Wikipedia:Reference desk/Archives/Mathematics/2006 November 12

From Wikipedia, the free encyclopedia

Mathematics desk
< November 11 << Oct | November | Dec >> November 13 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


Contents


[edit] November 12

[edit] How to use a lever to weigh a 1950 pound device with a 300 pound bathroom scale.

I would like to know the correct way to use a lever or series of levers,to weigh a 1950 pound device using a 300 pound maximum bathroom scale.

Thank you for the solution. Bill, not to swift at math.

Place one end of the lever on the ground, place the device so it's CG is 1/7th of the distance along the lever arm, and place the far end of the lever on the scale, so that the end of the lever is over the center of the scale (a small vertical extension will need to be attached to the lever arm, so the edge of the scale doesn't hit the lever). Note that this method would allow you to weigh objects up to 2100 pounds, but would be highly inaccurate, due to the weight of the lever arm and difficulties in getting the distances exactly right. StuRat 06:24, 12 November 2006 (UTC)
Here's a method with levers but next to no maths. Use a beam balance with equal length arms to balance the device against a big container of water with equal weight. Then bale the water out of the big container into smaller containers and weigh the water 250 pounds or so at a time. Add the weight of the water to the weight of the big container and you are done. However, you will have to bale around 900 litres of water, so this may take some time ! Gandalf61 12:02, 12 November 2006 (UTC)
If the container is graduated, you could find the mass of the water without weighing it by simple conversions (1 litre per kilogram, 1 cubic metre per metric ton etc.). I don't know the Imperial equivalents, but I'm guessing one fluid ounce = one ounce Laïka 09:55, 17 November 2006 (UTC)
If you know that it is 1950 lbs why are you weighing it?? ... Sorry ;) Vespine 04:49, 13 November 2006 (UTC)

[edit] Numbers in other bases

Two questions about dealing with numbers in other bases.

1. How do i multiply numbers in other bases? So for example - i have the number 95 (in base 10), which is 235 in base 6. Now, if i wanted to work out what 952 is in base 6, i would go 95*95=9025 (in base 10). I can then covert the base 10 number "9025" into the base 6 number "105441". However, the quicker way would be to directly square in base 6, so that is...go 235*235 in base 6.

I'd assume long multiplication in any base is the same as long multiplication in base 10. So to do 235*235 in base 6, i'd have to do 235*5 + 235*30 + 235*200, all in base 6.

And this is where i lose it. How the hell do i work out 235*5 in base 6? (note that "235" and "5" in this case are both base 6 numbers)? I start with 5*5 - but what does 5*5 in base 6 give? With something like binary, it's simple enough since all you have is 1s and 0s, and multiplicating 1s and 0s will never give you anything except more 1s and 0s.

But here, 5 by 5 is 25, which is 41 in base 6. But i'm sure that's not how it's supposed to work, since 5*5=25 only in base 10. So when my 5s are in base 6, what is 5*5 supposed to give?

2. How do i convert between bases? I can manage base 10 into any other base, and any other base into base 10. But i can't quite seem to get my head around going from one wierd base to another wierd base. Say if i wanted to convert 10110 (binary) to a base 4 number, what do i do? And what about something like base 3 to base 7? --`/aksha 04:06, 12 November 2006 (UTC)

Five times five is, in any base, twice ten plus five. Remember that the numbers are the underlying thing; they don't care what symbolic notation we write them in. The results of arithmetic operations are always the same. Then the base you're using determines how you write the result. --Trovatore 04:11, 12 November 2006 (UTC)
So you're saying 5*5 in base 6 is indeed 41? then if i was trying to evaluate 235*235 in base 6. I end up with 5235 (which is 235*5) + 31330 (which is from 235*3, shifted left by one decimal place) + 210200 (which is from 235*2, shifted left by two decimal places) = 251205. Which isn't the answer. So what am i doing wrong? --`/aksha 04:46, 12 November 2006 (UTC)
The problem is that you're multiplying by 10 in base 6 in order to get a shift left or right, which is wrong. Remember that 41 base 6 = 4 * 6 + 1, not 4 * 10 + 1.

Ahh, crap. I see what you mean now. Shifting a number to the left is in actual fact multiplying it by 10. Where as in fact, i need to 'shift' it by multiplying it by 6, and then 6^2, and so on.

