Talk:Multiplication

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For every assortment (unique or otherwise) of numbers there is a unique number called the product. Any two numbers in such an assortment can be replaced with their product without effecting any change in the product of the assortment. Any number of ones can be added or removed with no change in the product. Assortments with products other than zero contain only numbers other than zero.

The word multiplication also is used to refer to reproduction.

"Any two numbers in such an assortment can be replaced with their product without effecting any change in the product of the assortment."

This isn't a property of multiplication. This is a property of algebra that states for B = A, B can be substituted for A in any expression without effecting the value of the expression. The note on reproduction should probably be re-added, though.--BlackGriffen

No, no, you're both missing the point of the original article (which is not mine, but I know enough about group theory to understand it). The text above describes the meaning of "multiplication" in group theory, which is any operation (such as traditional multiplication, which the article now describes) that has the properties noted. The specific property mentioned above is not simple one-for-one substitution. Read it again: it's two-for-one substitution. For any collection of numbers, any two can be removed and replaced by the one number which is the product of those two numbers, and the product of the collection will stay the same. --LDC

Would it be too much trouble to motivate how multiplication gets defined for the rationals and reals? Also, the first paragraph is problematic in defining multiplication as repeated addition; how do you add 2.5 to itself 3.7 times? --Ryguasu 01:56 Feb 25, 2003 (UTC)

I think there is an error, when defining the infinite products from -oo to +oo as the sum of two limits, instead of the product. -Marçal

I corrected the 2 x 3 x 5 = 15 and the 2 x 2 x 2 x 2 x 2 = 16 examples please revert them if I did it wrong Markr9 (talk) 17:52, 6 September 2008 (UTC)[reply]

Could anyone write about how to define multiplication for non-integers in the article? (Current, it says one can define multiplication for real numbers but does not say how.) Or more like is it impossible? --Taku 18:45, Apr 2, 2005 (UTC)

I will try to do so, or at least put an adequate link. -Unknown

The idea is that rationals are obtained from integers by localization, and reals are obtained from rationals as factor ring, idem for complex numbers from reals. Both operations involve an injective ring morphism, so that the result of multiplication remains the same for elements of the previous subset. MFH 17:42, 5 Apr 2005 (UTC)

In other words, one first defines the multiplication of integers, then the multiplication of rationals in the usual way, by multiplying the numerators in denominators. After that, if one wants to multiply to real numbers, one approximates them by rationals and multiplies the rationals instead. Of course, to make this rigurous, you create two sequences of rational numbers converging to the two real numbers, then the product of that pair of rationals converges to the product of the pair of reals.

I could go in more detail if necessary. Oleg Alexandrov 17:58, 5 Apr 2005 (UTC)

There is another way to address multiplication for the rationals which doesn't require localization. Define non-zero rationals to be pairs of non-zero integers and with the multiplication induced from Z(+)Z quotient the set by the equivalence relation (a,b)~(c,d) iff ad = bc. If you want a more algebraic sort of quotient, you can describe the multiplicative structure of Z\{0}(+)Z\{0} as a semigroup.

As for the reals in my opinion the best way to characterize them is as the fraction field of power series in one variable with coefficients in Z/pZ quotiented by (x-1/p). As we exploit the fraction field construction, we might as well use localization in defining the rationals. TJSwaine 12:10 3/24/2006

Perhaps there should be another class of numbers called "unrepresentable numbers," which are real (or complex!) numbers that are defined that there is no way, in a finite amount of space, for them to be represented. All representable numbers can be multiplied together. Indeed123 22:12, 8 May 2007 (UTC)[reply]

There is a notion of representable numbers -- see Computable numbers, although this may not be what you are thinking about. Unfortunately with this representation you can only decide when two numbers (as represented by programs) are different and may not be able to positively determine when they are the same. They are, however, closed under multiplication. TooMuchMath 22:49, 9 May 2007 (UTC)[reply]

• It seems a mistake, to me, to flatly define multiplication with multiplication of whole numbers, since very early in grade school that will break on fractions. As this seems to have been mentioned, but not addressed, I'll take a shot at it. Pete St.John (talk) 23:27, 28 November 2007 (UTC)[reply]

I added some words to the intro; and a new section for multiplication of different kinds of numbers. I don't think this wins the $100 prize :-) but I hope it helps a bit. Pete St.John (talk) 00:38, 29 November 2007 (UTC)[reply]

