Where math comes alive


To find the GCF for three or more numbers,  follow these steps:

1)  Determine which of the given numbers is smallest, then find the smallest difference between any pair of numbers.

2)  See what is smaller:  the smallest number, or the smallest difference. Whichever one  is smallest, that number is the GPGCF (Greatest Possible GCF). That means that this is the biggest number that the GCF could possibly be. Or, more formally we would say:  The GCF, if it exists, must be less than or equal to the GPGCF.

3)  Check if the GPGCF itself goes into all of the given numbers. If so, then it is the GCF. If not, list the factors of the GPGCF from  largest to the smallest and test them until you find the largest one that does divide evenly into the given numbers. The first factor (i.e., the largest factor) that divides evenly into the given numbers is, by definition, the GCF.

EXAMPLE:

Problem:  Find the GCF for 18, 30,  54.

1)  Note that the smallest number is 18, and  the smallest difference between the pairs is 12 [54 - 30 = 24;  54 - 18 = 36;  30 - 18 = 12] .

2)  Of those four quantities (the smallest number and the three differences), 12 is the least. This means that the
GPGCF = 12.

3) Check if 12 divides evenly into the three given numbers: 18, 30 and 54. In fact, 12 doesn’t divide evenly into ANY of these  numbers. Next we check the factors of 12, in order from largest to smallest. Those factors are: 6, 4, 3 and 2. The first of those that divides evenly into all three numbers is 6. [18 ÷ 6 = 3;  30 ÷ 6 = 5;  54 ÷ 6 = 9]. So the GCF = 6. And we are done.
MORE CHALLENGING PROBLEM:

Find the GCF for 24, 148, 200.

1)  Note that the smallest number is 24, and that the smallest difference between the pairs is 52 [200 - 148 = 52;  200 - 24 = 176;  148 - 24 = 124] .

2)  Of those four quantities (the smallest number and the three differences), 24 is the least. This means that for this problem, the GPGCF = 24.

3) Check if 24 divides evenly into the three given numbers: 24, 148 and 200. While 24 does divide evenly into 24, it does not divide evenly into 148 or 200. So next we check the factors of 24, in order from largest to smallest. Those factors are: 12, 8, 6, 4, 3 and 2. The first of those that divides evenly into the three given numbers is 4. [24 ÷ 4 = 6;  148 ÷ 4 = 37;  200 ÷ 4 = 50]. So the GCF = 4. And, once again, we are done.

The process may seem a bit long, but once you get used to it and start doing it in your mind, not on paper, you should find that it actually is quite fast. And you’ll find yourself figuring out the GCF for three or more numbers all in your mind — with no need for pencil and paper — while everyone around you will be making prime factor trees or using calculators. And surely that is a good feeling.

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly  like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

Making Sense of Inequalities


OK, teachers, homeschooling parents and tutors … raise your hand if you’ve ever felt uncomfortable when students pose that question about inequalities?

That question being:  why do we flip the inequality symbol when we multiply or divide by a negative number?

I’d have to, sheepishly, raise my hand.

So when I got asked that question once again last week, I decided to figure it out and come up with an answer that would help students understand this point.

What I came up with is that it’s easiest to explain this through a combination of examples and logic. First, the examples.

Let’s break the situation up into three cases. We could have inequalities in which the numbers on the two sides are A) both positive, B)  both negative, or C) one number positive, the other negative.

Let’s start with Case A. Suppose we start with the statement, 2 < 4

Now, multiply both sides by a positive number (let’s use 3), and we get:  6 < 12. Still true, right?

But take the original inequality and multiply it by a negative number (let’s use – 3), and we get: – 6 < – 12. Not true, right? But if we flip the sign, we do get a true statement:  – 6 > – 12

Case B. Now let’s start with two negative numbers in our true inequality:  – 4 < – 2 If we multiply both sides by a positive number (3 again), we get:  – 12 < – 6, which is again true.

But if we multiply this inequality by a negative number (– 3 again), we get: 12 < 6, which is obviously false. However if we once again flip the sign, we get a true statement:  12 > 6.

