Here's another alternative for an algebra 1 course for homeschooled students: Algebra 1 by Paul Foerster (the textbook), and accompanying Home Study Companion video lessons made by David Chandler.
I've just written an in-depth review of both.
Jumat, 29 Februari 2008
Selasa, 26 Februari 2008
A bar diagram to solve a ratio problem
Dave at MathNotations had an interesting ratio problem:
He asked if it can be solved using "Singapore" style bar model.
I'm not sure if this is exactly how they'd do it, but this is how I'd do it... so here goes.
After I made the diagrams, I soon saw that Dave's numbers are two awkward; the bar diagram drawing would get too messy because we'd need to divide it into too tiny parts to see anything.
BUT... you probably know about the PROBLEM SOLVING STRATEGY called "solve an easier problem". My agenda is therefore:
1. Here's a bar diagram representing the fact that the ratio of the number of juniors to seniors is 7:5.
As you can see, the "whole" ends up divided into 12 parts (7 + 5).
We can use this diagram to solve problems such as:
If there are 768 juniors and seniors in total, how many juniors are in the school?
(Divide that into 12 parts, and then take 7 of those parts.)
Or...
If there are 95 seniors, how many juniors are there?
(Divide the number of seniors by 5, then multiply by 7.)
Or..
If there are 49 Juniors, how many students are there in all?
(Divide the number of juniors by 7, then multiply by 12.)
The diagram makes all this dividing/multiplying by 5/7/12 all crystal clear.
2. Let's change the numbers from the original problem to make them "friendly" for this approach. Let's solve this instead:
Our diagram becomes, first of all, like before:
But then we add the additional information into it. The ratio of junior males to females is 4:3, which means juniors as a whole are divided into 7 parts (4 + 3). Similarly, seniors need divided into 10 parts.
Voila! It just so happens (wonder why?) that our original division into 7 and 5 parts works beautifully to give us juniors divided into 7 parts (they already are!) and seniors into 10 parts if we only split each part into 2.
So, what is the ratio of junior males to senior females?
In the diagram, the WHOLE is now divided into 24 parts. Junior males are 8 of those parts, and senior females are 3 of those parts. Their ratio is therefore 8:3.
3. Solving the original problem, with those unfriendly numbers.... the diagram looks sort of like this:
It gets messy... I'd prefer using the bar model as a stepping stone and to illustrate the basic situation, but eventually using the "least common denominator" or algebraic methods.
In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 3:2.
The ratio of senior males to senior females is 4:3.
What is the ratio of junior males to senior females?
He asked if it can be solved using "Singapore" style bar model.
I'm not sure if this is exactly how they'd do it, but this is how I'd do it... so here goes.
After I made the diagrams, I soon saw that Dave's numbers are two awkward; the bar diagram drawing would get too messy because we'd need to divide it into too tiny parts to see anything.
BUT... you probably know about the PROBLEM SOLVING STRATEGY called "solve an easier problem". My agenda is therefore:
- show how to solve a few related simple ratio problems using the bar diagram
- solve a variant of the original problem (with friendly numbers)
- solve the original problem.
1. Here's a bar diagram representing the fact that the ratio of the number of juniors to seniors is 7:5.
As you can see, the "whole" ends up divided into 12 parts (7 + 5).
We can use this diagram to solve problems such as:
If there are 768 juniors and seniors in total, how many juniors are in the school?
(Divide that into 12 parts, and then take 7 of those parts.)
Or...
If there are 95 seniors, how many juniors are there?
(Divide the number of seniors by 5, then multiply by 7.)
Or..
If there are 49 Juniors, how many students are there in all?
(Divide the number of juniors by 7, then multiply by 12.)
The diagram makes all this dividing/multiplying by 5/7/12 all crystal clear.
2. Let's change the numbers from the original problem to make them "friendly" for this approach. Let's solve this instead:
In Virtual HS, the ratio of the number of juniors to seniors is 7:5.
The ratio of (the number of) junior males to junior females is 4:3.
The ratio of senior males to senior females is 7:3.
What is the ratio of junior males to senior females?
