken from a given set.
UNIVERSE: the set of all objects being considered.
NULL SET: a set that is empty, a set with no members, objects, etc.
NB: If two sets have no common elements, they are said to be "disjoint".

OPERATION ON SETS
Below are some terms and examples
Union ( U )
A U B means gather all the objects in a set A with all the objects in set B.
Union ( U ) means "or".

Compliment (A )
This means everything that is outside set A.
Compliment (A’) means "not".

Exclusion ( / )
A/B means everything that is in set A excluding those things that are also in set B.
Exclusion ( / ) means "excluding".
Intersection ( n )
A n B means to gather all the objects that are in both set A and B.
Intersection ( n ) means "and".

Do your homework ... Practise makes Perfect!

Good luck to all!

The next scribe is Eric.

Monday, December 15, 2008

logic

After that we finished the geometry unit. Today we started a new unit called logic.We started by looking at what the word "logic" means. we then watched a videos about the word argument, what it means on http://www.youtube.com/watch?v=teMlv3ripSM
We later studied some important words and definitions in logic


Logic: (from Classical Greek λόγος logos; meaning word, thought, idea, argument, account, reason, or principle) is the study of the principles and criteria of valid inference and demonstration. en.wikipedia.org/wiki/Logic

The argument: A fact or statement used to support a proposition; a reason:; A verbal dispute; a quarrel; A process of reasoning; A series of statements organized so that the final statement is a conclusion which is intended to follow logically from the preceding statements, which function as premises... en.wiktionary.org/wiki/argument .
The argument consists of two (2) parts:
1. a set of premises
2. a conclusion

We then started to make a difference between premises and a conclusion, valid argument and invalid argument
A premise is a valid truth
A conclusion is a result of an argument
An argument is called valid if its result is direct consequence of its premises. In a valid argument, if the premises are true, then the conclusion must also be true.
An argument is called invalid when its conclusion is not a direct consequence of it’s premises even if its conclusion are true.
Sound argument: an argument is sound if it is both valid and true

After learning the validity and the invalidity of an argument we learned the induction and deduction.
Induction: when we observe several particular examples of same things, and then conjecture that the “the discovered” pattern must always be true

Deduction: a process of reasoning in which a conclusion follows necessarily from the premises presented, so that the conclusion cannot be false if the premises are true.

We later did a drag and drop exercise on valid argument and invalid argument, inductive and deductive reasoning.

Mr. K gave us a homework which exercise number 45

The next scribe is Zawadi.

Today's Slides: December 15

Here they are ...



Wednesday, December 10, 2008

Niwatori-san (reflection)

Yes, well this unit was pretty fun if you know the rules/theorems it helps more. I actually had trouble in the beginning but hey I lasted to the end with results. The more ways to find the clues the easier the investigation for sure. Well finally who knew that we would struggle so much but cover a lot as well >.< wonder why lol.

Let's all do well on the test shall we?

Reflection !

hi everyone!!!!

Geometry unit was really fun. specially constructing the angles and finding the patterns and figure out the theorem. But it was a little bit hard to memorize the theorems. But i felt better when Mr. K summarized the theorems in four parts. it was easier than the first time.

Reflection

This unit for me was pretty straight forward. I liked drawing the circles and finding all the angles. I would have to say my weak spot was writing my statement and reason, I just cant seem to get the work in my head onto words on the paper. Also I tend to over think the question and make it harder than it really is, thats what I need to work on. We have a test tomorrow and im pretty sure im ready for it.

That is all.. lol :)

Pre-Test

In today's class, we had a pre-test. We had at least 20 minutes to finish the work. Then we were off to groups.

Here were the questions:
This question can be proved by the tangent chord theorem.

"The angle between a tangent and a chord is congruent to the inscribed angle subtended by the opposite side of the chord."

As you can see we have the AB as our tangent. We also have angle r and angle d as our angles between the chord and the tangent. Angle a is in the opposite side of the chord of angle r, therefore angle r is congruent to angle a.


Next question:

Since OB and OD are radii, they must be congruent. Therefore, we can form an isosceles triangle through ΔOBD. We then formed a right angle through angle OBC because OB is perpendicular to AC. We can prove this by the tangent-radius theorem. Now that we've formed a right angle through angle OBC, we can now find angle OBD, which is 50° by subtracting 40°(given) from 90°(right angle). We've also said that the triangle was an isosceles triangle so angle OBD must be congruent to angle BDO. Therefore OBD and BDO are 50°. We can now find angle x through the sum of angles in ΔDBO, 180° - (50° + 50°) = 80°.


Next question
This one is pretty straight forward. We can find angle x by doing Thales' theorem.

"If an inscribed angle is subtended by the diameter then the inscribed angle is a right angle."

Now that we know, there is a right triangle, we can now easily find angle x. We can now find x by subtracting 40°(given) from 90°(right angle) which is equal to 50°. Therefore x is equal to 50°.


Next question:
It shows that we've got 2 isosceles triangle, ΔRSU and ΔSTU. Lets focus on eliminating y. In ΔRSU, we have y + 2a = 180°. In a cyclic quadrilateral RSUT, we have y + b = 180°. We can isolate y here, so we get y = 180° - b. Now we can substitute the y in therefore we get (180° - b) + 2a = 180°. We now get 2a = b.

We can also say that this cyclic quadrilateral is an isosceles trapezoid. We can prove this by saying RS is congruent to UT therefore we can say that RU and ST are parallel. Since they are isosceles we can say that x = b - a. Since we said b = 2a then we can substitute b into 2a. Now we get x = 2a - a. Then we get x = a.


We can now find x through b + 2a + x = 180. By substituting what we found the equation would turn out to be 2a + 2a + a = 180. This would equal to 5a = 180. Therefore we've got a = 36°. We said that a = x, therefore x is also 36°.


Reflection

Hiya (:

I find the Circle Geometry unit easier now than the first time that I saw it. The drawing part was fun, but understanding the theorems was a quite hard. I had difficulties on how to state and reason, and up until now, I still use the "paragraph/sentence" thingy. But I know that I will get used to the more formal one. The theories, as I said, were hard to know which one to describe with. Thanks to the summaries of the theories, I already know which theory to use. The extra days that we took to review helped me understand more. The pre-test today made me confident that I was ready for the actual test. I'm just a little nervous about the test tomorrow, though... But I hope that we all do well on the test tomorrow. Good luck, my fellow classmates! ^-^

--CharChar :D

reflection.

This unit was a little fun, all because of drawing circles and other stuff. Anyway, Circle Geometry was pretty good. I could understand most of it, especially the theorems. It's just kind of confusing or hard when we need to visualize the question and look at it carefully and think of which theorems to use for solving the problem. My weakness for this unit was the Statement and Reason part, it's just hard for me to state my reason on how I got the answer for a question in mathematical ways. I hope we'll all do well on the test tomorrow :) Good Luck !

Today's Slides: December 10

Here they are ...



Tuesday, December 9, 2008

Circle Geomtry Wordle

See any common concerns? Any patterns?

Hi, y'all

For me our topic now in Geometry is kinda easy than those previous one....but sometimes I’m getting confused to those theorem, I mixed them up sometimes....good thing Mr.K help us to solve this problem by classifying each theorem.... ohh...in reasoning i always have mistakes, so I think I need to practice more.....GOOD luck everyone....:>

reflection

Hi there!

I am quite ready for the pre-test tomorrow. Pretty much the theorems are straight forward. I think to succeed in this unit, I must be able to visualize the diagrams in many angles as possible because every time you find out something, its a clue to the the proof. I just need to continue on practicing to be better in geometry and proving.

Honestly, Proving is one of my weakness but I've been through this unit before, so i should be optimistic for the upcoming test!

Good luck everyone!

reflection

hi to everyone!!

Geometry is really fun unit, because we get to draw different diagrams. As for my opinion I'm kinda scared and nervous for this upcoming test, but hopefully tomorrow's pre-test will tell me if I really need to work twice as hard!..

Before I really don't get those theorems and when Mr. K started to teach them in class it actually making more sense and I like the way he do workshops in class because he makes you think hard and get to share our own knowledge in class!..So I just want to say good luck to everyone!..I hope everyone will do their best to get a better mark:P..

And don't forget always study notes and do home works:P...

good luck!!..:)

reflection...

hello!! everyone...
tomorrow will be our pre-test in geometry and I think I'm ready for it but not in actual test. This unit is kind a hard for me because in order to solve the problem I should know what kind of theorem is it and if I got it right I still have to prove it. Proving and finding the angles are my weakness so I hope I'll be able to work on it before the test...

that's all..
good luck everyone....

Reflection

hi! tomorrow will be our pretest and test on Thursday...Hope we'll pass this test..I'm sure you all will^_^..anyways this unit..I'm a little confused on somethings..like how do i know which theorem to state in every diagram...but hope tomorrows pretest will help me figure things out.. Good luck on our test^_^...

Today's Class

For today's class, we had a workshop class instead of a pre-test because we decided we needed one more day to prepare for the pre-test and test.
We were to work on the following questions and were to solve for Angle 1 and Angle 2. Here are the questions. I'll post the answers to each of them separately. I'll also provide an explanation for each.

For circle one:

To find Angle 2 you just need to know if two angles are opposite of each other in a polygon then the sum of those two angles equal to 180° by the cyclic quadrilateral theorem. Since 100° is given, and it's opposite angle is 2 you just need to subtract 100° from 180° to get angle 2, which is 80°


We had a discussion for finding Angle 1. We were to find the reason why lines AD and BC are parallel. We used the Parallel Chords Theorem in a way to explain why AD and BC are congruent. Since AB and DC are congruent, we can say that the Arc's subtending lines AB and DC are congruent. That means if the Arc's are the same, the lines AD and BC must be parallel by the Parallel chords Theorem. Now we can solve for Angle 1. Since lines AB and DC are congruent and lines AD and BC are parallel, we get an Isosceles Trapezoid. Meaning two base angles are the same, therefore two sides are also the same. Since it's given lines AB and DC are congruent, we can now say Angle 1 is 80°.