What about the second problem, of converting between bases? Is there a methodology for say, converting "23412" (base 5) into a base 7 number? (without doing the convert-to-base-10-then-re-convert thing). --`/aksha 06:43, 12 November 2006 (UTC)

No, shifting the base-6 representation of a number one to the left multiplies it by six, not ten.
I'm afraid I can't follow your figures at all, though. Working in base 6, 235*2 should be 514. Where do you get 210200? --Trovatore 06:47, 12 November 2006 (UTC)
The 210200 came from the fact that 235*2 in base 10 gives you 470. And 470 written in base 6 is 2102. And then i shifted it left by 2. But that really doesn't work, since "235" was in base 6 in the first place. My problem is basically that i can't understand how you multiply in another base. Can you like, explain where you pulled the "514" from? Exactly how'd you manage to get that from 235*2. --`/aksha 07:07, 12 November 2006 (UTC)
Try not to convert between bases so much - go back to the basics of how the algorithm works in the first place. I hope this helps: Let's multiply 2356*2.
  • Start with the ones digit, 5 ones * 2 = ten, which breaks down into four ones and a six, so we write down "4" and carry the six.
  • 3 sixes * 2 = six sixes, plus the carried six, gives us seven sixes. That breaks down into one six and a carried thritysix.
  • 2 thirtysixes * 2 = 4 thritysixes, add the carried thirtysix to get 5 thirtysixes.
  • 5 thirtysixes + 1 six + 4 ones = 5146.
I was taught New Math-style (in the 1990s, when it wasn't new anymore), and it gave me a knack for this stuff. :) —AySz88\^-^ 07:21, 12 November 2006 (UTC)
If you understand C... base.c --wj32 talk | contribs 07:47, 12 November 2006 (UTC)
Otherwise: to convert a decimal number to another base:
Say the decimal number is 100 and I want to convert it into base 4.
  1. Divide the number by the base, and take note of the remainder (100 / 4 = 25, remainder 0)
  2. Write down the remainder to the left of the previous number you wrote using this step.
  3. Take the quotient as "the number" and goto step 1 again. Repeat until you have a quotient of zero.
So, with 100 I would have:
  • Divide by 4 = 25, remainder 0. Number is 0
  • Divide 25 by 4 = 6, remainder 1. Number is 10
  • Divide 6 by 4 = 1, remainder 2. Number is 210
  • Divide 1 by 4 = 0, remainder 1. Number is 1210
So, 100 in base 4 is 1210.
Hope that helps, --wj32 talk | contribs 07:54, 12 November 2006 (UTC)

Okay, yeah...i think i get what i was doing wrong before. Thanks for all the responses. --`/aksha 08:17, 12 November 2006 (UTC)

I'll work the problem Yaksha stated, "converting "23412" (base 5) into a base 7 number":
23412/12 = 1442 remainder 3
1442/12 = 120 remainder 2
120/12 = 10 remainder 0
10/12 = 0 remainder 5
so 234125 = 50237.
All the numbers here are in base 5, except the remainders are in base 7. I hope I haven't blundered, doing base 5 long division in my head! —Tamfang 10:26, 12 November 2006 (UTC)

For Q1, I would multiple 235(b6) to any other number including 235(b6) in exactly the same way as in primary school except the multiplication table is replaced from base 10 to base 6, for example 3*5(b6)=23(b6) and 5*5(b6)=41(b6). It is quick if you have to multiply hundreds of them.

For Q2, say from base-7 to base-3, set up a table 1->1, 2->2, 3->10, 4->11, 5->12, 6->20, 7->21. Denote x->k(x) as a formula. If abcd is a number with base-7, then {[k(a)*21+k(b)]*21+k(c)}*21+d is the required number in base-3 using base-3 mutliplcation table: 1*1=1, 1*2=2*1=2, 2*2=11 and also the addition table: 1+1=2, 1+2=2+1=10, 2+2=11. It is quick if you have to convert hundreds of them. A computer program can be written accordingly.

