Wikipedia:Reference desk/Archives/Mathematics/2014 April 26

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April 26

1/2÷1/2

In mathematics, 1/2÷1/2=1, but if you take an apple, and divide it in half, you get half an apple. Then if you take that half and divide it half again, you get 1/4 an apple. So then why in mathematics is 1/2÷1/2=1? ScienceApe (talk) 17:08, 26 April 2014 (UTC)

Dividing in half is not the same as dividing by one half. If you divide one half in half, you get (1/2)/2=1/4. 1/2 divided by 1/2 gives you the answer of how many times 1/2 is 1/2, which is 1. —Kusma (t·c) 17:44, 26 April 2014 (UTC)

Dividing in half is the same as dividing by 2 (or multiplying by a half). This is the multiplicative inverse of dividing by a half (or multiplying by 2). Dbfirs 18:17, 26 April 2014 (UTC)

"dividing in half" is x divide 2.

"dividing by half" is x divide (1/2) or x times 2

24.192.240.37 (talk) 20:09, 26 April 2014 (UTC)

Could you say that in English, please? Dbfirs 20:44, 26 April 2014 (UTC)

You don't speak Reverse Polish? --ColinFine (talk) 21:54, 26 April 2014 (UTC)

There's a similar English problem with percentages:

• $10 increased by 150% = $25.

• $10 increased to 150% = $15. StuRat (talk) 16:42, 28 April 2014 (UTC)

Now to give a real world example. Say you have half a pizza, and ask "how many times can that half a pizza be divided up into half a pizza ?". The answer, of course, is one time. StuRat (talk) 12:53, 27 April 2014 (UTC)

Yes, I see the problem now: dividing an apple "in two" and "by two" are both the same as dividing it "in half", but dividing "by half" is the inverse. It's just a quirk of the language, but it must confuse many people, including some native speakers of English. Dbfirs 08:42, 28 April 2014 (UTC)

The truth is that there is no reason that it has to make any sense. If you are paying your friend a debt of 4 (you have -4) then when you are done paying half, you have -2 since (-4 * (1/2) = -2). If you are dividing a pizza in 2, (1 ÷ 2 ) then if you only do a quarter of the division you should have 2 pizzas (1 ÷ (2/4)) since 2/4 is a quarter of 2. But that means that when you are a quarter of the way dividing a pizza in half, you temporarily have two whole pizzas. When you are halfway through dividing a pizza in half, you have a full pizza. These are obviously absurd results that don't make sense.

We therefore realize that the symbolism is arbitrary. What do you think of this reasoning, guys? 91.120.14.30 (talk) 14:32, 28 April 2014 (UTC)

I don't see what's so absurd about the first one, if you take negative money as representing a debt (which you did). Then you owe your friend 2 instead of 4 after your partial repayment, and hence have −2.

Could you explain what you meant by doing only a quarter of the division? I'm not sure if I understand you. Double sharp (talk) 14:52, 28 April 2014 (UTC)

While advanced math often no longer has any relation to the physical world, I don't see a problem with dividing by fractions. It takes a bit more thought, but it's quite doable. In your (1 ÷ (2/4)) pizza case, the way to phrase that in English is "Divide one pizza up into fourths. How many groups of two/fourths do you now have ?". StuRat (talk) 16:38, 28 April 2014 (UTC)

I'm rather surprised that someone who asks intelligent questions elsewhere is so confused by fractions. May I politely ask 91.120.14.30, are you just trolling us for fun, or would you like us to find a website that explains division of fractions for you? The concept of a division only a quarter completed seems nonsense to me. Dbfirs 17:24, 28 April 2014 (UTC)