I think in the Discussion section "Add m to itself n times" should actually read "Add m to itself n-1 times". Adding m to itself once is m+m. --Heycam 08:45, 9 Apr 2005 (UTC)

In my opinion n x m, which reads as n times m, should be interpreted as m+m+...+m, rather than n+n+...+n. I can imagine laying 5 times a pound on a table, but hardly laying a pound times 5. 130.89.220.52 21:19, 16 Apr 2005 (UTC)

• We say "add m to itself n times" because there are n summands, not n pairwise additions. "Add up n instances of m" might be better (as defining an algorithm) but is less natural language. Pete St.John (talk) 17:22, 29 November 2007 (UTC)[reply]

There are many tricks out there, we've all seen those mofo's who use their fingers to multiply things like primative abacii and envied them terribly. I recently learnt the trick to the nine times table one, there are many more; further there are many simplistic methods of dealing with even large number multiplication. If I were to do a summary of the nine times in a seperate article and link it through to here, would anyone else be willing to take on the challenge of equalising for all the geeky kids out there who went through the same mathmatically repressed childhoods that us non-finger-jedi-masters went through with me? :P Jachin 12:23, 11 May 2006 (UTC)[reply]

Does anyone know if this has a specific name, or even a wiki page? 64.179.161.110 22:57, 24 November 2006 (UTC)[reply]

• What you're suggesting is a good way to introduce someone to the concept of multiplication, but multiplication is about scaling a number, not just repeated addition. While multiplication may have originated as a method of "speed counting", as you say, that definition does not capture the full meaning of multiplication.

As a simple example, you may be able to represent 2x3 as 2+2+2, but how would you represent 2x3.5? 2+2+2+(2x0.5)? What does it mean to count something 0.5 times? Without the concept of scale behind multiplication, we would be unable to proceed.

Moreover, the convention of saying "four times three" to represent 4x3 reflects the concept of scale: in the same way that one might say "twice three" as verbal shorthand for "twice as much as three" to denote the scaling of 3 by a factor of 2, "four times three" or "four times as much as three" denotes the scaling of 3 by a factor of 4. So in fact, saying "four three times" has just the opposite meaning of scaling 4 by a factor of 3. A subtle, but important distinction.

That said, I agree with you that the memorization of "times tables" and "multiplication tricks" is somewhat regressive. In fact, the only place they are found in common application, outside of education (unfortunately), is in computer science and in nonstandard algebraic systems. That's why they are not and should not be in this article. 74.178.45.239 (talk) 18:06, 3 August 2009 (UTC)[reply]

Not sure where to add this but shouldn't we have a link or say something about how computers mulitply? There are generally two ways of doing it. The simpler processors only add and let multiplication be done in macro instructions, the other - and today more common way - is to implement a multiplication unit. The algorithm such a unit might use may vary but often go along the lines of multiplying an n bit number A by another n bit number B to produce a 2n bit result. This is in the simplest form done by having a counter and repeat the same process n times. For example by having a 2n bit register with the high n bits called H and the low n bits called L and the least significant bit of L called L0 (a single bit) you start by setting L equal to A, H equal to 0 and then add B to H if L0 is 1 and do not add if L0 is 0. Then shift all bits of H:L down by one so old L0 is dropped and the bit that used to be bit 1 becomes L0 and the carry if any from adding B to H is made most significant bit of H. Then the process is repeated until you have done it n times and all bits of old A is gone and H:L contains the result of the multiplication. Schemes to make this go faster should probably also be mentioned.

The algorithms as described above can thus be written like this:

[Initialize] H := 0 L := A Carry (a single bit storage) := 0

repeat n times.

  if L0 then H := H + B, Carry := Carry from H + B operation.
  [shift down H:L]
  Shift down L, set Lmax (most significant bit) equal to H0.
  Shift down H, set Hmax equal to Carry.
  Carry := 0

end repeat -- H:L contains result of A * B with H holding the n most significant bits and L holding the n least significant bits.

This algorithm works for unsigned multiplication. For signed multiplication you add an initialization step prior to this multplication and also a step after it.

First, check signs of A and B and compute sign of result based on it. If A and B have the same sign, the result is positive while if A and B have different sign the result should be negative. Thus, a sign_result := sign(A) == sign(B).

Compute absolute values of A and B (this is done in parallell and can also be done in parallell with the sign computation above. I.e. if either A or B are negative, negate them.

