January 19

This is probably a stupid question but never the less. The definition of multiply is to make many or manifold; increase the number, quantity, etc., of. So when you, for example, multiply 1 car by no cars how do you end up with less then what you started if it's supposed to increase in number? So to me 1x0 doesn't make complete sense, yet 0x1 does. Same goes with 1x1. Once again, I apologize if this is a supid question. —The preceding unsigned comment was added by 203.40.139.85 (talk) 01:38, 19 January 2007 (UTC).[reply]

Think of it like this: you can think of 2x3 as (to use your example) taking 2 cars three times, which is 6. As you said, 0x1 is taking no car, one time, which would be 0 cars. 1x0 would then be taking 1 car no times. Get it? So you would have zero groups of one car each. That would be zero cars, right? Oskar 02:08, 19 January 2007 (UTC)[reply]

Sorry but I'm finding it hard to follow you. Maths isn't a strong point for me. I don't get it when you say 2 cars three times or zero groups of one car. Where do groups come into it? Isn't it as simple as having 1 car then multiplying by 0 cars to still have 1 car because there's nothing to multiply it by?

You are confusing yourself because it doesn't make sense to multiply cars by cars. Even saying “one car multiplied by zero” doesn't sound right to my ear. Try saying it “one car zero times” as in “I wanted to test drive my friend's Porsche but I never got around to it – I wound up driving that one car zero times (i.e., never).” If you are talking about cars, you probably just want to be mulitplying by integers. (To be pedantic/flippant, 1 car muliplied by 0 cars would be “zero cars²” but that probably isn't what you want to know.) —Ben FrantzDale 02:45, 19 January 2007 (UTC)[reply]

In a room are 7 people. Each has 10 fingers. How many fingers are there all together in the room? 7 × 10 = 70 fingers.

Now what if there are no people in the room? In other words, if you count how many people are in the room, the answer is zilch, nada, 0. OK, now how many fingers are there in the room? 0 × 10 = 0 fingers: there are no fingers in the room. Unless someone from the bunch of 7 was careless with a knife.

For the rest, the mathematical meaning of "multiplying", which is very precise, is related to, but not the same as, the imprecise informal notion of multiplying. The reason why 0 × 10 = 0 is that mathematicians have chosen to define it so. Why have they chosen the definition that way? Because it is the most useful definition, which gives you nice properties like (x × y) × z = x × (y × z) and (x + y) × z = (x × y) + (x × z).  --Lambiam^Talk 02:50, 19 January 2007 (UTC)[reply]

It's a semantic issue. In the general sense, "multiply" does indeed mean to make many. If I talk of my crops or my bank balance or my number of children multiplying, it's always an increase I'm talking about, never a decrease, and never even staying level. But in the mathematical sense, "multiply" refers to a defined action involving two numbers, either of which can be positive, or negative, or zero. Depending on the numbers involved, the original quantity can increase, stay the same, or decrease. If your starting number is 2, some of the possibilities are: 2 x 4 = 8 (increase); 2 x 1 = 2 (no change); 2 x -1 = -2 (decrease); 2 x 0 = 0 (decrease). JackofOz 03:14, 19 January 2007 (UTC)[reply]

So it all comes down to the mathematical definition of multiply? Makes sense. But I can't seem to find the definition you mentioned anywhere.

See dictionary.com: “Arithmetic. to find the product of by multiplication.” and “Mathematics To perform multiplication on.” —Ben FrantzDale 04:20, 19 January 2007 (UTC)[reply]

If someone examines mathematics rather than take it for granted, we see hope for a promising new mathematician. :-)

Here when you say "the definition of multiply is …", you use an everyday or dictionary definition. Mathematics has a habit of taking familiar words and using them in precisely defined ways that may conflict with the familiar meaning. In the case of multiplication, we have an example even more extreme than 1×0 or 0×1; we multiply (−2)×(−3) and get +6.

One formal definition of multiplication for "counting numbers" (the numbers 0, 1, 2, …) is as repeated addition. Addition itself is defined as repeatedly "increasing by one". For formal, technical work, this is what we use. However, it may be helpful to have a simple mental model.

I think of forming rectangular patterns of dots. The product 2×3 would be two rows, each three dots long:

${\displaystyle {\begin{matrix}\bullet &\bullet &\bullet \\\bullet &\bullet &\bullet \end{matrix))\qquad 2\times 3=6}$

We can turn this on its side to see that we get the same result as 3×2.

${\displaystyle {\begin{matrix}\bullet &\bullet \\\bullet &\bullet \\\bullet &\bullet \end{matrix))\qquad 3\times 2=6}$

In this model 1×0 would be one row with zero columns, so no dots; while 0×1 would be zero rows with one column, and still no dots. The appropriate everyday term for this concept might be "replicate", not "increase.