What I said is based on my intuition. I did not prove whether the method is correct. What do you think? Twma 12:51, 12 November 2006 (UTC)

To do arithmetic with the algorism you learned in school but adopted to base 6 you first have to learn the tables of addition and multiplication in base 6:
1 + 1 =  2    1 + 2 =  3    1 + 3 =  4    1 + 4 =  5    1 + 5 = 10
2 + 1 =  3    2 + 2 =  4    2 + 3 =  5    2 + 4 = 10    2 + 5 = 11
3 + 1 =  4    3 + 2 =  5    3 + 3 = 10    3 + 4 = 11    3 + 5 = 12
4 + 1 =  5    4 + 2 = 10    4 + 3 = 11    4 + 4 = 12    4 + 5 = 13
5 + 1 = 10    5 + 2 = 11    5 + 3 = 12    5 + 4 = 13    5 + 5 = 14

1 x 1 =  1    1 x 2 =  2    1 x 3 =  3    1 x 4 =  4    1 x 5 =  5
2 x 1 =  2    2 x 2 =  4    2 x 3 = 10    2 x 4 = 12    2 x 5 = 14
3 x 1 =  3    3 x 2 = 10    3 x 3 = 13    3 x 4 = 20    3 x 5 = 23
4 x 1 =  4    4 x 2 = 12    4 x 3 = 20    4 x 4 = 24    4 x 5 = 32
5 x 1 =  5    5 x 2 = 14    5 x 3 = 23    5 x 4 = 32    5 x 5 = 41
To multiply by a single digit, for example, 235 x 5, you get something like: 5 x 5 = 41, write 1 carry 4, 5 x 3 + 4 = 23 + 4 = (since 3 + 4 = 11) 31, write 1 carry 3; 2 x 5 + 3 = 14 + 3 = 21; thereby producing 2111 from left to right. For 235 x 235 you get then this:
   235
   235
   --- x
  2111
 1153
 514
------ +
105441
 --LambiamTalk 14:11, 12 November 2006 (UTC)
If you can hold "3*4=12" in your head as "twelve objects" instead of "ten and two", you can just directly go to 3*4 = two groups of six = 206, and don't need to memorize or look it up in tables or use any other sort of conversion. —AySz88\^-^ 00:18, 13 November 2006 (UTC)
That is a big if; see The Magical Number Seven, Plus or Minus Two. Otherwise we might not to memorize the base-10 tables either for arithmetic by hand.  --LambiamTalk 00:27, 13 November 2006 (UTC)
I didn't mean it in the sense of the Magic Number ("ten" is not within that range either), my point was to avoid thinking about the number as if it were intrinsically in base 10, but rather think of the number as just the number, to turn the obscure "I have to convert 1210 to something6..." step into a simpler "how many sixes are there in twelve?". Of course, this works only in the fairly low numbers that don't get above the cube of the base, but that's all that's needed in the multiplication algorithm. (And.... I could be the odd one out here, but is base 10 arithmetic tables usually taught as things to memorize? o.O ) —AySz88\^-^ 05:37, 13 November 2006 (UTC)

"my point was to avoid thinking about the number as if it were intrinsically in base 10, but rather think of the number as just the number" <<< this is rather...difficult, considering almost all maths is taught in base 10. Just about everything we learn in maths builds up from the basic mathematical principals which are in base 10. It's the way we are taught to think. Which is what i guess makes this fascinating, and at the same time confusing when i'm trying to work out exactly which bits of my mathematical knowledge are things which are the result of thinking in base 10, and which are things which should be true no matter what base i'm trying to work arithmetic in. --`/aksha 07:12, 13 November 2006 (UTC)

[edit] Logarithmic equations

I'll confess, this is part of a homework problem, but I'm not asking the helpful folks at the reference desk to do it for me, so hopefully you won't mind giving me a hand.

The question is "Find all the critical numbers in the interval (0, \infty) of the function y = x[ln(x)]2."

I've done the following:

  1. Taken the first derivative and simplified — y' = 2ln(x) + [ln(x)]2
  2. Set y' = 0.

Now I'm a little lost on how to proceed in solving for x. Any help would be appreciated. Thanks! --12.206.147.115 23:40, 12 November 2006 (UTC)

Your approach is absolutely right on the money. Try factoring the equation 0 = 2ln(x) + [ln(x)]2 using lnx as the variable. If it makes it easier to think of, replace the lnx with a u and solve for u, then substitute back and solve for x. --TeaDrinker 23:53, 12 November 2006 (UTC)
Aha, that works like a charm. Thanks for the quick response! --12.206.147.115 23:59, 12 November 2006 (UTC)