Mathematics desk
< November 29	<< Oct \| November \| Dec >>	December 1 >

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.

November 30

Joint Account Bafflement

Dave and Angela have a joint account. They both pay in the same amount each month and pay the bills out of it. On one sunny Sunday, Angela buys herself a nice coat out of the account. There was £1000 before, the coat costs her £100. Now there is £900.

If Dave decides that the coat can be his gift to her for their Gold-Pressed Latinum anniversary - how much does he need to put in the account? If he put in £50, would that not cover 'her half' of the cost, given that he had already put in £50? Or would he have to put in the full £100?

When I comtemplate this, parts of my brain battle, one part saying he should obviously put in the full £100, the other convinced that he is then, somehow, paying twice.

And what if it was a bill that Dave was covering for Angela? When a £100 electric bill is due, taking the funds down to £900, and Dave says he will cover Angela's contribution, does he have to pay in the full hundred, or only £50? If it is £50, would she owe the money directly to Dave, or would she owe it to the joint account?

Thanks 195.60.13.52 (talk) 13:51, 30 November 2009 (UTC)[reply]

A good way to approach these problems, if they seem too complicated, is to arrange the figures in a table. Have you done this? --PS T 14:03, 30 November 2009 (UTC)[reply]

This sort of confusion can be solved by double-entry bookkeeping. Imagine there are three "accounts": the checking account (C), an account that totals the amount Angela has spent (A), and an account that totals the amount Dave has spent (D). Accounts such as (A) and (D) aren't really bank accounts, they just keep running totals of the money each person would owe the checking account to restore it to its initial balance. So, at every moment, C + A + D should equal the initial balance of the checking account. This means, for example, that if Dave deposits $25 then he should add $25 to C and subtract $25 from D. Similarly, when they pay a $100 electricity bill, the remove $100 from C and add $50 each to A and D.

With that accounting setup, when Angela bought the coat, she removed $100 from account C and added $100 to account A, to record that she was the one responsible for the coat. So if Dave wants to take responsibility for the coat, he should remove $100 from account A and add $100 to account D. There is no need to actually deposit any money in the bank. — Carl (CBM · talk)

But doesnt that mean that Dave has only bought Angela a hypothetical anniversary gift, if he has not recharged the bank account? 195.60.13.52 (talk) 14:21, 30 November 2009 (UTC)[reply]

Yes. With this sort of accounting, whenever funds are running low, Angela and Dave should each pay the amount listed in their own personal account into the checking account, to restore the initial balance. So by adding $100 to his account (D), Dave is committing himself to pay $100 into the checking account the next time it is recharged. He could go ahead and deposit the $100 and decrease his account by the corresponding amount, if he wants to. — Carl (CBM · talk) 14:35, 30 November 2009 (UTC)[reply]

P.S. Note that a more thorough accounting scheme might have a ledger like this:

	Transaction	Checking account	Amount Angela owes the checking account	Amount Dave owes the checking account	Total amount Dave has deposited into checking	Total amount Dave has deposited and spent	Total amount Dave has deposited but not spent
0	Starting values	1000	0	0	0	0	0
1	Angela buys coat	900	100	0	0	0	0
2	Dave takes responsibility for coat	900	0	100	0	0
3	Dave deposits 100 into checking account	1000	0	0	100	100	0
4	Dave preemptively deposits an extra 200 into checking	1200	0	-200	300	100	200
5	Dave buys Angela dinner for 200	1000	0	0	300	300	0

Here the sum of the first three columns will always be 1000, and the fourth column will always be the sum of the fifth and sixth. The second and third columns are for "short term" accounting: they should return to 0 very frequently, say at the end of the month when Angela and Dave recharge the account. The firth, fifth, and sixth accounts are for "long term" accounting (say to reconcile things at the end of the year). You can see how this sort of scheme is set up to make it more straightforward for a business to account for all the money it has and all the money it has spent. — Carl (CBM · talk) 14:54, 30 November 2009 (UTC)[reply]

Does that mean this is the scenario for the electric bill? 195.60.13.52 (talk) 15:18, 30 November 2009 (UTC)[reply]

	Transaction	Checking account	Amount Dave owes the checking account	Total amount Dave has deposited into checking	Total amount Dave has deposited and spent	Total amount Dave has deposited but not spent
0	Starting values	1000	0	0	0	0
1	Electric bill comes out	900	0	0	0	0
2	Dave takes responsibility for Angela's half of the bill	900	50	0	0
3	Dave deposits 50 into checking account	1000	0	50	50	0
4	Dave preemptively deposits an extra 200 into checking	1200	-200	250	100	200
5	Dave buys Angela dinner for 200	1000	0	250	250	0

Not quite. When electricity bill of 100 is deducted from the account, both Angela and Dave owe the account 50. When Dave takes responsibility for Angela's half of the biil he now owes the account 100 and Angela owes it 0. Remember, first three columns must always sum to 1000. Gandalf61 (talk) 15:43, 30 November 2009 (UTC)[reply]

For the electric bill, assuming it is shared equally, the first three accounts would work like this. — Carl (CBM · talk) 16:21, 30 November 2009 (UTC)[reply]