Then perform the unsigned multiplication described above. If sign_result is negative you then negate the result of the multiplcation.

For sign magnitude reperesenation a negation above can be as simple as ignoring the sign bit but for 2 complement representation you essentially invert and add 1.

Multiplication of floating point numbers are also based on integer multiplication. You extract the exponent part and the mantissa separately and then you add the exponents and multiply the mantissa. In this case you typically preserve the more significant bits of the result and round or truncate the less significant bits. I.e. you only keep H and forget about L except perhaps for the most significant bit of L which is used in rounding.

As I said, these are the basic algorithms - various ways to optimize them exist so that a computer can multiply faster. One such optimization to bear in mind is a "tree structure" way of doing it. Multiply the n bits by n bits to produce a 2n bit result can be done by considering half of n (say n = 2m) and each value A and B can then be considered to be split in two - a high part and a low part.

A = AH * F + AL, B = BH * F + BL

Here AH, AL, BH and BL are each m bits (half of n) and F = 2**m.

Multiplying A and B is then the same as:

(AH * F + AL)(BH * F + BL) = AH*BH * F*F + (AH*BL + AL*BH)*F + AL*BL

Thus a m by m multiplication producing a 2m = n result can be used to implement a n by n multiplication producing a 2n bits result. This can be repeated with half of m etc until you get down to single bits.

A single bit multiplier - i.e. 1 bit multiplied by 1 bit producing a 2 bit result is easy enough. The high bit is always 0 and the low bit is simply AND of the two input bits.

By combining this into a tree structure you can then perform a very fast multiplication. The problem is that you use an awful lot of transistors, so a middle way can be found where you find a lower n which is done by the algorithm described earlier and then you can use the latter method to build up 2n, 4n, 8n etc multiplication circuits by combining the two methods.

Does anyone know of additional procedures? Can anyone describe the methods used in typical modern day computers such as Intel series or others?

I added some links to the See also section to articles on computer multiplication. As a whole they are no t that great. More is neded. Also see category:Computer arithmetic--agr 12:27, 1 December 2006 (UTC)[reply]

The algorithm used in computers is the same as used by Ancient Egyptian multiplication and its explanation is much simpler. --82.141.61.150 12:16, 20 May 2007 (UTC)[reply]

It's essentially the same, but I think the Egyptians used big-endian multiplication and most hardware implementations are little-endian. However, it is certainly a point worth making. Silly rabbit 12:27, 20 May 2007 (UTC)[reply]

if 1 = O and 2 = OO as object counts how much is .05 = ? multiplication has several uses, not just as size values. auctioneers and banktellers when counting 20 dollar bills or 10's use multiplication speed counting still today. An auctioneer does so when he counts 5,10,15,20 do i have a 25. I believe a bjorked . (point) decimal system is a large culprit of your problem. cheers on figuring that out as i have yet to find a mathematician that can and can tell what various decimal values represent as physical objects. math was made for doing inventory and counting stuff.

The fact that on the front page of wiki has incorrectly states that 3 times 4 means 4+4+4=12 is a prime example. 3x4 means 3+3+3+3=4 the 4 is the number of times you add the 3. but oh well. here we go again lol. —Preceding unsigned comment added by 75.128.44.182 (talk) 09:06, 7 August 2009 (UTC)[reply]

In the section on properties, should it be mentioned that commutativity doesn't hold over ${\displaystyle \mathbb {H} }$?

I'm not sure either way. This article's scope isn't terribly clear to me. Noncommutativity is mentioned at Product (mathematics), at least. Melchoir 18:37, 29 June 2006 (UTC)[reply]

Never mind, I'm stupid; ArnoldReinhold already worked it in! Melchoir 18:41, 29 June 2006 (UTC)[reply]

I made a change to the example of multiplying fractions in the introduction. It used to say "a/b × c/d = ac/bd" and I changes it to "a/b × c/d = (ac)/(bd)" meshach

Definitely the parentheses in the denominator should be there to avoid what would be ambiguity at best. Michael Hardy 00:42, 3 July 2006 (UTC)[reply]

Re: http://en.wikipedia.org/wiki/Multiplication the following proof is shown:

(−1) × (−1)
= (−1) × (−1) + (−2) + 2
= (−1) × (−1) + (−1) × 2 + 2
= (−1) × (−1 + 2) + 2
= (−1) × 1 + 2
= (−1) + 2
= 1