Finally, Case C. Now we start with an inequality that has both a positive and a negative number:  – 2 < 4. If we multiply both sides by  positive 3, we get:  – 6 < 12, which is still true.

But if we multiply both sides by our – 3 again, we get:  6 < – 12, which is once again false. And again we need to flip the sign to make it true:  6 > – 12.

So far so good, but this lacks the logic of an explanation. How can we bring in some logic and reasoning, to help students see why all of this stuff happens?

Here’s my — granted, informal — way of explaining this. When we multiply or divide a number by positive numbers, we don’t change its sign; if the number was positive, it stays positive, and if it was negative, it stays negative. But when we multiply or divide a number by a negative number, we do change its sign … either from positive to negative, or from negative to positive.

So the reason that we flip the inequality symbol must be related to the fact that — by multiplying or dividing both sides of the inequality by a negative number — we are changing the signs of both numbers in the inequality. But how exactly does this work?

The answer, it turns out, is rooted in the relationship between the absolute value of numbers and their relative sizes. For numbers that are positive, there’s one way to tell which number is larger … the number with the larger absolute value is the larger number. For example, comparing 4 and 12, we know that 12 is larger than 4 because the absolute value of 12 is larger than the absolute value of 4. But for numbers that are negative, the exact opposite is true. For two negative numbers, the number with the larger absolute value is actually the smaller number. For example, compare – 4 and – 12. Their absolute values are 4 and 12, respectively, but the number with the larger absolute value is in fact the smaller number, not the larger number. In this example, – 12 (with the bigger absolute value of 12), is in fact smaller than  – 4 (with the smaller absolute value of 4).

So the point to remember here is that there are two different relationships between the absolute values of numbers and the relative sizes of numbers. For positive numbers, the greater the absolute value, the greater the number. But for negative numbers, the greater the absolute value, the smaller the number.

This fact has an impact on inequalities where we change the signs of the numbers. Before changing the signs of the numbers, the numbers on the two sides of the inequality had one size relationship; one number was larger than the other (let’s say that Number A is, at this stage, larger than Number B). But when we multiply or divide both of these numbers by a negative, we flip the signs of both numbers. And by flipping the signs of both numbers, we change the size relationship of the numbers to each other. The one that was the larger one ends up being the smaller one, and vice-versa. So in our abstract example, if Number A was larger than Number B before their signs were changed, after both signs are changed, Number A will be smaller than Number B.

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly  like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 


As you’re probably aware, I’m a big believer in using stories to bring math to life. Especially when you’re teaching tricky concepts, using a story can be the “magic switch” that flicks on the light of understanding. Armed with story-based understanding, students can recall how to perform difficult math processes. And since people naturally like stories and tend to recall them, skills based on story-based understanding really stick in the mind. I’ve seen this over and over in my tutoring.

Stories from My Tutoring Work

The kind of story I’m talking about uses an extended-metaphor, and this way of teaching  is particularly helpful when you’re teaching algebra. Ask yourself: what would you rather have? Students scratching their heads (or tearing out their hair) to grasp a process taught as a collection of abstract steps? Or students grasping  a story and quickly seeing how it guides them in doing the math? I think the answer is probably pretty clear. So with this benefit in mind, let’s explore another story that teaches a critical algebraic skill: the skill of  “unpacking” terms locked inside parentheses.

To get the picture, first imagine that each set of parentheses, weirdly or not, represents a corrugated cardboard box, the kind that moving companies use to pack up your possessions. Extending this concept, the terms inside parentheses represent the items you pack when you move your goodies from one house to another.  Finally, for every set of parentheses (the box), imagine that you’ve hired either a good moving company or a bad moving company. (You can use a good company for one box and a bad company for a different “box” — it changes.) How can you tell whether the moving company is good or bad? Just look at the sign to the left of the parentheses. If the moving company is GOOD, you’ll see a positive sign to the left of the parentheses. If the moving company is BAD, you’ll spot a negative sign there.

