Reason why multiplication gives fraction of a number

[mtanti]

Why is it that when you do A * b/c the answer is one (b/c)th of A? What is the operation of multiplication doing to the two numbers so that this happens? Also how does this logic work when A is also a fraction?

 

[DaveC426913]

Multiplication is mathematically equivalent to the word "of".

So 4 of 1/3 gives you 4 1/3rds - or 4/3rds.

Think of poker chips as counters. You've got a pile of poker chips, all with the value of "1/3" printed on them. If you have a poker chip, you have 1/3. Now what if you have 4 of them? Well, you literally have four(4) of the pile of thirds don't you? The chips you have value 4/3rds.

And if you have only 1/2 of a chip (half of a third), you have chips that value 1/6.


So, when you see the symbol 'x' (times), think 'of'.

 

[mtanti]

Yes but why is it that you use multiplication to findthat (1/3)rd or 4? From what you say you are just showing that you are finding 4 thirds and not (1/3)rd of 4.

 

[chroot]

Why is it that when you do A * b/c the answer is one (b/c)th of A?

It isn't.

<br /> A \cdot \frac{b}<br /> {c} = \frac{A}<br /> {{c/b}}<br />

In other words, it's one "(c/b)th" of A.

In more detail,

<br /> A \cdot \frac{b}<br /> {c} = A \cdot bc^{ - 1} = \frac{{Ab}}<br /> {c} = \frac{A}<br /> {{b^{ - 1} c}} = \frac{A}<br /> {{c/b}}<br />

- Warren

 

[arildno]

Yes but why is it that you use multiplication to findthat (1/3)rd or 4? From what you say you are just showing that you are finding 4 thirds and not (1/3)rd of 4.

1/3 of 4 is the same as 4 thirds, by commutativity of multiplication.

 

[DaveC426913]

...you are finding 4 thirds and not (1/3)rd of 4.

What is the difference?

Q: If I have 4 pies, each cut into three, and I eat 8 of the peices, how much pie do I have left?

a] 4 'one-third' pie slices?
b] 4/3rds of a pie?
c] 1/3 of 4 pies?

A: all of the above

 

[mtanti]

The question is why is it that when you find 4 1/2s of a pie you are also finding 1/2 of 4? Is it simple commutivity? If so explain by logic what multiplication is doing instead of just adding 1/2 to itself.

In another thread it was concluded that multiplication is just a fast way of solving this problem (finding a fraction of a number) and that there is a longer and more logical way to do it. I was wondering if you guys knew something about it...

 

[chroot]

Yes, multiplication is commutative. That's the only answer. If you multiply some numbers in one order, you get the same result as multiplying in any other order.

- Warren

 

[FrogPad]

Do you want a mathematical operation of how multiplication works?

 

[mtanti]

I know how to multiply, but I seem to misunderstand what it does in practice to physical quantities. Why is it that when you multiply 2 apples by 1/2 you get 1/2 of 2 = 1 apple.

 

[chroot]

Because multiplication is commutative. Could you think of any other sensible property?

- Warren

 

[DaveC426913]

I don't know about the rest of you, and particularly mtanti, but simply repeating "multiplication is commutative" does not really provide a concrete understanding of the issue.

You're talking about mathematical theory while he's asking about apples. (OK, to be fair, he was talking theory at first too, but clearly he's looking for a more "grokable" answer.)

 

[DaveC426913]

I confess mtanti, like everyone else, I'm not sure what exactly you're having difficulty with.

I know how to multiply, but I seem to misunderstand what it does in practice to physical quantities. Why is it that when you multiply 2 apples by 1/2 you get 1/2 of 2 = 1 apple.

What about the above apple-chopping does not make sense to you?

 

[chroot]

Yes, Dave.. I don't get it either. I don't see any other way multiplication could possibly work. I also don't understand what's confusing about "half of two apples = one apple" and "two halves of one apple = one apple."

- Warren

 

[mtanti]

OK let me rephrase the question. Forget about the apples.

Why is it that to find a fraction of a quantity (say 1/2 of 100), you multiply that quantity by the fraction. What does multiplication have to do with finding fractions of quantities?

Better or worse explanation?

(I asked this question in the maths forum but they said that it was a philosophical question.)

 

[chroot]

Halving an apple means dividing it by two, yes? Multiplying it by 1/2 means the same thing, because 2 and 1/2 are multiplicative inverses.

I really have no idea what you're asking us to explain, or what's not clear to you.

- Warren

 

[arildno]

mtanti:
Are you at all familiar with the idea that you actually have to DEFINE what you want to talk about?

 

[eieio]

mtanti:

Look at it like this:

1 * A = A

I think it's clear this makes sense to you concretely.

0 * A = 0

This also makes intuitive sense.

Doesn't it also make sense that if you choose a number smaller than 1 but greater than 0, that the result of multiplication should be between A and 0?

 

[radou]

Debates based on topics such as this one really do not seem to have an end.

 

[arildno]

mtanti:

Look at it like this:

1 * A = A

I think it's clear this makes sense to you concretely.

Does it really?
Why?
Does that string of symbols mean anything independently from definitions you set up?

 

[chroot]

Go easy on him, arildno, he's a buddy of mine, and is only trying to help.

- Warren

 

[eieio]

Does it really?
Why?
Does that string of symbols mean anything independently from definitions you set up?

Based on observations of earlier posts, it seems that mtanti understands and accepts the idea and sense of those two "strings of symbols." I thought it made sense to point out the line between those points.

And no, they don't mean anything independently of the definitions. But then again, I don't think that's the issue here.

 

[arildno]

Well, with all respect (and I do understand you wanted to start out with something mtanti might feel to be intuitively true), his main problem, as I see it, is that he goes around with all sorts of vague, unexpressed ideas of "what is multiplication" "what is a fraction", and these ideas somehow don't fit together properly.
Thus, I wanted to force him to re-think and express what he actually is thinking, and comparing these thoughts of his by the clear-cut definitions of mathematics.
Unless he is willing to speak the language of others, then it is rather futile trying to help him out.

 

[DaveC426913]

Why is it that to find a fraction of a quantity (say 1/2 of 100), you multiply that quantity by the fraction. What does multiplication have to do with finding fractions of quantities?

I still think that my explanation in https://www.physicsforums.com/showpost.php?p=1083624&postcount=2" explains it.

multiplication means 'of'

So, 1 of 2 means:
Of the two apples, you have one. How much of the apples do you have? You have 1/2 of them. So, 1/2 x 2 = 1.

Likewise, 1/2 of 100 is the same as 1/2 x 100.

 

[mtanti]

You know actually it's true, I do have problems expressing myself, even at school. Teachers just explain the surface of my questions and so the heart of the question remains unanswered. Still, there are questions which one cannot answer on his own so I need to ask for help when I need it...

I'll state what I think so that we'll straighten things up.

A * a/b = (A/b) * a
which means that you first break A into b equal pieces and then add the magnitude of 'a' of those pieces.

Is this it? b does not enter the multiplication process, it is just divided. But division is a process of multiplication. Therefore the question of what multiplication is doing when multiplicating A by 1/b to find (1/b)th of A is still present. What is happening to the 2 numbers so that the result is actually (1/b)th of A?

The part of the answer being between A and zero makes sense but how are you sure that the actual answer is the one you're expecting? I'm just trying to be certain of the basics instead of just accepting things as they are.

 

[arildno]

Division is not a separate process from multiplication.

The so-called 4 disciplines of arithmetic are all concerned with:
How to find the denary (or decimal) representation of a number?
For example, the process of adding 37 to 59 is to find the denary representation of the number 37+59 (that is a NON-denary representation of the number).
Dividing 45 by 9, that is perform the division 45:9, is to find the denary or decimal representation of the number given as the product 45*(1/9) (this is a NON-denary representation of the number).

The number 1/9 is the number that, by definition of it, fulfills 9*1/9=1.

Now, a bit more on division:
1/9, which is the standard form of writing this number, is not a decimal reprentation of itself.
Thus, it seems kind of hard how to find the decimal representation of the PRODUCT 50*1/9 when one of the FACTORS has not been written in decimal form!
Thus, we do the following trick:
Let C be the decimal representation of 50*1/9, that is, 50*1/9=C

Multiplying this with 9 yields:
50=C*9.

Thus, we may find C by rephrasing our question a bit:
What number (in decimal form), when multiplied by 9, yields 50?

The answering strategy to this question is what is commonly called the discipline of division.

 

[Hurkyl]

A * a/b = (A/b) * a
which means that you first break A into b equal pieces and then add the magnitude of 'a' of those pieces.

(For nonzero b) the definition of (a/b) is that it's the unique number such that:

(a/b) * b = a

 

[mtanti]

Is multiplication by a fraction a/b just defined to be divide into b equal parts and take a of them? Is it just a definition and that is why you multiply?

 

[Hurkyl]

(Talking about the rationals)

No, the definition of multiplication by (a/b) is:

(a/b) * (c/d) = (a*c) / (b*d)

(and, don't forget that the integer n corresponds to the fraction (n/1))

 

[mtanti]

But that is just a derived short cut right?

But anyway is multiplication by a fraction to attain a fraction of a magnitude just a 'man made' definition? It can't be logically shown that when you multiply a magnitude by a fraction you get the fraction of the magnitude?

It's just divide the magnitude by the denomenator (and thus start a process to do that) and then multiply by the numerator.