For circle two:
To find Angle 1 we first need to find angle BCD. Since angle BCE is given, and DE is a straight line, we need to find the supplementary angle to angle BCE. To find angle BCD we subtract 100° from 180° to give us angle BCD, which is 80°. We can now find Angle 1 by using the Cyclic quadrilateral theorem. Angle 1 is opposite of angle BCD, therefore the sum of those two angles add up to 180°. Angle 1 one is 100° by subtracting 80° from 100°.

For circle three:

The first step to do is to find Angle CDA. We find this angle by subtracting the sum of the known angles from 180°. Angle ACD is a right angle and angle CDA is given ( 25° ), Angle CDA is 65°.

Since ABCD is a Cyclic quadrilateral, we can use the Cyclic quadrilateral theorem to find Angle ABC. Opposite to angle ABC is angle CDA, which is 65°. ABC and CDA must me supplementary, therefore Angle ABC is 115°.

It's given that lines AB and BC are congruent, therefore an Isosceles triangle is formed, triangle ABC. That also means Angles BAC and BCA are congruent. So all we have to do now is to subtract 115° from 180° and divide by 2 to get those 2 angles. Leaving us with 65° divided by 2 which equals to 32.5°. 1 = 32.5°

We also had a second question. We didn't have much time to solve it, but I'll post the question for those who need it.

I'll also post my solution to the question:


Note: My solution may contain errors. I've provided my solution to give an idea / jump start to those who may be stuck. Don't take it as the 100% correct answer.

That sum's up the class for today. Remember we have a pre-test tomorrow and the test on Thursday. Study hard, and good luck!

Next Scribe: jdizon

Today's Slides: December 9

Here they are ...



Sunday, December 7, 2008

In Friday's class, we started off by doing Quiz Number 2. We had thirty minutes or so to finish it. Then time was up! We passed them up so Mr. K could pass them out so we could go over them and correct them.

Here is question one! We had to find the arcs. of AB, BC, CD and AD.













It starts us off with the Angle BOC, which is 115 degrees.
So, Arc BC will also be 115 degrees since its a central angle.
* An angle in a circle with the vertex at the circle's center*

Angle AOD would also be 115 degrees because its an opposite angle from Angle BOC.
So, Arc AD will also be 115 degrees because its a central angle.

Angle AOB will be 65 degrees because its supplementary to angle BOC.
*Two Angles that add up to 180 degrees*
So, Arc AB will be 65 degrees because its a central angle.

Angle DOC will also be 65 degrees because its an opposite angle to Angle AOB.
So, Arc DC will be 65 degrees because its a central angle.







For this Question we had to figure out how far from the centre of the circle is to the chord DC.
















It gives us:
Chord AB is 14 cm
Chord DC is 7 cm

AB runs through the centre of the circle, which makes it the diameter.
So the Radius is 7 cm.

Line EO bisects chord DC. Which makes it a right angle and cuts chord DC in half and makes each half will be 3.5 cm.

Then, make a radius OC and you then have a right triangle.

So far you will have:
Chord AB is 14 cm
Chord DC is 7 cm

Radius = 7 cm

OE = 3.5 cm


The next step is to use the Pythagorean theorem to find the length of line OE.

The answer will be 6.06 cm.








For this Question it is very similar to question two, where we have to find the length of from the centre of the circle to the chord. This time, we don't have a picture. So, we have to draw it.














They gave us:
Chord DC is 8 cm
Chord AB (diameter) is 10 cm
Radius is 5 cm

Line OE bisects chord DC, which makes it into a right triangle.

So we use the Pythagorean theorem.

Which gives us the answer 3.









Finally! For the last question! It is the same as the previous question, just switched around. We have to find out the diameter of a circle. Plus we have to draw the picture also.















They give us:
Chord AB is 16 cm
Line OE is 15cm (bisects chord AB, which makes a right triangle)
Half of Chord AB is 8

To find the answer you use the Pythagorean theorem.

Using that you will find the radius being 17 cm. Times that by two and you get the diameter 34 cm.


*The main thing that you have to remember though is NOT to write in paragraph form when you state the statement.*


p.s i cant explain well... so hopefully its somewhat helpful


The next blogger is Dianne..

Thursday, December 4, 2008

December 4 Math Class

Today in math we started off the class with circle problems. We were put into groups and were told to solve each of the problems in 4 minutes. The group who got the answer the quickest put their solution on the smart board and we went over it as a class, these are the problems:



After that we got Investigation 8 and a homework assignment to do. We also got assigned Exercise 32 in our text books.

We will be having a work class tomorrow and finishing up Circle Geometry.

The Next Scribe is... Lani

Today's Slides: December 4

Here they are ...



Monday, December 1, 2008

December 1st Class

Today in class Mr. K had to leave so we were left with Mr Watt to watch over us. We were assigned Investigations 5 and 6 for circle geometry and exercise 32 in our text books.

Remember to finish up your homework :)