You're right - I just wasn't thinking about it clearly. There is nothing suprising. "How many quarters of a pizza are there in a pizza" is a division problem: 1 ÷ (1/4) and clearly has the answer 4 - just not "4 pizzas" but "4 pieces-of-a-pizza." So is "How many quarters of a pizza are there in 2 pizzas" which is 2 ÷ (1/4). Neither result is surprising. The only mistake I made is in calling the latter result "8 pizzas" rather than "8 pieces of pizza". 91.120.14.30 (talk) 09:08, 29 April 2014 (UTC)

This is a very common type of set problem about fractions, and probably as many people get it wrong as the Monty Hall problem and need it explained. Luckily quite a few eventually agree with the result rather than still disputing it like that famous problem! Anyway the first question to ask so one knows there is a potential problem with reasoning is if dividing a half by two gives a quarter, then is it likely that dividing it by a half which is a totally different number will give the same result? Dmcq (talk) 19:11, 28 April 2014 (UTC)

If you want to talk about doing a fraction ("a quarter") of a multiplication ("of the division"), you need to work in logarithmic space to get the natural symmetries. Then doing a quarter of halving is multiplying by $2^{-1/4}\approx 0.841$. Do that twice and you will have multiplied by ${\sqrt {1/2}}\approx 0.707$; do it four times and you will have halved the original. --Tardis (talk) 02:07, 29 April 2014 (UTC)

True, but in "real space", that's just multiplying by the square root of the square root of a half. (Are we not confused enough already?) Dbfirs 07:18, 30 April 2014 (UTC)

Variant of the modulo operation

Is there a name for this function?

$f(a,n)={\begin{cases}a\,{\bmod {\,}}n,&n\nmid a\\n,&n\mid a\end{cases}}$

Jackmcbarn (talk) 20:59, 26 April 2014 (UTC)

Not this one per se, but if you fix n, then the function $f(a):=f(a,n)$ is a lift of the canonical projection map $\mathbb {Z} \to \mathbb {Z} _{n}$ that associates to each integer in $\mathbb {Z}$ the corresponding element of the residue class group $\mathbb {Z} _{n}$. There are many lifts. The modulo map is one, and the one you wrote down is. All lifts are of the form

$f(a)=(a\,{\bmod {\,}}n)+k(a)n$

for some (arbitrary) function $k:\mathbb {Z} _{n}\to \mathbb {Z}$. Sławomir Biały (talk) 21:31, 26 April 2014 (UTC)

You can get rid of the cases by writing $f(a,n)=(a-1\,{\bmod {\,}}n)+1$. —Kusma (t·c) 12:52, 27 April 2014 (UTC)

I hadn't seen that particular notation before, with $\nmid$ and $\mid$, but Kusma's simplification makes everything much clearer. Is this notation common? -- The Anome (talk) 15:25, 27 April 2014 (UTC)

@Kusma: That's not the same function. In my function, f(10.5, 10) = 0.5, but in yours, f(10.5, 10) = 10.5. (Perhaps I should have specified that I wanted this to work on non-integers, since it seems that's not always the case with mod). I guess $f(a,n)=\lim _{x\to 0^{+}}(((a-x)\,{\bmod {\,}}n)+x)$ would work, but at that point it's just as complicated as the piecewise definition. Jackmcbarn (talk) 17:31, 27 April 2014 (UTC)

(ec) I think people here were probably assuming you were working over the integers in which case Kusma is correct. If your working with reals you probably want something like

$r=a-n\left\lceil {a \over n}\right\rceil +n.$

using the ceiling function. Although you want to check you get the right answer for negative values.--Salix alba (talk): 18:12, 27 April 2014 (UTC)

I had assumed this, since the notation $\mid ,\nmid$ certainly suggests that the arguments are integers. But my response only requires a small change. Then the residue class group is replaced by the quotient group $\mathbb {R} /n\mathbb {Z}$ of the reals by the lattice $n\mathbb {Z}$. There is a natural projection $\mathbb {R} \to \mathbb {R} /n\mathbb {Z}$ and the function f is a lift that is continuous at every point except the identity. Sławomir Biały (talk) 00:28, 28 April 2014 (UTC)