Here’s how this idea looks:

+ (    )     The + sign here means you’ve hired a GOOD moving company for this box of stuff.

– (    )     This – sign means that you’ve hired a BAD moving company to pack up this box of things.

Now let’s put a few “possessions” inside the boxes.

+ (2x – 4)  This means a GOOD moving company has packed up your treasured items: the 2x and the – 4.

– (2x – 4)  Au contraire! This means that a BAD moving company has packed up the 2x and the – 4.

[Remember, of course, that the term 2x is actually a + 2x. No sign visible means there's an invisible + sign before the term.]

What difference does it make if the moving company is GOOD or BAD? A big difference! If it’s a GOOD company, it packs your things up WELL.  Result: when you unpack your items, they come out exactly the same way in which they went into the box. So since a good moving company packed up your things in the expression:  + (2x – 4), when you go to unpack your things, everything will come out exactly as it went in. Here’s a representation of this unpacking process:

+ (2x – 4)

=      + 2x – 4

Note that when we take terms out of parentheses, we call this “unpacking” the terms. This works because algebra teachers fairly often describe the process of taking terms out of (   ) as “unpacking” the terms. So here’s a story whose rhetoric  matches the rhetoric of the algebraic process. Convenient, is it not?

Now let’s take a look at the opposite situation — what happens when you work with a BAD (boo, hiss!) moving company. In this case, the company does such a bad job that when you unpack your items, each and every item comes out  “broken.” In math, we indicate that terms are “broken” by showing that when they come out of the (  ), their signs,  + or – signs, are the EXACT OPPOSITE of what they should be. So if a term was packed up as a + term, it would come out as a – term.  Vice-versa, if it was packed up as a – term, it would come out as a + term. We show the process of unpacking terms packed by a BAD moving company, as follows:

– (2x – 4)

=      – 2x + 4

And that pretty much sums up the entire process. Understanding this story, students will be able to “unpack” terms from parentheses, over and over, with accuracy and understanding.

But since Practice Makes Perfect, here are a few problems to help your kiddos perfect this skill.

PROBLEMS:

“Unpack” these terms by removing the parentheses and writing the terms’ signs correctly:

a)  – (5a + 3)

b)  + (5a – 3)

c)  – (– 3a + 2b – 7)

d)  + (– 3a + 2b – 7)

e)  6 + (3a – 2)

f)  6 – (3a – 2)

g)  4a + 6 + (– 9a – 5)

h)  4a + 6 – (– 9a – 5)

ANSWERS:

a)  – (5a + 3)   =   – 5a – 3

b)  + (5a – 3)  =  + 5a – 3

c)  – (– 3a + 2b – 7)  =  + 3a – 2b + 7

d)  + (– 3a + 2b – 7) = – 3a + 2b – 7

e)  6 + (3a – 2)  =  + 3a + 4

f)  6 – (3a – 2)  =  – 3a + 8

g)  4a + 6 + (– 9a – 5)  =  – 5a + 1

h)  4a + 6 – (– 9a – 5)  =  + 13a + 11


Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly  like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 


Hi everyone,

I have some exciting news.

I will be conducting a workshop this Tuesday morning, and the workshop is with a school in Nigeria.

Thanks to my good friend Ibraheem Dooba (aka Professor Brainy), and also thanks to the marvels of modern technology, this Skype workshop has been all set up and is ready to go.

The school whose teachers are receiving the workshop (not sure if that’s the right way to phrase it, or if there is a way to phrase that!) is the Esteem International School, in Nigeria’s capital city, Abuja. This is a wonderful school for elementary age children. Like students in the United States, students at the Esteem School are learning aspects of algebraic thinking in elementary school. And since teaching algebra to young students is a challenge, that will be the focus of my workshop.

If anyone here or anywhere in the world is interested in receiving a workshop on this topic, please let me know, and we’ll set one up.


You know, I’m a good math tutor. I think pretty much everyone who comes to work with me says that. And today I realized something about being a tutor.

Stories from My Tutoring Work

Tales from the Tutoring Experience

A big part of it — maybe as much as half it — involves nothing more than  …   being nice.

By that I mean being kind.

By which I mean that if someone looks at you, as a young man did today, shaking his head and saying, “It’s crazy … I don’t know what 3 x 6 is,” I don’t laugh or chuckle or say anything remotely mean or mocking. Instead I just say, “It’s o.k. Look, I tutor people every day who don’t know what 3 x 6 is. Who cares, really? Let’s just try to figure it out … or use a calculator, as long as your teacher doesn’t mind.”

Really. That is a lot of what being a math tutor is about. Being nice. Really nice. Really understanding. And being there to be accepting of people no matter how much mental pain they may be in about math. Because there’s a lot of pain out there. Many people are carrying loads of pain about math. They feel dumb. They feel like it’s some huge reflection on whether or not they can make it in the world.

And so it is my job, as a tutor, to listen to their worries and to assure them that they will get better. And that even if they keep struggling, as they probably will, to some extent, it is ok. They can still live good lives, and math is not going to define or confine them,” to quote Bob Dylan a bit.

Don’t get me wrong in terms of what I said up above. A math tutor has to know the material … extremely well. And he needs to know how to teach the material and the skills of math. But once he gets that down, once he gains in competence, he can really open up his heart and help people with the emotional struggles they go through with math as well.

I generally like being a tutor. It feels satisfying. I love seeing people go through the gradual transition from hating to have to see me, to feeling somewhat ok about it, to starting to feel good because their grades are going up and they are starting to get it better. They start to walk taller, literally as well as figuratively. They come right into the office, after several weeks or months of working with me, and they tell me exactly what they need help with. They become their own best advocates. And they get over that horrible feeling that math is holding them back.For the most part, to be honest, the students I work with don’t end up loving math. For the most part, they go from hating it to feeling ok about it. And that is ok with me. I just want students to feel like they can understand many of the parts of math and to feel competent in relation to math. Seeing that transition occur is the best reward I can get, and it actually happens more often than not.

So if you or someone you know needs a math tutor, I suggest you find one. It can make a big difference in a person’s life. A good tutor can really help a young person grow.


If you or someone you know struggles when combining numbers with opposite signs — one positive, the other negative — this post is for you!

To be clear, I’m referring to problems like these:

 – 2 + 7 [first number negative, second number positive], or

+ 13 – 20 [first number positive, second number negative]

To work out the answers, turn each problem into a math-story. In this case, turn it into the story of a tug-of-war battle. Here’s how.

In the first problem, – 2 + 7, view the – 2 as meaning there are 2 people on the “negative” team; similarly, view the + 7 as meaning there are 7 people on the “positive” team.

There are just three things to keep in mind for this math-story:

1)  Every “person” participating in the tug-of-war is equally strong.

2)  The team with more people always wins; the team with fewer people always loses.

3)  In the story we figure out by how many people the winning team “outnumbers” the other team. That’s simple; it just means how many more people are on that team than are on the other team. Example: if the negative team has 2 people and the positive team has 7 people, we say the positive team “outnumbers” the negative team by 5 people, since 7 is 5 more than 2.

Now to simplify such a problem, just answer three simple questions: 

1)  How many people are on each team?
In our first problem, – 2 + 7, there are 2 people on the negative team and 7 people on the positive team.

2)  Which team WINS?
Since there are more people on the positive team, the positive team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the positives have 7 while the negatives have only 2, the positives outnumber the negatives by 5.

Now ignore the answer to the intro question, Question 1, but put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  +

ANSWER TO QUESTION 3:  5

ANSWERS TOGETHER:  + 5

All in all, this tells us that:  – 2 + 7 = + 5

For those of you who’ve torn your hair out over such problems, I have good news …

… THEY REALLY ARE THIS SIMPLE!

But to believe this, it will help to work out one more problem:  + 13 – 20.

Here, again, are the common-sense questions, along with their answers.

1)  How many people are on each team?
In this problem, + 13 – 20, there are 13 people on the positive team and 20 people on the negative team.

2)  Which team WINS?
Since there are more people on the negative team in this problem, the negative team wins.

3) By how many people does the winning team OUTNUMBER the losing team?
Since the negatives have 20 while the positives have only 13, the negatives outnumber the positives by 7.

Just as you did in the first problem, put together your answers to Questions 2 and 3.

ANSWER TO QUESTION 2:  

ANSWER TO QUESTION 3:  7

ANSWERS TOGETHER:  – 7

All in all, this tells us that:  + 13 – 20  = – 7

Now try these for practice:

a)  – 3 + 9

b) + 1 – 4

c)  –  9 + 23

d)  – 37 + 19

e) + 49 – 82

Answer to Practice Problems:

a)  – 3 + 9 = + 6

b) + 1 – 4 = – 3

c)  –  9 + 23 = + 14

d)  – 37 + 19 = – 18

e) + 49 – 82 = – 33

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like the way Josh explains these problems, you will very likely like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

The Log Blog


Tales from the Tutoring Experience

Tales from the Tutoring Experience

It’s that time of year again when Algebra 2 students are all studying logs — not the lumberjack things, but ‘logarithms’ — so I’d like to present a concept that helps students work with logs.

I call it the “Whole-Part-Equality Principle” (as I’ve never seen it named by anyone else … is there a name for it? Anyone know?) But I prefer to call it the “Peter-Paul-Pennies-in-the-Pocket Principle.”

Here’s how it works. There’s Peter, and there’s Paul. We are told that Peter and Paul have no money except pennies, and they transport their pennies only in their right and left pants pockets (if anyone can think of a way to pack this story with even more p’s, please let me know).

Anyhow, we know three additional facts:

1)  The number of pennies that Peter is transporting equals the number of pennies that Paul is transporting.

2) Peter and Paul each have three pennies in their right pants pockets.

3)  Peter and Paul transport their pennies NOWHERE but in their pants pockets.

QUESTION:  What can we conclude about the number of pennies that Peter and Paul have in their left pants pockets?

ANSWER:  It’s obvious, right? While we don’t know how many pennies Peter and Paul could be transporting in their left pants pockets (it could be any number, right?), it is nevertheless clear that they must have the same number of pennies in their left pants pockets.

WHY?  View it like this … If the wholes are the same (the total number of pennies that Peter and Paul each has), and if one of two key parts are the same (the number of pennies that Peter and Paul have in their right pants pockets), then the other parts must also be equal (the number of pennies they have in their left pants pockets).

Why am I bringing this up? To point out an important principle.

This same principle — if the wholes are equal, and if one of their two parts are equal, then the other parts must also be equal — can be used to solve many log and exponent problems.

EXAMPLE 1:  Suppose you have this equation:  log x = log 7.2. What can we conclude? Well, the wholes are equal (meaning the left and right sides of this equation are equal), and the bases of the logs are equal (logs are always base 10 unless another base is given), therefore the remaining parts, the ‘arguments,’ also must be equal. The ‘argument’ is the term after the word ‘log,’ so for this equation the arguments are x and 7.2, and they must be equal … meaning that  x = 7.2.

EXAMPLE 2:   Suppose you have the equation:  log 2^x = log 16. Again, the wholes are equal, and the logs have the same base, so the arguments must be equal. That means that 2^x = 16. Since 2^4 = 16, x = 4, and that’s the answer.

EXAMPLE 3:  Suppose you have the equation:   a^log x = a^log 12.9. Since the wholes (the left and right sides of the equation) are equal, and since the bases are equal as they are both ‘a,’ therefore the only remaining parts, the exponents, must also be equal. So this means that log x = log 12.9. Following the same logic as we used in Examples 1 and 2, this means that x = 12.9.

Any questions? If so, please post as a comment. If not, please use this principle, and enjoy its profound practicality. (OK, I’m done.)

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com

Tag Cloud

Follow

Get every new post delivered to your Inbox.

Join 208 other followers

%d bloggers like this: