Is Mr. Wizard's Statement About Ambiguity True?

[evagelos]

In browsing other forums ,i came across the following statement:



Mr. Wizard 48/2y is kind of ambiguous, since it could mean (48/2)y or 48/(2y), depending on what the author means.

Cap'n Refsmmat
SFN Administrator

Could that under any circumstances be true??

 

[micromass]

Hi evagelos!

I suggest you read this: https://www.physicsforums.com/showthread.php?t=494675

To answer your question short: there is only one meaning to 48/2y and that is (48/2)y. This is the result you get when following the mathematical rules.

I agree that there are a lot of people that actually mean 48/(2y), but that is bad notation. It's ok if everybody knows what they mean, but it's still incorrect.

Also, if you feel that a notation could be misinterpreted, then use brackets. It's no coincidence that you'll NEVER see things like 48/2y turn up in science papers, even if the meaning is clear.

 

[Hurkyl]

Also, if you feel that a notation could be misinterpreted, then use brackets. It's no coincidence that you'll NEVER see things like 48/2y turn up in science papers, even if the meaning is clear.

Unfortunately, I have seen it in some math papers -- in both the correct and incorrect fashion. But usually as you say, people have the foresight to avoid this issue.

 

[micromass]

Unfortunately, I have seen it in some math papers -- in both the correct and incorrect fashion. But usually as you say, people have the foresight to avoid this issue.

My God I have lost faith in the mathematical community now...

 

[evagelos]

Hi evagelos!

I suggest you read this: https://www.physicsforums.com/showthread.php?t=494675

To answer your question short: there is only one meaning to 48/2y and that is (48/2)y. This is the result you get when following the mathematical rules.

I agree that there are a lot of people that actually mean 48/(2y), but that is bad notation. It's ok if everybody knows what they mean, but it's still incorrect.

Also, if you feel that a notation could be misinterpreted, then use brackets. It's no coincidence that you'll NEVER see things like 48/2y turn up in science papers, even if the meaning is clear.

How is a/b defined in algebra??

 

[micromass]

How is a/b defined in algebra??

a/b is defined as the unique number c such that a=bc. Not sure what you're getting at here...

 

[Mute]

Also, if you feel that a notation could be misinterpreted, then use brackets. It's no coincidence that you'll NEVER see things like 48/2y turn up in science papers, even if the meaning is clear.

As Hurkyl said, abuses of notation are not uncommon. For a common example, in statistical mechanics, one often sees $\mbox{stuff}/k_BT$, as $k_B$ and T always come in that combination (when the author hasn't decided to set $k_B = 1$ or write $1/\beta = k_BT$), so it's generally unambiguous in that context. However, I have also definitely read papers in which it's not so clear what is meant, so typically I must assume that if I see an expression abcd.../efgh... that it's really (abcd...)/(efgh...), otherwise the author would have written it abcd...fgh.../e. An annoying disregard for BEDMAS when working with variables, but a common one.

 

[coolul007]

Is there anyway to write 2y where it does not mean 2 times y or is the coefficient of y? Since when did "programmer" notation cause ambiguity in algebraic notation. This would clearly be written with a horizontal line not with a slash or virgule.

 

[evagelos]

a/b is defined as the unique number c such that a=bc. Not sure what you're getting at here...

Then 48/2y is defined as the unique number c such that 48 =2y.c

And c can be only 24/2y.

Hence 48/2y = 24/y ,only

 

[micromass]

Then 48/2y is defined as the unique number c such that 48 =2y.c

And c can be only 24/2y.

Hence 48/2y = 24/y ,only

No, what you have is 48/2*y. The rules that govern this explicitely say that such an expression should be evaluated from left to right. Thus it should be read as (48/2)*y, this is by convention.

If you mean 48/(2y), then you should write it as such. (but as you can see above there are horrible papers where they tend not to do this...)

The same thing arises with 24/2+y, the rules say that / must be evaluated first, and then +. Thus the expression becomes (24/2)+y. If you wish to divide 24 by 2+y, then you should write 24/(2+y).

 

[evagelos]

No, what you have is 48/2*y. The rules that govern this explicitely say that such an expression should be evaluated from left to right. Thus it should be read as (48/2)*y, this is by convention.

What rules? I only know of axioms in algebra and none of them speeks about convensions

You could also write a/b as a/1*b ,then a/b = ab

If you mean 48/(2y), then you should write it as such. (but as you can see above there are horrible papers where they tend not to do this...)

Lets clear one thing: When we write 48/2y do we mean: $$\frac{48}{2y}$$??

The same thing arises with 24/2+y, the rules say that / must be evaluated first, and then +. Thus the expression becomes (24/2)+y. If you wish to divide 24 by 2+y, then you should write 24/(2+y).

 

[coolul007]

$${\frac{48}{2y}}$$ What is ambiguous with that?

 

[evagelos]

$${\frac{48}{2y}}$$ What is ambiguous with that?