Try as I might, I cannot follow the transition from
(-1) x (-1) + (-1) x 2 + 2 ------------ Line 1
to
(-1) x (-1 + 2) + 2 -------------------- Line 2

Comments, please. -Mark jager 23:16, 19 November 2006 (UTC)[reply]

Addition distributes over multiplication: a (b+c) = ab + ac. EdC 19:27, 20 November 2006 (UTC)[reply]

OK I have some knowledge retained from highschool maths some 38 years ago,
and my maths may be hazy but I 'read' Line 1 above as

((-1) x (-1)) + (( -1 ) x 2) + 2

and I can see no way to extract a common factor so that Line 1 may be expressed as Line 2.
I am now as much intrigued as to why I cannot understand the proof as I am by the fact that 'two negatives make a positive' when multiplied.

The common factor is (-1). Here's how it goes:

((-1) x (-1)) + (( -1 ) x 2) = (-1) x ((-1) + 2)

Another way to see this is to replace (-1) by "a" every place it occurs in line 1:

(-1) x (-1) + (-1) x 2 + 2 ------------ Line 1

Becomes:

a x a + a x 2 + 2

which is equal to

a x (a + 2) +2

--agr 04:44, 21 November 2006 (UTC)[reply]

Hello. Concerning the proofs for {1} -1*x = -x and {2} -1*-1 = 1:

• in {2} ... = (−1)·(−1) + (−2) + 2 = (−1)·(−1) + (−1)·2 + 2 = ...

Here we use {1} for (-2) -> (-1)·2, but {1} can be applied just at the beginning: -1*-1 = -(-1) = 1

• in {1} ... = 1·x - 2·x = -x

Can we do this directly? I think that this involves {1} itself, leading to incorrect proof. —Preceding unsigned comment added by 85.130.107.83 (talk) 20:17, 31 January 2008 (UTC)[reply]

Good catch. To fill the gaps in the proof of {1} would have taken several more lines; I have replaced it by a proof along different lines. I have also used the above as a much simpler proof of {2).  --Lambiam 07:35, 1 February 2008 (UTC)[reply]

The problem with the proof you gave is that it uses distributive law for subtraction which itself uses (-1)x = -x. 85.130.107.83 (talk) —Preceding comment was added at 15:07, 2 February 2008 (UTC)[reply]

Has anyone come across this phrase? In googling 'times less' there's actually a high incidence of it, but I think it's actually a horrible misconception. You can have 'times more', which is, multiplied by whatever number precedes it. Ala, 10 times more is whatever it's being compared to, multiplied by 10. How do you have times less though? What would 10 times less be?

In actuality, 'times more' is an incorrect usage of 'times'. As in, I've 10 times the amount of cookies. 'times more' is really unneeded, so by extension, 'times less' should be as well. What does it mean? It can't mean division, because we would simply use fractions, such as 'a tenth'. So I'm thinking anyone who uses it is simply not understanding mathematical language.

For example, I read this in a newsletter:

Females have about 10 times less anabolic hormones in their bloodstream than men do.

Now, would that mean they have 10%? Why don't people write that? I don't understand it. Tyciol 23:35, 2 January 2007 (UTC)[reply]

Well, evidently it does; and it is established usage, so saying it's "incorrect" is a stretch. I guess people prefer the "n times less" usage because it's concise, if not very clear. –EdC 04:53, 3 January 2007 (UTC)[reply]

Yes, I agree that it is improper usage of the term "times" but it is standard usage. However, I have wondered that same question before as well. In my experiences with the phrase, Females have about 10 times less anabolic hormones in their bloodstream than men do. means, Males have about 10 times the anabolic hormones in their bloodstream as females do. "Times less" appears to be just a more convenient albeit incorrect way of expressing the inverse. Zrs 12 (talk) 00:57, 31 January 2008 (UTC)[reply]

It may be beyond the normal means of this encyclopedia, but it seems odd to me that this article only discusses multiplication in the sense of R x R -> R. Given that there are other multiplications that exist for other rings and what not, shouldn't this article discuss a multiplication in general? Or perhaps I'm just strange :) 67.142.130.18 04:11, 23 February 2007 (UTC)JSto[reply]

See Product (mathematics). Perhaps the links to that article could be improved. –EdC 04:31, 23 February 2007 (UTC)[reply]

The section on infinite products calls the things being multiplied "terms". This is correct for summation, but surely it is wrong when we are multiplying? What do we call these components? Factors or multiplicands, maybe? nadav 08:30, 12 April 2007 (UTC)[reply]

Which symbol is more common - the cross or the dot? I was hoping that'd be answered in the article. As it is now I can only assume the list of common notations is ordered by which is most common, in which case the answer would be the cross. Since I received a telling off from one of my German maths teachers once for using the cross and defending it as the more-in-use symbol, being told that's bogus, I'm a bit curious what, exactly, is true. Trivial question, I realise, and it's not life-altering, but I do wonder who was 'right'. :) Maybe that could be added to the article. -pinkgothic 14:34, 9 July 2007 (UTC)[reply]

I dunno... I haven't used the cross since the beginning of 7th or 8th grade. Looks kinda stupid, besides being easily confused with "x" as a variable-- of course, the dot looks like a period just as badly. When you think about it, it's a matter of personal preference, but it seems to say "I go through so many multiplications that I have to use a shortcut-- therefore I'm smarter and am a mathematician!!!" when you use the dot.75.36.45.94 04:18, 10 August 2007 (UTC)[reply]

In my experience (with numerals) the dot is used more commonly. (It seems to me that the cross is used more commonly in elementary school when variables are out of the question). However, as the article states, when multiplying variables, they should be juxtaposed (i.e. xy).

Zrs 12 (talk) 00:53, 31 January 2008 (UTC)[reply]

I'm facing the same problem: decide what to use in a global company. My idea is to follow ISO. They use the cross, but only if a sign is really necessary. —Preceding unsigned comment added by 163.157.254.25 (talk) 14:00, 14 March 2008 (UTC)[reply]

Mathematicians normally don't use either of these symbols in their mathematical writings, but instead juxtaposition; they write 2xe^−2x, and not 2×x×e^−2×x or 2·x·e^−2·x. The exception is when this results in something that is unclear or ambiguous, as for products of numerals, in which case the more common solution is to use centred dots, as in 5! = 1·2·3·4·5 (and not 5! = 12345). When the subject matter is not itself a mathematical topic, and the target audience is not assumed to be familiar with the conventions of mathematicians, the cross symbol may be used in formulas. See for examples Percentage.  --Lambiam 21:43, 15 March 2008 (UTC)[reply]

I live in Sweden and I've been taught to only use the centre dot ·. (The only exception seems to be when giving the dimensions of a rectangle, e.g. a screen resolution of 640 x 480, where an x is used rather than a cross.) This is what the swedish article has to say on the topic:

The multiplication sign is a dot · placed at the same height as the plus + and - minus sign, alternatively a cross ×. However the cross should be avoided since it also has a different meaning, namely the cross product.

Looking through some of the american math and computer science literature on my shelf I notice they tend to use the dot as well. My impression though is that neither sign is automatically wrong, they're simply used in different contexts. The cross is used in simple everyday maths and the dot is used in more "scientific" context. Some countries (like Sweden) have decided to promote only one standard. (The same goes for division by the way. When I was in elementary school the horizontal bar was the only way to write a division/fraction. Neither / nor ÷ were accepted.)

I believe both symbols should be give the same status in the article, at least until the topic has been discussed further. In the Wikipedia Manual of Style both seem to be given the same status (with the consequence that both might be mixed within an article - just take a look at this one).

Another thing that might be worth mentioning in the article is the relation to the signs used for the dot and cross products. As long as ordinary multiplication is represented by a dot "v · w" there's no problem telling it from the cross product "v × w". However, sometimes an author might find the need for a separate sign for the dot product. Here I've seen two solutions, one is the dingbat "v•w", the other is the more lengthy "(v|w)". Tasnu Arakun (talk) 02:53, 12 April 2008 (UTC)[reply]

Use of the × symbol is common in primary school, when the students are learning how to multiply numbers, like in the tables of multiplication, something they have to unlearn if they continue to study mathematics. I don't know about elementary education in Sweden; perhaps they use centred dots from the start.  --Lambiam 17:06, 12 April 2008 (UTC)[reply]

Very nice discussion! I too have problems in reading some articles in English Wikipedia, because of that abundant use of "x" for mere multiplications. At first read, I thought what do Factorials have to do with vectors? ("5x4x3...") But then I got to know that they use the "x" for multiplication here. I don't like this at all; because this is the English Wikipedia and this language is mostly spoken in the U.S. and in the UK, so we should (!) stick to the dot. My 2c -andy 92.229.70.233 (talk) 01:40, 27 January 2009 (UTC)[reply]

the answer is the product —Preceding unsigned comment added by 72.211.208.77 (talk) 03:35, 9 October 2007 (UTC)[reply]

