To wonder if division notations are really not all equivalent (Telegraph maths challenge)

[Theoretician]

In isolation, I would say all the following are equivalent expressions:-

1

• = 1/3 = 1÷3

3

In a Telegraph maths puzzle, the correct answer depends on whether you think:-

3 ÷"one over three" = 1 or 9.

(For "one over three" please substitute the first of the three notations I mentioned, I can't type it other than at the start of a line because the forum software removes the leading spaces that keep the 1 and the 3 in the correct position.)

I initially said 9, which is apparently correct, but there is no explanation in the video the Telegraph links to as to what rule requires the horizontal bar that separates numerator from denominator in the fraction to have a higher precedence than the other division symbol that appears in the wider expression. If it doesn't have higher precedence, if the three expressions I posted at top are always completely equivalent, then the correct answer is 1.

www.telegraph.co.uk/news/2016/05/15/can-you-solve-the-simple-maths-question-confusing-adults/

I wouldn't.

I would say that 1÷3 is a binary operator acting on two integers

1
_
3

is a rational number, one third, which is atomic

and

1/3 is a new-fangled hacky notation which might be the one, might be the other and might mean one shilling and thruppence (for older readers).

So the answer is quite clearly 9: a binary operation (÷) acting on two operands, one an integer (3) and the other a a rational (one third).

The ambiguity arises because you're so used to / being both part of a rational and a division operator you assume the same is true of other things.

How many times does one third fit into three?
nine times

No, you've just misunderstood the notation.

Three divided by "one over three" is equivalent to three multiplied by "three over one" which is nine.

Do you mean you should be expanding the expression to "three divided by one divided by three"? That's not what that means although it's more obvious if you re frame the question:

"I have three pizzas. Each toddler will eat a third of a pizza. How many toddlers can I feed?"

or

"3 ÷ 0.333333"

Yes, it depends on whether the expression is interpreted at "three divided by one third" or "three divided by 1, then divided by 3". The former is correct - you are dealing with a fractional denominator, so flip it and then multiply is the rule.

The notation

x
_
y

is also used where both x and y can be any (possibly quite complex) expressions. So for "1 over 3" to only be a number, and not a (vertical) expression with a binary operator, there must be a special rule that says if numerator and denominator are integers then the notation means something different to when they are not.

"it depends on whether the expression is interpreted at "three divided by one third" or "three divided by 1, then divided by 3""

It would never be interpreted as the latter, given the notation in the OP.

No.

Only what appears immediately above and below the line (in this case 1 and 3 respectively) are part of the expression. You could put brackets in to clarify matters, but you don't have to because of notation convention.

If it were:

1 ÷ 3 + 6
3

we shouldn't try to shove the "÷3+6" on to the top of the first expression, because they don't belong there.

Three divided by "one over three" is equivalent to three multiplied by "three over one" which is nine.

See, in your mind you (and some others who've posted) are treating "one over three" as a single subsidiary unit in the wider expression, but the question was what rule dictates that you should do that?

Is there a rule that says "x over y" is (for precedence purposes) a number if x and y are integers, but means "x divided by y" otherwise?

Isn't it more l likely that the notation always means "x divided by y" but there is an implicit parenthesis round the whole thing when used in a wider expression?

What rule? Fucking notation.

ARGH the width of the line limits the extent of the expression. Put brackets round it if it helps.

If you wanted to make it say what you wanted, you'd need to write:

3÷1
..3

But it doesn't say that.

we shouldn't try to shove the "÷3+6" on to the top of the first expression

In that example, the equivalent wouldn't be putting it on top, but the bottom, i.e. we would be asking why your expression shouldn't be rewritten as:-

1/3/3 + 6

I'm not disagreeing with any of you that we should treat "one over three" as a unit, but given that the problem is about precedence, and the general meaning of "x over y" is "x divided by y", and there are two other possible notations for "x divided by y" what rule dictates that this notation should deliver a different result to the other two, when used in wider expression?

It doesn't!

I'm not explaining this at all well.

If it's bounded by the line, it's part of the dividing. The numbers versus algebra stuff is irrelevant.

3+4×6+23
.8+1

4x-9+6xy
7-3y

BIDMAS
indices one third equals three to the minus one
so
do one divided by three and then three divided by the result of the bolded calculation

There is an implicit set of brackets at the immediate end of the line.

So

a + b
---- +1
c + d

that is, ((a + b) / (c + d)) + 1

is not the same as

a + b + 1
------
c + d

that is (a + b + 1) / (c + d)

nor is the same same as swinging the line anti-clockwise and making it into a / and writing it as

a + b / c + d + 1

which would be equivalent to

a + b
-----
c + d + 1

or

a + (b / c) + d + 1

and would be wrong.

So

a + b
---- +1
c + d

is

( a + b )

( --- ) + 1
( c + d )

where the stacked brackets are one large bracket the height of the whole expression.

And that's the key point: that there's an implicit set of brackets around the algebraic fraction, and you go inside those brackets with care.

BIDMAS
indices one third equals three to the minus one

So would BIDMAS say that "x over y" must be interpreted as "x times y to the minus 1"?

If I can find a link that supports that, I think that would be the answer.

Yes. That's what it means.

math.berkeley.edu/~gbergman/misc/numbers/ord_ops.html