Nothing .$${\frac{48}{2y}} = \frac{24}{y}$$

 

[micromass]

What rules? I only know of axioms in algebra and none of them speeks about convensions

The axioms in algebra talk about binary operations. For example, * and / are binary operations. If you go by the axioms, then things like

$$2*2*2$$

don't exist. Here's where conventions come in. Indeed, by associativity, it turns out that (2*2)*2 and 2*(2*2) are the same. And we denote this quantity by 2*2*2. So far no trouble.

However, when arguing in algebra, things like 48/2*2 are not defined. And there is no way to give them a meaning. We could try (48/2)*2 and 4/(2*2), but they turn out to be different. That's why mathematician have put in the convention that things like 48/2*2 are to be evaluated as (48/2)*2.

If you purely follow the axioms, things like 48/2*2 or 2*2*2 don't exist. However, to make life easier, we have put in such convention. And the convention is that 48/2y is to evaluated as (48/2)y. If you don't like it, then use brackets.

Lets clear one thing: When we write 48/2y do we mean: $$\frac{48}{2y}$$??

No, when we write 48/2y, we mean $\frac{48}{2}y$.

 

[micromass]

Is there anyway to write 2y where it does not mean 2 times y or is the coefficient of y? Since when did "programmer" notation cause ambiguity in algebraic notation. This would clearly be written with a horizontal line not with a slash or virgule.

Quite right, I would discourage everybody from using programmer notation and always use horizontal lines. However, things are not always ideal...

 

[evagelos]

The axioms in algebra talk about binary operations. For example, * and / are binary operations. If you go by the axioms, then things like

$$2*2*2$$

don't exist. Here's where conventions come in. Indeed, by associativity, it turns out that (2*2)*2 and 2*(2*2) are the same. And we denote this quantity by 2*2*2. So far no trouble.

However, when arguing in algebra, things like 48/2*2 are not defined. And there is no way to give them a meaning. We could try (48/2)*2 and 4/(2*2), but they turn out to be different. That's why mathematician have put in the convention that things like 48/2*2 are to be evaluated as (48/2)*2.

If you purely follow the axioms, things like 48/2*2 or 2*2*2 don't exist. However, to make life easier, we have put in such convention. And the convention is that 48/2y is to evaluated as (48/2)y. If you don't like it, then use brackets.




No, when we write 48/2y, we mean $\frac{48}{2}y$.

Do you know what is a well formed formula ??

Is a 48/2*2 a well formed formula? .Since you defined / as a binary operation??

 

[micromass]

Do you know what is a well formed formula ??

Yes, I know wff's.

Is a 48/2*2 a well formed formula? .Since you defined / as a binary operation??

No 48/2*2 is not a wff. Even 2*2*2 is not a wff. However, to make our lives easier, we have set a series of conventions to live by. However, we should always be aware what the conventions mean and what they represent. And in this case, 48/2*2 is by convention equal to (48/2)*2.

Why do we make conventions? To make our life easier. If we wouldn't have conventions and if we would be reasoning in wff's all the time then even simple statements like 0+0=0 become quite difficult:

$$\forall x\forall y\forall z((({}^\neg (\exists u(u\in x)))\wedge (({}^\neg(\exists u(u\in y))))\wedge(\forall u((u\in z)\leftrightarrow ((u\in x)\wedge (u\in y)))))\rightarrow ({}^\neg(\exists u(u\in z))))$$

Like you see, we don't like to work with wff's...

Edit: Hmmm, something is fishy with my formula. But in any case, it's not even a wff, since $\exists$ and $\leftrightarrow$ aren't allowed. So it becomes even more complicated...

 

[evagelos]

Yes, I know wff's.



No 48/2*2 is not a wff...

Why??

 

[micromass]

Because / is an operation:

$$/:\mathbb{R}\times\mathbb{R}_0\rightarrow \mathbb{R}$$

and * is an operation

$$\ast:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$$

So correct wff's are of the form *(/(48,2),2) or /(48,*(2,2)). Already we denote this as (48/2)*2 and 48/(2*2), so strictly spoken, not even these things are wff's.

But writing things like 48/2*2 is certainly not a wff.

In fact, a wff is formed by certain rules (simplified here):
1) constants are wff's, for example 48 and 2.
2) if a and b are wff's than so are (a*b) and (a/b).
3) nothing else is a wff.

As you can see, by 2, we have that (48/2) is a wff. Again by 2, we have that ((48/2)*2) is a wff.
However, there is no way you can construct 48/2*2 from the rules (1) and (2). So by (3) it isn't a wff...

 

[evagelos]

Because / is an operation:

$$/:\mathbb{R}\times\mathbb{R}_0\rightarrow \mathbb{R}$$

and * is an operation

$$\ast:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$$

So correct wff's are of the form *(/(48,2),2) or /(48,*(2,2)). Already we denote this as (48/2)*2 and 48/(2*2), so strictly spoken, not even these things are wff's.

But writing things like 48/2*2 is certainly not a wff.

In fact, a wff is formed by certain rules (simplified here):
1) constants are wff's, for example 48 and 2.
2) if a and b are wff's than so are (a*b) and (a/b).
3) nothing else is a wff.

As you can see, by 2, we have that (48/2) is a wff. Again by 2, we have that ((48/2)*2) is a wff.
However, there is no way you can construct 48/2*2 from the rules (1) and (2). So by (3) it isn't a wff...

How about:

if 2 and 2 are wwf so is 2*2. (by rule 2)

if 48 and 2*2 are wwf so is 48/2*2. (by rule 2)

And since 2 and 48 are wwf so is 48/2*2 (by rule 1)

 

[Hurkyl]

if 2 and 2 are wwf so is 2*2. (by rule 2)

That's not the conclusion of rule 2; read it again.

Have you heard of the term "fully parenthesized arithmetic expression"? That's essentially what he's describing.

 

[evagelos]

That's not the conclusion of rule 2; read it again.

Have you heard of the term "fully parenthesized arithmetic expression"? That's essentially what he's describing.

The right expression is (2*2) and (48/(2*2)), but we usually write a*b instead of (a*b) and a/b instead of (a/b).( convention)

For example we write : a*b = b*a ,for all real a,b

But as i said before if : 48/2y = (48/2).y =24y and not 48/2y = 24/y we go against the definition of division "/"

Here is the rule for the "fully parenthesized arithmetic expression"

1) Every constant is a term

2)If F is an n-place operation symbol , and $$t_{1},t_{2}...t_{n}$$ are terms (not necessarily distinct),then F($$t_{1},t_{2},...t_{n}$$) is a term.

3)Only strings are terms ,and a string is a term only if its being so follows from (1),(2)

 

[micromass]

48/2y = (48/2).y =24y and not 48/2y = 24/y we go against the definition of division "/"

We don't go against the definition of division at all, in fact, division has nothing to do with it.

How would you evaluate 2-2+3? Is it (2-2)+3 or 2-(2+3)? That's essentially the same thing. By convention, we choose to evaluate it the first way.
The same thing happens with 2/2*3, by convention we evaluate it as (2/2)*3.

Not that I believe you should trust it, but check most of the calculators and programming languages...

 

[coolul007]

Yoo hoo, algebra not programming or calculators. Calculators evaluate -2^2 as -4. So, now we can get into the discussion whether numbers carry a sign or not.

 

[micromass]

Yoo hoo, algebra not programming or calculators. Calculators evaluate -2^2 as -4. So, now we can get into the discussion whether numbers carry a sign or not.

Well, if calculators say $-2^2=-4$, then they're correct!

I don't say you should trust what your calculator says, but sometimes they give indications of what is true...

 

[evagelos]

We don't go against the definition of division at all, in fact, division has nothing to do with it.

How would you evaluate 2-2+3? Is it (2-2)+3 or 2-(2+3)? That's essentially the same thing. By convention, we choose to evaluate it the first way.
The same thing happens with 2/2*3, by convention we evaluate it as (2/2)*3.

Not that I believe you should trust it, but check most of the calculators and programming languages...

2-2+3 can only be (2-2)+3 . 2-(2+3) =2-2-3 because if you put parenthesis in front of a sum then you change the result.

Let me ask you a question .

When we write a/b ,do we mean b divides a ,only when a and b are just Nos ,or could they be algebraic quantities ,like a= 2x+3 ,b =2y e.t.c??

 

[micromass]

2-2+3 can only be (2-2)+3 . 2-(2+3) =2-2-3 because if you put parenthesis in front of a sum then you change the result.

The only reason that we evaluate 2-2+3=(2-2)+3 is because we made the convention to evaluate a sum from left to right.
Similary, we evaluate 6/2*3 as (6/2)*3 because we made that convention.

I don't get why you understand 2-2+3 and not 6/2*3, it's the same thing...

Let me ask you a question .

When we write a/b ,do we mean b divides a ,only when a and b are just Nos ,or could they be algebraic quantities ,like a= 2x+3 ,b =2y e.t.c??

[/QUOTE]

They could be quantities, of course. You can divide the polynomails 2x+3 and 2y and you'll get (2x+3)/(2y). But writing 2x+3/2y is wrong because that would mean

$$2x+\frac{3}{2}y$$

 

[coolul007]

When a = 2x + 3 we treat a as a container mainly ( ), therefore its expansion always implies parentheses.

 

[micromass]

Indeed, the same thing happens with negations. If a=2x+3, then -a=-(2x+3) and not -2x+3.

 

[Hurkyl]

Anyways, evagelos, this is a closed topic and you have a long history of being argumentative, so I'm going to close this before you get yourself into trouble. micromass has already linked you to the summary of the facts in the matter.

When you understand the summary and if still have questions (and if you can avoid being argumentative), you can reopen the topic.