	Transaction	Checking account	Amount Angela owes the checking account	Amount Dave owes the checking account
0	Starting values	1000	0	0
1	Electric bill comes out (a check is drawn on the account)	900	50	50
2	Dave deposits to cover his half of the bill	950	50	0
3	Angela deposits to cover her half of the bill	1000	0	0

I'm nearly there now. If we assume that Dave and Angela don't care about reinstating the account every time something is withdrawn, and they both deposit 1000 a month which will cover bills, when Dave takes responsibility for the electric bill, does he add in an extra 100 or an extra 50 to cover Angela? Or does he just give her 50 cash and leave the account as it is? 195.60.13.52 (talk) 16:30, 30 November 2009 (UTC)[reply]

if Dave takes responsibility for the whole electric bill of 100, he needs to deposit 100 into the account to cover the check that pays the bill. — Carl (CBM · talk) 16:43, 30 November 2009 (UTC)[reply]

Why is 3SAT hard ?

I don't understand why 3SAT is hard, and even less why it is NP.

In the Boolean_satisfiability_problem article, an example is given: E = (x₁ or ¬x₂ or ¬x₃) and (x₁ or x₂ or x₄)

But this category of problems seem easy to solve, by simply applying distributivity (and de Morgan if necessary, but not in this case):

E = (x₁ + ¬x₂ + ¬x₃) . (x₁ + x₂ + x₄)

= x₁x₁ + x₁x₂ + x₁x₄ + ¬x₂x₁ + ¬x₂x₂ + ¬x₂x₄ + ¬x₃x₁ + ¬x₃x₂ + ¬x₃x₄

= x₁ + ¬x₂x₄ + ¬x₃x₂ + ¬x₃x₄

Which clearly states all the solutions, and seems a scalable way to find the solutions of any similar problem.

I trust all the brilliant people who worked on this before me, so I understand there must be an error in my reasoning. But where is the error ? What is so hard in applying distributivity and simplification on a boolean expression ? — Jerome.Abela talk 16:28, 30 November 2009 (UTC)[reply]

If your 3-CNF has n clauses, the result after using distributivity to bring it to DNF as you are attempting to do may have as many as 3ⁿ terms, so this is not a feasible algorithm. It's no more efficient than the trivial exponential-time algorithm which tries all possible assignments and checks whether one of them satisfies the formula. — Emil J. 16:36, 30 November 2009 (UTC)[reply]

Resolved

– Thanks! Jerome.Abela (talk) 09:05, 1 December 2009 (UTC)[reply]

Solving for Exponents via Algebra

How to solve 10^x = 100 using algebra?

Reason I am asking is because, we can solve the above using Log or Ln function. But I can't remember how to solve without using Log/Ln.

Using Log:

Log10^x = Log100
xLog10 = Log100
x = Log100/Log10
x = 2

We just punch in log100/log10 on our scientific calculator to get answer.

What do we punch in our calculator to solve for x without using Log/Ln?

In other words: How to solve 10^x = 100 using algebra?

--33rogers (talk) 22:40, 30 November 2009 (UTC)[reply]

Log10^x = Log 100
Log10^x = Log 10²
x Log10 = 2 Log 10
x = 2

Readro (talk) 22:48, 30 November 2009 (UTC)[reply]

Read my question properly. It states I already know to solve it using Log/LN. I am looking for a way to solve it without Log/LN. Thanks. --33rogers (talk) 23:06, 30 November 2009 (UTC)[reply]

10^x = 100
10^x = 10¹⁰

Whenever you have an equation of the form a^b = a^c, for real numbers with a>0, you can assume b=c, so:

x=10

This solution doesn't really help you in general; if you can't rewrite the rhs as a power with the same base, then you have to use logarithms. --COVIZAPIBETEFOKY (talk) 23:14, 30 November 2009 (UTC)[reply]

Why are you using binary notation? -- Meni Rosenfeld (talk) 09:34, 1 December 2009 (UTC)[reply]

This is precisely what logarithms do, so if you find a way to do it without logarithms you are just finding a way to write logarithms in a different way. You can change the notation, but you can't change the underlying maths. --Tango (talk) 23:21, 30 November 2009 (UTC)[reply]

Thanks Tango, for the perfect answer. Now I know why I am studying Logarithms :)

Resolved

--33rogers (talk) 02:44, 1 December 2009 (UTC)[reply]

I shall make a few additional remarks. When one solves a linear equation (that is, $ax+b=c$ , one employs the following steps:

ax+b=c

ax=c-b

x={\frac {c-b}{a}}

Essentially, one is "bactracking"; that is, subtraction is the inverse of addition, and division is the inverse of multiplication (these facts enable one to separate x in the equation). With a polynomial equation of degree 2, one employs the following steps:

ax^{2}+bx+c=0

x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}

Notice the presence of a "square root". Since x is squared in the equation, one must apply the "inverse" of the "square operation"; that is, one must take the square root. The other terms are obtained by noting that the inverse of addition is subtraction and the inverse of multiplication is division. With an equation such as, $10^{x}=100$ , to separate x, one must apply the "inverse" of "powering by 10". As I described earlier, the inverse of this operation is precisely underpinned by the "logarithmic function". Just as to remove a "square" we take the square root, to remove a "power", we take "logs". Hope this helps. --PS T 07:47, 1 December 2009 (UTC)[reply]