7
$\begingroup$

When people write $x/yz$, do they usually mean $x/(yz)$ or $(x/y)z$?

For example, from Wikipedia

If $p\geq 1/2$ , then $$ \Pr\left[ X>mp+x \right] \leq \exp(-x^2/2mp(1-p)) . $$

Thanks!

$\endgroup$
9
  • 7
    $\begingroup$ It's ambiguous. In this case I would guess that it means the former, but you yourself should not use this notation. $\endgroup$ Oct 14, 2012 at 3:30
  • 4
    $\begingroup$ @Tim: Use $()$, like $\exp(-x^2/(2mp(1-p)))$ or $\exp(-(x^2/2m)p(1-p))$ (or even $\exp(-(x^2/2)mp(1-p))$, whichever is meant. If the $p(1-p)$ is supposed to be in the numerator you could also write $\exp(-x^2p(1-p)/(2m))$ and decrease confusion. In this case I'm not sure which it is; you'll have to do the calculations carefully and re-derive the proof, and check. (And then fix the Wikipedia article.) $\endgroup$ Oct 14, 2012 at 3:52
  • 3
    $\begingroup$ I've almost never seen $x/yz$ meaning $(x/y)z$. If you are thinking about $(x/y)z$, why would you write the ambiguous $x/yz$ instead of $xz/y$ ? $\endgroup$
    – mercio
    Oct 14, 2012 at 5:39
  • 1
    $\begingroup$ This seems relevant. $\endgroup$
    – user02138
    Oct 14, 2012 at 18:37
  • 2
    $\begingroup$ @user02138: Nice indeed. $\endgroup$
    – Tim
    Oct 14, 2012 at 18:50

6 Answers 6

22
$\begingroup$

A few weeks ago I did a search for this construction. I found three things:

  1. It was much less frequently used than I expected; I think this is because mathematicians are concerned that it might be unclear.
  2. The few examples I did find invariably meant $x/(yz)$, not $(x/y)z$.
  3. Complicated examples are extremely uncommon.

Here are some examples I found:

  • In the proof of theorem 4.2 at the bottom of the page there is $u_n/n\alpha_n$. (Alberto Torchinsky, Real-Variable Methods in Harmonic Analysis)
  • Page 3 of Walter Rudin's Real and Complex Analysis uses $y/2\pi$, and the bottom of page 99 has $\delta = \eta/2k$. However, on page 88 of the same book, Rudin writes $1/(2\pi)$, even though $1/2\pi$ always means $1/(2\pi)$, never $(1/2)\pi$.
  • Page 12 of George Simmons' Differential Equations says that the negative reciprocal of $rd\theta/dr$ is $-dr/rd\theta$. Page 45 mentions the use of $1/xy$ as an integrating factor. In a rare complicated example, he writes $(2n)!/(n!)^2 2^n$ (meaning $(2n)!\over(n!)^2 2^n$) on page 222.
  • I found no examples at all where $x/yz$ was used to mean $(x/y)z$.

However, Wikipedia articles are sometimes written by blockheads. From looking at related versions of the same formula, for example here (page 4) or here (theorem 1.5) I guess that that was the case here, and that the $p(1-p)$, if it belongs there at all, should have been in the numerator. I think you should be wary of taking the Wikipedia claims at face value, and instead refer to a source that is written by someone with a reputation.

You should also be wary of taking my claims at face value, since I know absolutely nothing about Chernoff bounds.