I'll leave a note on Silly Rabbit's talk page regarding his changes. Pete St.John (talk) 17:22, 23 January 2008 (UTC)[reply]

Why not discuss it here?  --Lambiam 23:06, 23 January 2008 (UTC)[reply]

because my remarks include the context of a previous edit revert (that had a happy ending). Of course if there are any content questions sure, they belong here. But I think it's a hastiness question. And surely, surely, I'm hasty sometimes myself. Pete St.John (talk) 23:11, 23 January 2008 (UTC)[reply]

As the original author of the passage in question, I am in a rather unique position to comment on the author's intent. To compute 13 × 21, you must double 21 three times. So, starting with 21 × 1 = 21 (which doesn't count as a "doubling"). Then, doubling again, 21 × 2 = 42. Then, again, 21 × 4 = 42 × 2 &= 84. Finally, 21 × 8 = 84 × 2 = 168. In other words, you need to go through all these steps. You are invited, of course, to write out an algorithm, and see that in fact all the intermediate stages are necessary to give a complete illustration of the technique. Silly rabbit (talk) 01:54, 24 January 2008 (UTC)[reply]

Please see also the article Ancient Egyptian multiplication, of which this section is only a summary. Note that all intermediate powers of two are included in the calculation, even if they have no direct relevance to the final result. Silly rabbit (talk) 02:04, 24 January 2008 (UTC)[reply]

In performing the algorithm (the ancient way), yes, the doubling that makes 42 would have been performed, in order to get the next doubling, 84. Then some of those intermediate calculations (the ones corresponding to 1, 4, and 8, which add up to 13) would be summands for the final calculation. It's fine to express it that way if it's clear. So for example:

1. Decompose 13 into a sum of powers of two, getting 1, 4, and 8.

2. Double 21 repeatedly, getting 21 (the vacous doubling, corresponding to the summand 1 which is 2 to the power zero), 42, 84, 168;

3. Choose the intermediate results that correspond to the addends 1,4, and 8; that would be 21, 84, and 168;

4. Add up those results, getting 273, which indeed is 13 times 21.

My objection to the latest wording is purely pedagogical. If we just distinguish the "doublings" which are intermediate results (2*21=42, computed so as to get 4*21 by doubling 42) from those which produce summands used in the final reckoning, then it's fine. Also, it's confusing to say "...doesn't count as doubling" but then "...doubling again..."; think how that reads to a schoolchild. You can't double "again" if you hadn't doubled previously. So I would call 1*21 the "null" doubling; it's indeed a doubling, it's just a doubling zero times (that is, 2^0). Pete St.John (talk) 17:51, 24 January 2008 (UTC)[reply]

OK, let me back off on this. The wording as it existed (and now, modulo the character chosen for multiplication) is quite good, and I shouldn't have reverted it. Pete St.John (talk) 22:23, 24 January 2008 (UTC)[reply]

I'm not sure that I understand the operators ${\displaystyle \sum }$ and ${\displaystyle \prod }$ so could someone tell me if these equations are correct?

${\displaystyle \sum _{n=m}^{k}{\frac {n}{2}}={\frac {m}{2}}+{\frac {m+1}{2}}+{\frac {m+2}{2}}+...{\frac {k-1}{2}}+{\frac {k}{2}}}$

${\displaystyle \prod _{n=m}^{k}{\frac {n}{2}}={\frac {m}{2}}\cdot {\frac {m+1}{2}}\cdot {\frac {m+2}{2}}\cdot ...{\frac {k-1}{2}}\cdot {\frac {k}{2}}}$

(sorry- signed Zrs 12 (talk) 00:42, 31 January 2008 (UTC))[reply]

I'm not sure this is the best place to ask, but yes, sure, you have it right. The Sigma is short for "summa" (summation, addition) and the Pi is short for Product (multiplication). Incidentally, the funky streched out S used for integration is a germanic 'S' and also comes from "summation" (a definite integral is the limit of a sequence of sums). (Incidentally, if that had been signed, I would have moved it to the talk page...) Pete St.John (talk) 20:21, 30 January 2008 (UTC)[reply]

Your understanding of the math is right, but for typesetting style I'd rather see

${\displaystyle \sum _{n=m}^{k}{\frac {n}{2}}={\frac {m}{2}}+{\frac {m+1}{2}}+{\frac {m+2}{2}}+\cdots +{\frac {k-1}{2}}+{\frac {k}{2}}}$

instead of

${\displaystyle \sum _{n=m}^{k}{\frac {n}{2}}={\frac {m}{2}}+{\frac {m+1}{2}}+{\frac {m+2}{2}}+...{\frac {k-1}{2}}+{\frac {k}{2}}}$

(with \cdots instead of "...", and a "+" BOTH before and after the \cdots). Michael Hardy (talk) 20:33, 24 February 2008 (UTC)[reply]

If you're going to contribute to this article with mathematics, please take the time to mark it up correctly in the TeX language (this link may help). Otherwise it not only looks poor but can be difficult to read. Thank you! Stephen Shaw (talk) 19:21, 8 May 2008 (UTC)[reply]

For inline expressions, HTML markup is often preferable to LaTeX-style markup. See also Wikipedia:Manual of style (mathematics). Neither looks good on all platforms, unfortunately.  --Lambiam 10:30, 9 May 2008 (UTC)[reply]

I was mainly referring to things like writing a2 instead of ${\displaystyle a_{2}}$, which are two entirely different things when written on paper. Since the manual of style states that mark-up changes of simple expressions from HTML to LaTeX or vice-versa are condoned provided the entire article is consistent, it is therefore necessary that any further mathematics posted to this article be marked-up in LaTeX, to keep everything consistent. Stephen Shaw (talk) 20:18, 9 May 2008 (UTC)[reply]

I am looking for the terminology. Is it "factor times factor equals product" or is it called "multiplier," or "coefficient" etc. --Tattoe (talk) 09:45, 23 June 2008 (UTC)[reply]

"Coefficient" has a different implication (i.e., that a variable, not a constant, is being multiplied by it. "Factor" or (inelegantly but accurately) "multiplicand" is fine. Bongomatic (talk) 09:52, 23 June 2008 (UTC)[reply]

So it's "presumptuous" to say that most people learn multiplication in elementary school (see the recent edit history)?? Will we next read that it's "presumptuous" to say that most elementary schools teach reading and the most people learn to tie their shoes before they're 40 years old? Michael Hardy (talk) 03:36, 18 November 2008 (UTC)[reply]

I removed it because I thought it might seem offputting or patronising to people coming here who have not had any sort of mathematical education. Kind of like: "Most people already know all this, so if you're looking here then you must be pretty dumb". And for those who have learned it at elementary school, well, they don't need to be told. But if you feel especially strongly that it should be included then please put it back in. Matt 21:15, 18 November 2008 (UTC). —Preceding unsigned comment added by 86.133.55.73 (talk)

I don't want to be offputting or patronising, but unless things have changed FAR more than I thought since I was in school, it would be extraordinarily unusual not to learn what multiplication is in elementary school. It's much simpler than learning to read and write, and if someone's looking at this page, they must have done that. Michael Hardy (talk) 18:25, 29 November 2008 (UTC)[reply]

This is beyond my level of expertise, but on the face of it, it does not appear to make much sense:

Multiplication for some types of "numbers" may have corresponding division, without inverses; in an integral domain ${\displaystyle x}$ may have no inverse "${\displaystyle {\frac {1}{x}}}$" but ${\displaystyle {\frac {x}{y}}}$ may be defined. In a division ring there are inverses but they are not commutative (since ${\displaystyle \left({\frac {1}{x}}\right)\left({\frac {1}{y}}\right)}$ is not the same as ${\displaystyle \left({\frac {1}{y}}\right)\left({\frac {1}{x}}\right)}$, ${\displaystyle {\frac {x}{y}}}$ may be ambiguous).

Why would 1/x not existing be expected to cause problems for x/y? For example, 1/0 is undefined yet 0/1 is fine.

Why would (1/x)(1/y) not being commutative be a problem for x/y? If the non-commutative example was x(1/y) then it would seem to make more sense.

Matt 21:22, 18 November 2008 (UTC). —Preceding unsigned comment added by 86.133.55.73 (talk)

Hi. What means :

${\displaystyle {\frac {F_{n}(x,c)}{\prod _{m}G_{m}(x,c)}},n:m}$

I do not know how to read Pi notation here.

it is from paper Andrey Morozov : Universal Mandelbrot Set as a Model of Phase Transition Theory

--Adam majewski (talk) 15:44, 21 November 2008 (UTC)[reply]

• You might get a quicker reply to this at Wikipedia:Reference_desk/Mathematics. —Preceding unsigned comment added by 86.134.53.253 (talk) 12:43, 27 November 2008 (UTC)[reply]
• You are right. thx.

