Category: division

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original photo from Richard Masoner Edited by Brian

Not sure what got me thinking about this today, but I was musing about the commutative property, and how it applies to addition and multiplication. (Yeah, it was a pretty boring morning.)

Specifically, I was thinking about the word, “augend”. The augend of an addition problem is the first of the series of addends. It’s not a word that is usually taught, and I was wondering why not.

You should be aware that addition and multiplication have the commutative property, and subtraction and division don’t. That just means 4+6 is the same as 6+4, and 3*7 is the same as 7*3, but 8-2 and 2-8 are not the same, nor are 9/3 and 3/9.

So it doesn’t really matter which of the numbers is placed where in addition, so you don’t need to specify which is the augend, and which is not. You can call them all addends. Same with multiplication - you can call them all multiplicands. (Not so with minuends, subtrahends, dividends and divisors, though - they don’t commute.)

But I think we should teach about augends. There are several good reasons why, and you can read about them at a lesson I just put up at Names of the numbers in basic arithmetic operations. That lesson is all about why we give the different parts of arithmetic problems their names, (like dividend, divisor, and quotient) and why it makes sense to learn them.

Now that I feel like I’ve cleared this all up for you and me, I’ve got something that I’m not so clear on. Maybe some kind reader has some insight about it she or he’d like to share. It’s this:

Since you must differentiate the names of the parts of subtraction or division problems, what happens if a problem has more than two terms, like 8-3-2? Is there a name for the third term? What if there are fourth or fifth terms?

I assume that they are called, “the second (secondary?) subtrahend, the third (tertiary?) subtrahend, and so on, but I’m not sure.

Anybody got any insights?

You may want to check Names of the numbers in basic arithmetic operations first, though.

By the way, if anybody can write me and tell me why I chose the image that I used for this post, I’ll send them a free Math Mojo e-booklet. (Use the contact me near the top right navigation bar on this page.)

Update: You don’t have to write about that anymore - we have a winner! Mark (see below) got the booklet.

To clarify: The big dog in the picture is “Doggy Daddy,” and the little dog at the door of the train is “Auggie Doggy.” (They are Hannah-Barbera cartoon figures.) They are about to enter a commuter train. Get it? Augend/Auggie, commutative property/commuter train? (Groan.)

Recently an interested reader (a teacher) wrote in a great question. I thought you might be interested in it, too. Here it is:

I ran across your website of mathematical terms. Is there a specific name for the division bracket? We are introducing 3rd graders to the vocabulary and symbols. Thank you.

Haven’t you ever wondered about things like that? They may not be earth-shattering like learning math concepts, but I think little things like that make math more interesting.

Whenever you introduce a little thing that makes a child (or anyone else go, “Yeah…I wonder why…” you’ve helped them get a bit more curious - and that’s what it’s all about.

So, if you’re curious to find out the answer, I’ve put up a little post where you can read more about it at MathMojo.com

Happy pondering!

The Professor

A few posts ago, I offered some tips about how to check large division problems without having to multiply huge divisors and quotients to get even huger dividends.

One of the drawbacks to using the “crunch” method, which I described, is that it is not 100% accurate.

Often, people who need to defend the status quo (you know who they are, they work in the principal’s office) insist that checking by crunching is not acceptable, because it is not foolproof.

Let me give an example:

You could divide 1206 by 18 and get 64

It looks like it works. But it doesn’t. The real answer is 67.

Sometimes you can transpose digits, or make a mistake, the crunch of which will work out to the crunch of the real answer. After all, there are only 10 digits which all integers can crunch to.

It is very seldom, though, that you will crunch mistaken digits and do the multiplication, and have the answer come out to a crunch number that still has the same crunch number as the real answer.

The reason that mistakes are so seldom, is that it is easy to add numbers like 1+8 and 6+4 and multiply the results.

Although mistakes still can be made, much less mistakes are made with this method than with the cumbersome method you probably learned in school. Consider this: What is easier to do,

(1+8)*(6+4)
=9*1
=0, (all of which you can do in your head, with no training, in seconds)
or
18*64? (Do you really want to multiply that mentally if you don’t have to?)

There are many small-minded people in education, who insist that methods other than theirs must be foolproof, when their own methods are even less foolproof.

The mission of MathMojo is not only to teach easier, more effective methods, but also to make math more meaningful to you. And one way to do that is to sharpen your critical thinking skills.

Here is a perfect opportunity to do just that. Can you see the flaw in the small-minded person’s argument? They set up conditions that new things must fulfill in order for them to consider using them. But they don’t put their own things under the same conditions.

That phenomenon is one of the most prevalent flaws of society. Catch it when you see it, and call them on it.

Hotcha!

We’ve been talking about using factors to make long-division problems easier, sometimes being able to turn them into a manageable sequence of short-division problems, in which no paper and pencil (and certainly no calculators!) are needed.

Want to try another one? How about

962/52 ?

Well, they’re both even, so that’s going to be a piece of cake to start. Divide both by 2 and turn the problem into:

481/26

Can you factor them further? You can tell that 2 won’t be a factor. And a quick look at 26 tells you that 3,4,5,6,7,8,9 and 10 won’t factor into it. It’s only factor, other than itself and 1 is 13. When you factor 13 into 21, you get 2.

Now all you have to do is test if 481 is divisible by 13. If you know the trick to test for divisibility by 13, you could try that, but let’s just assume you don’t, and go ahead and divide it in our heads.

13 goes into 48 three times, with 9 left over. Carry the 9 to the front of the 1 in 481, and get 91. Divide that by 13, and whaddyaknow, it goes in exactly seven times. That gives us 37.

We have reduced the problem from

without much trouble, and no writing. I think you can handle 37/2 on your own from here.

Right, it’s 18, r. 1.

Remember, that’ the answer to 37/2. But if you want to check it as the answer to 962/52, you’re going to have to re-factor in the 13 and the 2 to the remainder. When you multiply the remainder (1) by the factors (2 and 13) you get 1 x 2 x 13, which is 26.

Checking 962/52 = 18, r. 26

962/52 = 18 r. 26

So the crunch to the problem is 0, remainder 26.

So the crunch to the answer is 0, remainder 26.

The crunch to the problem matches the crunch to the answer, so the answer is very probably right.

In the last post we looked at the problem of 926/18, and we simplified it to 463/9, so we could make it a short division problem.

What if the problem had been 927/18?
Both numbers are not even this time, so it is not readily apparent if they have common factors.

If you know how to factor (if you don’t, you can get a lesson at The Pretty Good Guide to Prime Factorization at MathMojo.com.) then you factor both of these numbers by 9.

Here’s a hint: If a number can be crunched to 9 or 0, then nine is a factor of that number. If you want to know more about crunching, I refer you to “The See-Say-Write Method of Speed Addition“.

There are also many hints you can find about how to determine if numbers are divisible by other numbers. MathMojo will eventually cover this in depth, but I’m sure you can find info if you google “divisibility rules.”

Ok, so let’s factor 927/18.

Using short division by 9, we get 103/2. How easy is the problem now? Just cut 103 in half in your mind and get 51 remainder 1. But remember, like in the last post,
If you factor a division problem before you solve it, you must multiply the remainder by that factor after you are done.

So the answer to 927/18 is 51 remainder 9, (not 1).

Go ahead and check it. Remember how? If not, check out this post.

(Is that title an oxymoron?)

Imagine you have to do this division:

926/18

How would you do it? Would you rewrite it with that funny division symbol (“division bracket,” or “right parenthesis followed by a vinculum over the dividend”)? Would you use a calculator? (Please say “no” to that!)

After you rewrote it, would you start by trying to figure out how many times 18 would go into 92? If you did, you would be doing it the way most people learned in school, and you would be wasting a lot of time and effort.
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They way we are usually taught to check division problems in school is unnecessarily complex. There is a better way. I always wondered why, after thousands of years of mathematics, schools generally haven’t figured that out. But I’d rather try solving the Riemann zeta-hypothesis than figure out why schools teach the way they do.

An astute reader in Iceland (yes, we get readers from the coolest places!) asked the following question:

I have a minor problem regarding crunching division problem. How would you crunch a problem like 275 divided by 11 = 25?
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