Our diagram becomes, first of all, like before:
But then we add the additional information into it. The ratio of junior males to females is 4:3, which means juniors as a whole are divided into 7 parts (4 + 3). Similarly, seniors need divided into 10 parts.
Voila! It just so happens (wonder why?) that our original division into 7 and 5 parts works beautifully to give us juniors divided into 7 parts (they already are!) and seniors into 10 parts if we only split each part into 2.
So, what is the ratio of junior males to senior females?
In the diagram, the WHOLE is now divided into 24 parts. Junior males are 8 of those parts, and senior females are 3 of those parts. Their ratio is therefore 8:3.
3. Solving the original problem, with those unfriendly numbers.... the diagram looks sort of like this:
It gets messy... I'd prefer using the bar model as a stepping stone and to illustrate the basic situation, but eventually using the "least common denominator" or algebraic methods.
Senin, 25 Februari 2008
Notice
For those of you who're waiting... LightBlue 4th grade is just days from being finished. Right now I'm putting finishing touches on the lists of web resources I've added to each chapter. Just about everything else is ready.
Sabtu, 23 Februari 2008
Concept of average
I felt like giving away something again to help all of you who need math lessons... This one is taken from Math Mammoth Division 2 book and is a lesson on the concept of average... meant for initial teaching for 4th-5th graders who have mastered long division.
I've written it quite recently, and in fact improved it just last night by adding the bar graph problems into it. (That seems to be the "way of life" with these books: I constantly keep tweaking and improving them as I learn. Which means constant uploading of files here and there... keeping our bandwidth usage large...)
The lesson mainly uses word problems, BTW. Please post your comments below; I'm anxious to hear them!
Right click and choose save:
Average_from_Math_Mammoth_Division_2.pdf
I've written it quite recently, and in fact improved it just last night by adding the bar graph problems into it. (That seems to be the "way of life" with these books: I constantly keep tweaking and improving them as I learn. Which means constant uploading of files here and there... keeping our bandwidth usage large...)
The lesson mainly uses word problems, BTW. Please post your comments below; I'm anxious to hear them!
Right click and choose save:
Average_from_Math_Mammoth_Division_2.pdf
Selasa, 19 Februari 2008
Carnival of Homeschooling, Schoolhouse Edition
There's lots to read and enjoy at the newest carnival of homeschooling. I enjoyed the old schoolhouse photos as well! I know my dad attended a fairly small village school; maybe it was similar to these schoolhouses.
One math-related pick from the carnival:
Griddlers make great “paint by number” puzzles for kids of all ages - Sol from Wild About Math shares some great number puzzles where you color in a picture based on number clues.
One math-related pick from the carnival:
Griddlers make great “paint by number” puzzles for kids of all ages - Sol from Wild About Math shares some great number puzzles where you color in a picture based on number clues.
Sabtu, 16 Februari 2008
Learn to recognize numbers game for preschoolers
I just wanted to share a little "game" that helped my 3-year old preschooler to recognize her numbers here a few weeks back.
This game is SO simple that it's almost laughable; yet it went over VERY well with our tyke! I am still amazed. There is no strategy involved, it's so simple I almost hesitate to call it a game. But I did call it a "number game" to my 3-year old and she loved it.
I used foam numbers and plastic numbers, and just made a heap of them between us. I would pick one, hold it up high and call out loud its name, such as "Number five!" and put it to my personal pile.
She would then find the same number (I made sure there were at least two of each) and did the same, called out loud its name and gathered the number to herself.
Then it was her turn to pick any number from the pile, call out its name, and put it to her pile, and I had to find the same number.
After all the numbers in the middle pile were gone, her task was to arrange her numbers in order. That's it.
Actually we don't have to play that anymore because she learned to recognize them so quick. She's also learning to count, but just like all kids do, her fingers are sometimes faster than her mouth saying numbers, or vice versa, and so she doesn't always get it right. She enjoys watching big sister do computer games, such as at Time4Learning or UpToTen. Now they two can also play UNO together (since she recognizes the numbers!).
This game is SO simple that it's almost laughable; yet it went over VERY well with our tyke! I am still amazed. There is no strategy involved, it's so simple I almost hesitate to call it a game. But I did call it a "number game" to my 3-year old and she loved it.
