Wikipedia:Reference desk/Archives/Mathematics/2008 May 9

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May 9

Indeterminate relational operator

Is there an official mathematical relational operator that can be used when the equality - or lack of it - of two expressions is not known a priori? For instance, to determine which is the greater of ${\frac {3}{8}}$ and ${\frac {4}{9}}$ one might write,

${\frac {3}{8}}<>{\frac {4}{9}}$ and then by the ususal operations,

$27<>32\,\!$

Thus proving that ${\frac {4}{9}}$ is the greater. However, the symbol <> is not a good choice as this could be construed as "less than or greater than" (ie not equal to) rather than the meaning I intended. SpinningSpark 13:49, 9 May 2008 (UTC)

OT: 4*8=32 Algebraist 13:56, 9 May 2008 (UTC) oops! thankyouSpinningSpark 15:09, 9 May 2008 (UTC)

I don't think there is anything standard for this. However, operators like $\pm$ are generally taken to have a consistent instantiation - for example, $1\pm 2\pm 3$ means $1+2+3$ or $1-2-3$, but not $1+2-3$. Analogously, $2\gtrless 3\iff 4\gtrless 6$ could be taken to mean "$2>3\iff 4>6$ and $2<3\iff 4<6$". But this is non-standard. You can also just not write anything - put the two numbers in front of each other, and leave it blank until you figure out the correct relation. -- Meni Rosenfeld (talk) 14:18, 9 May 2008 (UTC)

Leaving it blank does not really fit the bill as the idea is not so much to determine the correct relation but rather to be able to manipulate the equation while at the same time showing that the inequality status is not known. The $\gtrless$ looks good though -what is its usual meaning? If it is "greater or less than" then it is undesirably excluding "equal to". SpinningSpark 15:09, 9 May 2008 (UTC)

Uhh, whatever it is you're doing, it's wrong. 4/9 exceeds 3/8. –King Bee ^{(τ • γ)} 15:03, 9 May 2008 (UTC)

I know - Algebraist already pointed that out above, but was kind enough to write it small. I have now corrected the original question. SpinningSpark 15:09, 9 May 2008 (UTC)

Too small for my eyes, apparently. Apologies. –King Bee ^{(τ • γ)} 15:14, 9 May 2008 (UTC)

I don't know of any standard way of writing that. Normally, you would work out which way round it was before writing it - writing it down is then just a proof, rather than the actual working. The actual working you can do on rough paper and it doesn't make any difference what notation you use as long as you understand it. --Tango (talk) 15:31, 9 May 2008 (UTC)

For my own notes I use ${\stackrel {?}{=}}$ or "vs.". The former emphasizes equality vs. non-equality more than I wish it did but I haven't found any better notation. Eric. 144.32.89.104 (talk) 17:15, 9 May 2008 (UTC)

One way that would avoid the need for such a notation would be to write it as:

${\begin{aligned}{\frac {3}{8}}={\frac {3\cdot 9}{8\cdot 9}}={\frac {27}{63}}\\{\frac {4}{9}}={\frac {4\cdot 8}{9\cdot 8}}={\frac {32}{63}}\\\implies {\frac {3}{8}}<{\frac {4}{9}}\end{aligned}}$

--Tango (talk) 18:03, 9 May 2008 (UTC)

For what it's worth, TeX has the operator \gtreqless:
$a\gtreqless b.$
The very first meaning that comes to my mind though is that there are instances for which term a is greater than b, some other instances for which $a=b$, and finally some cases for which $a<b$. Pallida  Mors 18:16, 9 May 2008 (UTC)

(Edit conflict) I've used similar for other operators. For example ${\overset {?}{<}}$. Note for this instance though you could probably just circumvent the issue by subtracting one from the other and going from there:

${\frac {3}{8}}-{\frac {4}{9}}={\frac {3\cdot 9}{8\cdot 9}}-{\frac {8\cdot 4}{8\cdot 9}}={\frac {27-32}{8\cdot 9}}=-{\frac {5}{8\cdot 9}}<0$

$\therefore {\frac {3}{8}}<{\frac {4}{9}}$

Ahem. 8*9. We so stink at arithmetic. Too many abstract operators swimming around in our heads I guess. --Prestidigitator (talk) 18:19, 9 May 2008 (UTC)

I am deeply regretting having given an example (and not just because I got the answer wrong). The question was not to find an alternative way of solving that particular trivial problem. The example was given just to make clear the meaning of the operator I required a symbol for. I guess that $\gtreqless$ is the closest I am going to get. Thanks everyone. SpinningSpark 18:30, 9 May 2008 (UTC)

All the methods given are precisely equivalent, they differ only in notation, which is what you were asking about. --Tango (talk) 19:49, 9 May 2008 (UTC)

It is exactly your misunderstanding that led to my regret at phrasing the question that way. I know the methods are equivalant but my question was not one of notation or method in general, but for the symbol for a specific operator. SpinningSpark 21:23, 9 May 2008 (UTC)

It doesn’t seem like he misunderstood – he specifically stated that the notation is what you were asking about.

Anyways, I’m personally biased toward blanks, filling in the correct information when you know it. Such as ${\frac {3}{8}}$ ___ ${\frac {4}{9}}$

I also like the question mark relations if there are only two options you might care about: the relation or it’s negation GromXXVII (talk) 23:15, 9 May 2008 (UTC)

Yeah, I always use the question mark over the less-than symbol. I doubt there's any official symbol, other than the standard variable symbols for relations (R, etc). Black Carrot (talk) 02:11, 10 May 2008 (UTC)

"Notation" is a superset of "symbols", thus "not... notation... but... the symbol" doesn't make much sense. Surely you know that this "operator" is not a real operator\relation, but rather a handwavish way to remind you what is the goal of your calculation. The reason people pointed out alternative approaches to solving your toy example is to demonstrate why the sought symbol isn't really necessary. -- Meni Rosenfeld (talk) 17:32, 10 May 2008 (UTC)