$\endgroup$
6
  • 1
    $\begingroup$ Thanks! I really appreciate that you have found so many examples! About the Chernoff bound mentioned in my question, I wonder how you know from the two pdf files that $p(1-p)$ should have been in the numerator instead of denominator? $\endgroup$
    – Tim
    Oct 14, 2012 at 3:59
  • 1
    $\begingroup$ I don't know. I was guessing. I wasn't even able to convince myself that the $p(1-p)$ should be there at all. $\endgroup$
    – MJD
    Oct 14, 2012 at 4:01
  • 1
    $\begingroup$ Well done the research. I'd like to add (with zero research) that if $1/2\pi$ is rare, $1/2/\pi$ is just not done, even though it is completely unambiguous (as nobody claims that '/' should associate to the right, as far as I know) and shorter than $1/(2\pi)$. $\endgroup$ Oct 14, 2012 at 12:33
  • 2
    $\begingroup$ Here is more data I found in Concrete Mathematics. P.64 ex 2.23 "Evaluate $\sum_{k=1}^n(2k+1)/(k(k+1))$ in two ways a. Replace $1/k(k+1)$ by..." ex 2.24 "What is $\sum_{0\leq k<n}H_k/(k+1)(k+2)$?". Note they are not systematic, but always mean $x/(yz)$ even when omitting parentheses around the denominator. Also p.504 in answer 4.40: $n!/p^{\lfloor n/p\rfloor}\lfloor n/p\rfloor!$ which was what I ran into before I started looking more systematically. Clearly GKP, who generally care a lot about notation, don't care parenthesising denominators. $\endgroup$ Oct 22, 2012 at 8:48
  • $\begingroup$ @MJD Wikipedia articles are often written by computer programmers, and $x/yz$ means $zx/y$ in every programming language I know, even in Wolfram's Mathematica (see my answer below). $\endgroup$
    – Reb.Cabin
    Sep 18, 2023 at 16:44
15
$\begingroup$

Because multiplication is expressed by the omission of an operator, simply by juxtaposing two factors, it should be treated as having a higher precedence than an operator whose presence is visible as an explicit glyph.

That is to say $abc/xyz$, should mean $(a\times b\times c)/(x\times y\times z)$. Basically we would like $abc/xyz$ to just be an in-line shorthand for ${abc}\over{xyz}$ with minimal fuss.

The interpretation $((((ab)c)/x)/y)/z)$ means we cannot do this. Expressing ${abc}\over{xyz}$ with a slash becomes cumbersome.

In computer programming languages, it is common for division and multiplication to have the same precedence and left-to-right associativity. But the situation there is different because multiplication is an explicit symbol like *. For instance in C a * b / c * d simply means (((a * b) / c) * d).

But if multiplication is notated by mere juxtaposition, a strong case can be made that it this juxtaposition should also denote "tighter binding" (higher precedence than any binary operator).

Mathematics notation is richly visual and two dimensional. Juxtaposition relationships matter. For instance the way an exponent binds with its base.

$\endgroup$
3
  • 1
    $\begingroup$ +1 Operator priority depends on what is written, not on what it means, so it is entirely natural that juxtaposition may have a different priority than a multiplication symbol. $\endgroup$ Oct 14, 2012 at 12:51
  • 1
    $\begingroup$ Spacing to indicate grouping is important: $x/y \,\, z$ is almost surely $(x/y)z$, and $x \, / \, yz$ is almost surely meant as $x/(yz)$. But $x/yz$ is lacking the spacing that would signify intent. $\endgroup$
    – user14972
    Oct 14, 2012 at 16:23
  • 1
    $\begingroup$ ^ definitely! But notice how / already comes with some space around it, when you have a decent font and decent typesetting: $abc/xyz$. $\endgroup$
    – Kaz
    Oct 14, 2012 at 17:07
5
$\begingroup$

The American Physical Society gives multiplication a higher order than division in the slash notation:

When slashing fractions, respect the following conventions. In mathematical formulas this is the accepted order of operations:

  1. raising to a power,
  2. multiplication,
  3. division,
  4. addition and subtraction.

(Styling applied) (Source: Page 6 of Section IV of Physical Review Style and Notation Guide, retrieved 2012.10.14)

This means that (for them) $$ x/yz = x/(yz) $$ However, keep in mind that this may represent only the view of one group of scientists, I don't know of any established standard.