I used foam numbers and plastic numbers, and just made a heap of them between us. I would pick one, hold it up high and call out loud its name, such as "Number five!" and put it to my personal pile.
She would then find the same number (I made sure there were at least two of each) and did the same, called out loud its name and gathered the number to herself.
Then it was her turn to pick any number from the pile, call out its name, and put it to her pile, and I had to find the same number.
After all the numbers in the middle pile were gone, her task was to arrange her numbers in order. That's it.
Actually we don't have to play that anymore because she learned to recognize them so quick. She's also learning to count, but just like all kids do, her fingers are sometimes faster than her mouth saying numbers, or vice versa, and so she doesn't always get it right. She enjoys watching big sister do computer games, such as at Time4Learning or UpToTen. Now they two can also play UNO together (since she recognizes the numbers!).
Minggu, 10 Februari 2008
Pi is a ratio, yet irrational?
Here's an interesting dilemma:
We know Pi is an irrational number; mathematicians have proven it to be so.
But its definition says that it is a RATIO of the circumference and the diameter of any circle.
Now, when you divide a rational number by another rational number, you get a rational number.
Doesn't this seem like a contradiction?
In the words of a certain visitor to my site:
In the equation where the circumference is divided by the diameter, when the circumference and diameter are rational values, why is it that the quotient can be an irrational quantity?
The solution
First of all, like a commenter pointed out, Pi being a ratio of two numbers does not mean it is rational. Pi has been established as irrational, and we know
Pi = C/d, where C is the circumference and d is the diameter of some circle.
It follows that either C or d or both have to be irrational!
This is kind of amazing to think about, but it's true: for any circle, either the circumference, or the diameter, or both are irrational (in the abstract world of mathematics...).
BUT (and herein lies the crux of the apparent contradiction)... in the real world where we have to use measurements, of course your measurements will always be rational numbers. ANY measurement you make is just an approximation anyway, and is rational, such as 14.52 cm or 8 1/16 in.
And, if you try to find the value of Pi using measurements from real world, all you get is a rational approximation to the value of Pi.
Even your calculator just gives you a rational approximation for Pi, to however many decimal digits it can show it. But that's all that is necessary for real world applications.
We know Pi is an irrational number; mathematicians have proven it to be so.
But its definition says that it is a RATIO of the circumference and the diameter of any circle.
Now, when you divide a rational number by another rational number, you get a rational number.
Doesn't this seem like a contradiction?
In the words of a certain visitor to my site:
In the equation where the circumference is divided by the diameter, when the circumference and diameter are rational values, why is it that the quotient can be an irrational quantity?
The solution
First of all, like a commenter pointed out, Pi being a ratio of two numbers does not mean it is rational. Pi has been established as irrational, and we know
Pi = C/d, where C is the circumference and d is the diameter of some circle.
It follows that either C or d or both have to be irrational!
This is kind of amazing to think about, but it's true: for any circle, either the circumference, or the diameter, or both are irrational (in the abstract world of mathematics...).
BUT (and herein lies the crux of the apparent contradiction)... in the real world where we have to use measurements, of course your measurements will always be rational numbers. ANY measurement you make is just an approximation anyway, and is rational, such as 14.52 cm or 8 1/16 in.
And, if you try to find the value of Pi using measurements from real world, all you get is a rational approximation to the value of Pi.
Even your calculator just gives you a rational approximation for Pi, to however many decimal digits it can show it. But that's all that is necessary for real world applications.
Kamis, 07 Februari 2008
The World Math Day Challenge
I got word about this nice math event.
In World Math Day Challenge (held on March 5), students from across the globe will be uniting online to play each other in real time mental arithmetic games. These are tailored to the students' levels so students of various ages can participate.
They are expecting more than 1 million students from over 100 countries! It is one of the world's largest participation events.
World Math Day is free of charge for both schools and students. All you need is Internet access. And, there are even prizes!
More information may be found at www.WorldMathDay.com.
This is a nice opportunity to do something fun with math, and at the same time connect with other kids, even across the globe, without leaving your home. You can register already; the event is on March 5.