[question and answers]. Maybe it can be included in article ? --Adam majewski (talk) 08:13, 29 November 2008 (UTC)[reply]

I've always thought 3 times 4 meant 4+4+4 as in "three times a lady". For the other meaning as in 3+3+3+3 I'd have to put in a pause after the 3 as in 3, times 4. This is in line with "Think of a number, multiply it by 4". Is there anything on the relative frequency of the two?, perhaps some elementary teaching gives a guide somewhere. Dmcq (talk) 09:06, 23 October 2009 (UTC)[reply]

I just did a quickl survey looking at the first few entries that came up with a google of "multiplication by repeated addition" and I got:

3*4 = 4+4+4 (Wikipedia) aaamath hartfordprimaryschool youtube

3*4 = 3+3+3+3 multiplication homeschoolmath grahamwroe

both teachingideas lessonplanspage cumbriagridforlearning

Neither: It Ain't No Repeated Addition maa devlin numberwarrior

Both or neither globaledresources

So there ain't no consensus that I can see. I'm rather surprised there's such an even match of divergence of opinion. Perhaps even there's some room in the article about it isn't repeated addition. Dmcq ([[Us

|}

Interesting. Wrong, but interesting. Your "reading" of multiplication as a "shorthand for repeated addition" is just wrong, although it might be an alternate definition in some contexts. Perhaps if you could supply a source that 3 × 4 is 3 + 3 + 3 + 3, rather than 4 + 4 + 4, it might justify some of your comments. As it stands, it shouldn't be in the article. — Arthur Rubin (talk) 05:57, 12 November 2009 (UTC)[reply]

Well certainly even saying that 3 miles × 4 miles is repeated addition adding up to 12 square miles is rather a long stretch. I'll have a look at The development of the concept of multiplication as it's something that can be cited. Dmcq (talk) 12:03, 12 November 2009 (UTC)[reply]

It is not wrong and real mathematicians do math, they dont argue with people for the sake of arguing. Quoting youtube sources needs to be invalidated as does quoting from blogs, etc. on wiki simply because you can write your own articles on youtube/blog and submit them as a wiki source. scaling numbers isnt what multiplication was used for, it was originaly done for counting objects quickly on an abbacus. objects are counted quicker in groups than one at a time.I dont need a source for verifying basic math, simple logic should suffice. If you would have asked me to quote a source for 1+1=2 you would be laughed at by the math commuinity. Also the flat earth society I can find quotable online sources from to "back that up" even though thats blatantly rediculous. This is how its taught by american math teachers and go visit a school in japan or the east where traditional counting is still done. Im done with wiki. Real scientists and mathematicians dont quote sources ^_^. Einstien, Newton were all originals not quoting various web publications/education sources. Real math and science uses logic and math proofs, not quotable sources. Invaladate that with whole numbers. simply take the first amount and repeatedly add the 2nd after the times sign proves the statement correct,using whole numbers. decimals are not standardized still esp when signifying object counts and Im not even going into it as Im not here to argue. example 1=O and 2=OO and 3=OOO and 4=OOO what does 4.23567 equal as an object count. good luck finding a source on that. Thats the proof and Im not going to embarrase you by giving 230,000 whole number examples. Im out and done with wiki and people arguing for the sake of arguing that cant think for themselves without quoting from books.a wise man knows nothing. Real mathematicians give math proofs, they dont quote sources for the sake of arguing. Multiplication was not developed originaly for scaling numbers it is used as a means of speed counting for counting beads quickly on an abbacus. Its faster to count beads in rows/groups than one bead at a time,speeding basic addition/counting. The code was designed to act as a memory aid and speed writing time and reduce calculation time for doing old fashioned counting, something that is missing in western school, having thrown out applied mathematics and sadly adopted pure mathematics, wich I call doing math for the sake of doing math. Mutliplication was originaly a form of speed inventory.I invite you to call various japanese and eastern school systems who still use abbacus's and traditional counting for verification rather than quote sources.

 —Preceding unsigned comment added by Nightgamer360 (talk • contribs) 19:58, 12 November 2009 (UTC)[reply]