When multiplication is extended to fractions, such as ²⁄₃×¹⁄₂, or to negative numbers, such an elementary model is awkward (though not impossible).

${\displaystyle {\begin{matrix}\bullet &\circ \\\bullet &\circ \\\circ &\circ \end{matrix))\qquad {\tfrac {2}{3))\times {\tfrac {1}{2))={\tfrac {1}{3))}$

We find ourselves extending the primitive definition and arguing from consistency.

distributive law

${\displaystyle {\begin{matrix}\bullet &\bullet &\bullet &{\color {Blue}\bullet }\\\bullet &\bullet &\bullet &{\color {Blue}\bullet }\\{\color {Red}\bullet }&{\color {Red}\bullet }&{\color {Red}\bullet }&{\color {Purple}\bullet }\\\end{matrix))\qquad (2+{\color {Red}1})\times (3+{\color {Blue}1})=12}$

implies (−1)×(−1) = +1

${\displaystyle {\begin{matrix}\bullet &\bullet &\bullet &\diamond \\\bullet &\bullet &\bullet &\diamond \\\diamond &\diamond &\diamond &\ast \\\end{matrix))\qquad (3-1)\times (4-1)=6}$

Still, this "geometric" model is powerful. When applied to calculating areas (counting squares instead of dots), it can be made to work even for products like √2×√3 or 2×π. --KSmrq^T 10:15, 19 January 2007 (UTC)[reply]

${\displaystyle \lim _{x\to 1}e^{\frac {1}{x^{2}-1))}$

• Don't give me the answer, I just need to know the first step or so, because this is totally baffling. My first thought was to split up the exponent, but I couldn't figure out a non-redundant way to do so. Help is appreciated. Thanks. --Ķĩřβȳ♥ŤįɱéØ 05:13, 19 January 2007 (UTC)[reply]

Ok, here's what I have now, I think this is the next step:

${\displaystyle e^{\lim _{x\to 1}{\frac {1}{x^{2}-1))))$

So I just have to evaluate ${\displaystyle \lim _{x\to 1}{\frac {1}{x^{2}-1))}$, which seems to not exist since the left-hand and right-hand limits are not the same. Am I doing this right? Thanks. --Ķĩřβȳ♥ŤįɱéØ 05:22, 19 January 2007 (UTC)[reply]

The limit is undefined. I don't think the steps are valid though. You took the limit inside the function ${\displaystyle e^{...))$; but I think you can only do that if ${\displaystyle e^{x))$ were continuous at infinity (the limit of the inside) on the projective real line, which it is not. --Spoon! 06:03, 19 January 2007 (UTC)[reply]

There are two different limits, both well defined. Begin with the denominator of the exponent, x²−1. When x is, say, ¹⁄₂, we get a negative value; this is true for all x with magnitude less than 1. So approaching 1 from below, the denominator approaches zero but is always negative. When x is, say, ³⁄₂, we get a positive value; this is true for all x with magnitude greater than 1. So approaching 1 from above, the denominator approaches zero but is always positive. Now look at the fraction in both cases, then at the exponential. It is really quite common for limits to depend on the direction of approach. In the complex plane, we can approach from many directions, not just two! So this problem teaches an important lesson. --KSmrq^T 10:38, 19 January 2007 (UTC)[reply]

We're trying to display the first number ${\displaystyle g_{1))$ in the recursive sequence by which Graham's number is defined, but I couldn't get it to display right. I doctored the output from my experiments and produced this, but if possible I'd prefer to get it rendered properly so we can put it in the article. Does anybody know how to get it to display how the graphic shows? Any alternative ways of displaying ${\displaystyle g_{1))$ neatly are also welcome. Maelin (Talk | Contribs) 06:51, 19 January 2007 (UTC)[reply]

Wrap it around a matrix before using \right }, alike so:

${\displaystyle g_{1}=3\uparrow \uparrow \uparrow \uparrow 3=\left.{\begin{matrix}\underbrace {3^{3^{\cdot ^{\cdot ^{\cdot ^{3)))))) &\ \\\underbrace {3^{3^{\cdot ^{\cdot ^{\cdot ^{3)))))) &{\text{threes))\\\underbrace {\vdots } &\vdots \\\underbrace {3^{3^{\cdot ^{\cdot ^{\cdot ^{3)))))) &{\text{threes))\\3^{27}&{\text{threes))\end{matrix))\right\}3^{27}{\mbox{ towers))}$

And there you go! — Kieff 07:07, 19 January 2007 (UTC)[reply]

You have 24 wheels for a three-wheeled model car. Some of the wheels spin more easily than others. You can use the car to run time trials using any three wheels you want for each trial. What is the minimum number of trials you must run in order to find the three wheels that will make the car run the fastest? dryguy 17:33, 19 January 2007 (UTC)[reply]

I can find the fastest wheel in 25 trials. Pick two wheels, then try each remaining wheel with this pair. Set aside the fastest of the 22 and pick 2 of the remaining. Try this pair with the current fastest and with each of the original pair used for testing. The fastest of these 3 trials is the fastest wheel. You could repeat the process with the 23 remaining, and again with the 22 remaining for a toal of 25+24+23=72 trials. Is there a way to do this in fewer trials? dryguy 18:11, 19 January 2007 (UTC)[reply]

On further thought, I can get it down to 31 trials. The first 25 for the first wheel as above. Three more to test the 2nd fastest wheel of the original 22 + three more for the 3rd fastest. Anyone have a way to do it fewer in trials? dryguy 19:37, 19 January 2007 (UTC)[reply]

After even further thought, I can get it down to 27 trials. 22 initial trials as above, followed by testing the initial two fixed wheels and the three fastest so far with two of the remaining wheels. dryguy 22:17, 19 January 2007 (UTC)[reply]

One thing I'm unclear on here is if the wheels are actually independent in respect to their effect on the speed of the car. It may be that the wheels act in combination with each other in a dependent relationship, meaning that for example wheel A might work well with wheel B but not as well with wheel C. There is also the possibility that different wheels work different in different wheel slots. In a three wheel car, for example, it might be that some wheels work well in the front wheel slot, but not as well in the back-right or back-left slot, or vice versa.

So if the slot of the wheel makes a difference, and the wheels act slightly differently in different combinations, then you'd theoretically have to check all possible ordered 3-tuplets of wheels to find the optimal wheel set (24x23x22 orderings). Dugwiki 20:40, 19 January 2007 (UTC)[reply]

Assume the slot position has no effect and that the wheels act independently. Also assume that the speed of the car is only limited by the worst of the three wheels. Assume that the effect of each wheel is additive. In other words, the "drag" from each wheel is summed to get the total drag on the car. dryguy 22:14, 19 January 2007 (UTC)[reply]

I would start out by trying 8 sets of 3 different wheels each and timing each run. (8 tests) The three fastest wheels will each be in one of the three fastest runs, so you've now reduced 24 wheels to 9. Now that you have 9 wheels, repeat the process, 3 timed tests (up to 11 now), which will tell you which of the three sets contains the -fastest- wheel. 3 more tests will now find the fastest wheel (up to 14) by substituting a slow wheel from one of your earlier runs for each in turn. At this point it gets trickier to be efficient... you've got 8 wheels, and you can certainly find the fastest two with 8 more timed runs on the remaining 8 wheels (up to 22 now), but can it be done better than that? - Rainwarrior 04:27, 20 January 2007 (UTC)[reply]

The three fastest wheels will each be in one of the three fastest runs... bad logic there. :) Jeff Carr 04:36, 20 January 2007 (UTC)[reply]

Actually, never mind, I was thinking in terms of there being 3 good wheels and 21 bad ones, the bad ones being uniform and the good ones also being uniform. If there's a continuum of wheel-goodness then a good wheel could be accompanied by two very bad wheels and make the run not as good as maybe a run with three average wheels, etc... So... never mind what I just said. - Rainwarrior 04:35, 20 January 2007 (UTC)[reply]

Funny I added my above comment just as you were also commenting. Never happened to me before. Cool that it worked. (mediawiki is really smart I guess). Jeff Carr 04:38, 20 January 2007 (UTC)[reply]

Looking again, the 22 + 5 = 27 tests is pretty good, but by recording the times accurately I think you could do even better. If after 22 test you had two fixed wheels A and B and three "so far best" wheels C D and E, making the assumption that the run times are more or less related to the wheel's properties in an additive way (probably not physically correct, but might be good enough for your purposes):

A + B + C = x

A + B + D = y

A + B + E = z

How many more measurements do we need to be able to solve for all unknowns A, B, C, D, and E? I'm thinking two more (would give you 5 equations with five unknowns each), giving a total of 24 tests in all. - Rainwarrior 05:04, 20 January 2007 (UTC)[reply]

Nicely done. Thanks! dryguy 13:31, 20 January 2007 (UTC)[reply]

There are two different kinds of inverses for powers (single arrow): logarithms and roots. Are there corresponding inverses for Knuth's up-arrow notation? And if so, could they be used to get something like x^^(3/2)? --204.56.135.2 18:32, 19 January 2007 (UTC)[reply]

Looks like the inverses are called Down Arrow Notation. As to your second question, I do not know at this time. I'll think about it, and hopefully someone else can answer you. --Sopoforic 22:38, 19 January 2007 (UTC)[reply]