Selasa, 05 Februari 2008
Shorter-wider table problem
I failed to publish this the other day since Blogger was acting up. I'll try it now.
1) You can solve this using algebra: set width to be x, the length is then 3x. We know these two will be equal if the former is increased by 3, and the latter is decreased by 3:
But since this is supposed to be a 6th grade problem, surely we can find another way to solve it, as well.
2) Think of the two quantities length and width as just numbers. If you reduce one by 3, and increase the other by 3, they will "meet" or be the same. Below, I've drawn the two numbers as lines; you could use bars.
length |---------------------------------|
width |----------|
If I decrease the length by 3 and increase the width by 3, they'll be equal:
length |---------------------|<------------|
width |----------|--------->|
So... the DIFFERENCE between these two is 6. The difference of 6 is also TWO TIMES the original width, because remember the length is 3 times the width.
So the original width is 3 m and the original length is 9 m (a giant table!).
3) Solving it geometrically is also quite easy. First draw a table that looks about 3 times as long as it is wide. Then trace the sides if it was a square, making the long side shorter, and short side longer:
It sure looks like that the side of the square should be double 3, or 6. The original sides would then be 3 and 9. And you can easily check out that those work.
--------------------------------------------------------------
I feel like the numbers in this problem are almost too easy. But maybe this problem could be used as a stepping stone for a harder version of the same.
A rectangular kitchen table is three times as long as it is wide.
If it were 3 m shorter and 3 m wider it would be a square.
What are the dimensions of the rectangular table?
1) You can solve this using algebra: set width to be x, the length is then 3x. We know these two will be equal if the former is increased by 3, and the latter is decreased by 3:
x + 3 = 3x - 3
2x = 6
x = 3.
The dimensions are 3 and 9.
But since this is supposed to be a 6th grade problem, surely we can find another way to solve it, as well.
2) Think of the two quantities length and width as just numbers. If you reduce one by 3, and increase the other by 3, they will "meet" or be the same. Below, I've drawn the two numbers as lines; you could use bars.
length |---------------------------------|
width |----------|
If I decrease the length by 3 and increase the width by 3, they'll be equal:
length |---------------------|<------------|
width |----------|--------->|
So... the DIFFERENCE between these two is 6. The difference of 6 is also TWO TIMES the original width, because remember the length is 3 times the width.
So the original width is 3 m and the original length is 9 m (a giant table!).
3) Solving it geometrically is also quite easy. First draw a table that looks about 3 times as long as it is wide. Then trace the sides if it was a square, making the long side shorter, and short side longer:
It sure looks like that the side of the square should be double 3, or 6. The original sides would then be 3 and 9. And you can easily check out that those work.
--------------------------------------------------------------
I feel like the numbers in this problem are almost too easy. But maybe this problem could be used as a stepping stone for a harder version of the same.
Sabtu, 02 Februari 2008
Longer-wider problem to solve
I took another problem from the same collection as before, this time for 6th grade. It reads:
I'll let you try it first, but don't post the solution as a comment into this post. Let others try solve it too.
(You can get an extra point for answering this: Is this a realistic kitchen table?)
A rectangular kitchen table is three times as long as it is wide.
If it were 3 m shorter and 3 m wider it would be a square.
What are the dimensions of the rectangular table?
I'll let you try it first, but don't post the solution as a comment into this post. Let others try solve it too.
(You can get an extra point for answering this: Is this a realistic kitchen table?)
Blogged math blogs
I got word that folks at Blogged.com had rated a bunch of math blogs, and placed mine in the top ten, with rating 8.6 (out of 10). They evaluate blogs based on the following criteria: Frequency of Updates, Relevance of Content, Site Design, and Writing Style.
So here is a list of mathematics blogs, rated in a scale 0-10. The list does resemble the blogrolls that I've seen on other math blogs... If you're a math blogger, go find yours!
So here is a list of mathematics blogs, rated in a scale 0-10. The list does resemble the blogrolls that I've seen on other math blogs... If you're a math blogger, go find yours!
Langganan:
Postingan (Atom)