$\endgroup$
4
$\begingroup$

By the usual order of operations, it means $(x/y)z$. But people do often mean $x/(yz)$ instead, so you have to work out the intention from the context. :(

$\endgroup$
15
  • 3
    $\begingroup$ I would be very interested to see any example from a reputable source that uses $x/yz$ to mean $(x/y)z$. $\endgroup$
    – MJD
    Oct 14, 2012 at 3:43
  • 3
    $\begingroup$ (x/yz) = (x/y)z in most calculators. It's also the order given by "PEMDAS". $\endgroup$ Oct 14, 2012 at 4:01
  • 3
    $\begingroup$ Which calculators let you use juxtaposition for multiply? What if you're entering digits? 5/42 is (5/4)2? :) $\endgroup$
    – Kaz
    Oct 14, 2012 at 6:41
  • 1
    $\begingroup$ The "usual order of operations" (UOOO) applies to arithmetic like $42\div 7\times 3$. Algebraic notation defies UOOO. I would tend to agree that $x\div y\times z$ means $(x/y)z$, yet disagree that it has anything to do with $x/yz$. :) $\endgroup$
    – Kaz
    Oct 14, 2012 at 6:53
  • 2
    $\begingroup$ How the calculator got that is that there is no operator between the $2$ and the $\pi$. I naively just hit the 2 button followed by the pi button, which causes $\pi$ to be interpreted as a correction which replaces the operand $2$. The result $0.636...$ is then just $2/\pi$. $\endgroup$
    – Kaz
    Oct 15, 2012 at 23:58
3
$\begingroup$

I was told once by a senior engineer that the last explicit symbol is what counts, it cancels the action of the previous one. So, for example, xyz/abc means $(x*y*z)/(a*b*c)$ and xyz/ab*c means $(x*y*z)/(a*b)*c$, as expressed by any standard computer language.

I was a little upset by that implicit convention, but soon found out that it worked well and was able to understand all old handbooks, where equations were written straight in a single line.

I suppose aesthetics is what counts here. I have the feeling that our need for explicit symbols came with a generation which had computer programming lessons at college, so a relatively new thing starting at what, the 60's, the 70's?

$\endgroup$
3
$\begingroup$

The link in the answer above by "@dualed" to the American Physical Society seems to be a dead link. Here is a current link, with some more information:

Section E.2.e of the Style Guide of the American Physical Society states that multiplication has higher precedence than division, implying that one should read $x/yz$ as $x/(yz)$.

I know I have read something similar from the American Mathematical Society in the past, but I am unable to find a current citation for corroboration.

I know of no programming language that treats x/y*z like x/(y*z). Rather, programming languages commonly treat x/y*z like (x*z)/y. Mathematica is one of these programming languages:

Wolfram Language 13.0.0 Engine for Mac OS X ARM (64-bit)
Copyright 1988-2021 Wolfram Research, Inc.

In[1]:= FullForm [ x / y z ]                                                                                                                                              

Out[1]//FullForm= Times[x, Power[y, -1], z]

In[2]:= x / y z === x z / y                                                                                                                                               

Out[2]= True

I know of a case, which shall remain anonymous, where this anomaly caused a failure in a robotic controller. An engineer wrote mathematics and a programmer transcribed it verbatim into code. It was in a corner case not caught in testing.

Some software companies mandate that all arithmetic and logical expressions be fully parenthesized.

$\endgroup$
2
  • 1
    $\begingroup$ +1 Rarely can I upvote a new answer to an old question. $\endgroup$ Mar 30, 2023 at 23:37
  • $\begingroup$ Unfortunately, we had a case recently in a Physics code where this problem cropped up again and caused a waste of money. The code ran for days and produced answers that were /slightly/ off some checks inserted via the personal initiative of a paranoid physicist. Without those checks, the incorrect answers would have been published. The subsequent damage to reputations and to policy decisions by those "following the science" is unknown but probably big. $\endgroup$
    – Reb.Cabin
    Sep 18, 2023 at 16:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .