binary can denote any number from 0+. Now i dont remember what an integer is, so i hope that answers your question.
And when you ask "Is there some form of reasoning behind it", what exactly do you mean?
Joined: 8/1/2004
Posts: 2687
Location: Seattle, WA
Yes, binary can denote any positive integer. The 'reasoning' you seek is more or less a conversion function, which isn't really applicable to high digits without the help of a computer.
You can use binary to denotate any real number. If you're talking about with a computer though, the int type is 16 bits, and it can be signed or unsigned. Meaning that you can use it to either reference 0 to 65535 or -32768 to 32767. There's also float types which are used for 7 (?) significant digits plus an exponent.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
You can use binary to represent any finitely representable number whatsoever - not just integers. Repeating decimals are still a problem using normal encodings (0.000110011... is the binary equivalent of 1/10), but you can always use something like Binary Encoded Decimal to help alleviate that problem. The IEEE Floating Point standard (and others as well) even reserves room for encoding NANs and Infinity in binary. For 32 bit numbers:
0 11111111 00000000000000000000000 = Infinity
1 11111111 00000000000000000000000 = -Infinity
0 11111111 00000100000000000000000 = NaN (right bits just can't be all 0's - doesn't matter)
The problem with IEEE floating point numbers is that you don't necessarily get 7 significant digits, as seen with binary 1/10 where you can't even get 1 decimal digit significant. You do have 23 binary significant digits though if you don't mind thinking like that :p
(2 weeks later)
AAAAAAAAAAAAAH
I've lost so much hair over this thing, I keep getting the quotient between the two to be -2 instead of 2, and I just can't find that elusive mistake.
Bleh.
"We observe the behaviour of simple folk, and derive pleasure from their defects."
-Aristotle - Book of Humour
You sound like you almost got it... hmmm, did you screw up somewhere and position P and Q wrongly? Q should be on top while P below... i think.
EDIT: Hah... managed to solve it finally :P
This works :P
For my solution now:
Consider forces parallel to string for Q
CF1 sin(alpha) - T - 2mg cos(alpha) = 0 (forces balance)
2m(w^2)(QB sin(alpha)) sin (alpha) - T - 2mg cos(alpha) = 0
(2 QB) m(w^2) sin^2(alpha) - T - 2mg cos(alpha) = 0
Consider forces parallel to string for P
CF2 sin(alpha) + mg cos(alpha) - T = 0 (forces balance)
m(w^2)(BP sin(alpha)) sin (alpha) + mg cos(alpha) - T = 0
(BP) m(w^2) sin^2(alpha) + mg cos(alpha) - T = 0
Solving simultaneously
(BP) m(w^2) sin^2(alpha) + mg cos(alpha) = (2 QB) m(w^2) sin^2(alpha) - 2mg cos(alpha)
m(w^2)(2QB - BP) sin^2(alpha) = 3mg cos(alpha)
w^2 = {3g cos(alpha)} / {(2QB - BP) sin^2(alpha)}
Pretty close now, just need to figure out the inequality:
2QB - BP = 3QB - length of string ; given as 'a' in question
since QB < a
3QB < 3a
3QB - a <2a>
Therefore, {(3QB - a)^ -1} > {(2a)^ -1}
=> w^2 > {3g cos(alpha)} / {2a sin^2(alpha)}
For part (ii)
when (w^2) --> infinity,
{3g cos(alpha)} / {(2QB - BP) sin^2(alpha)} --> infinity
hence denominator --> zero
Therefore, 2QB - BP --> zero <=> BP/QB ~= 2
Suddenly got the inspiration to figure it out :P hope its understandable
I got to the First equations, but decided to solve it as a complete system, rather than solving for each ring individually, then combining, I think that's what threw me off.
But good problem none-the-less :)
"We observe the behaviour of simple folk, and derive pleasure from their defects."
-Aristotle - Book of Humour
Here's a probability exercise I can't solve. It's supposed to be kinda easy, but I just can't think of the right way to solve it.
Here goes:
Of all born children, 51.2% are male. What's the percentage for those families that have 7 children and all of those children are
a) male b) female
Any help would be appreciated.
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Guybrush wrote:
Here's a probability exercise I can't solve. It's supposed to be kinda easy, but I just can't think of the right way to solve it.
Here goes:
Of all born children, 51.2% are male. What's the percentage for those families that have 7 children and all of those children are
a) male b) female
Any help would be appreciated.
a) The possibility that all those 7 children are male is 0.512^7 = 0.922%
b) The possibility that all those 7 children are female is (1-0.512)^7 = 0.659%
You can think of it this way. The possibility that one of the kids is a male is 0.512. The possibility that the second is male given that the first is male is 0.512*0.512. You can then repeat this as many times as you want. A problem I can give you is: What is the percentage that 3 out of these 7 children are male/female? Think about that for a while ;)
Here's a probability exercise I can't solve. It's supposed to be kinda easy, but I just can't think of the right way to solve it.
Here goes:
Of all born children, 51.2% are male. What's the percentage for those families that have 7 children and all of those children are
a) male b) female
Any help would be appreciated.
MALE: Probability is 0.512 that if they have one child, it'll be a boy. Let's say they have two kids; that would be 0.512*0.512, since they're independent events. So, for seven, all boys, it's 0.512^7= ~0.0092233
FEMALE: Same exact thing, but with the probability for girls (which would be 1-0.512=0.455). 0.455^7= ~0.0040371
Not hard at all if you work it out a little.
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
I would say that the illegal operation occurs here:
I can't exactly say why, but it has to do with the properties of the exponential function. It's just... wrong. :P Perhaps someone can explain better why this is an illegal operation?
I would say that the illegal operation occurs here:
I can't exactly say why, but it has to do with the properties of the exponential function. It's just... wrong. :P Perhaps someone can explain better why this is an illegal operation?
It's with taking the implicit logrithm of an imaginary value.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Randil, it is actually correct. The rule goes like this: (a^m)^n = a^mn = (a^n)^m
However, it isn't actually correct, because to prove that lemma you need to use logrithms. However, when imaginary numbers are involved logrithms to weird stuff. The logrithm function isn't actually a function in the imaginary plane.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
You can't do that. The identity a^(bc) = (a^b)^c only holds when the following condition is met:
-п < (b ln a)Im ≤ п
However, when imaginary numbers are involved logrithms to weird stuff. The logrithm function isn't actually a function in the imaginary plane.
No, every function is a function. To prove this, if there was a function f that wasn't a function, there would be numbers a, b and c where b ≠ c, f(a) = b and f(a) = c, thus giving b = c (an impossibility), indicating that the function that wasn't a function, had to be a function after all.
The probability that 3 of 7 children are male is 0.512^3*0.488^4*7C3 = 0.2664141. And the probability that they are female is 0.2795164.
Sorry if they've already been posted.
1.
Disprove this: 64 = 652.
Find a 10-digit number which uses each of the digits 0-9 only once, such that the number created by the first two digits is divisible by 2, the number created by the first 3 digits is divisible by 3, et cetera, all the way up to the entire number being divisible by 10.
3.
A state lottery drawing involves picking X numbers out of a pool of Y numbers. Each number can only be picked once per drawing. A lottery ticket contains X numbers. A contestant wins if all X of the numbers on his ticket were drawn in any order. How many tickets must be purchased in order to guarantee winning the lottery?
Joined: 5/23/2006
Posts: 361
Location: Washington, United States
CtrlAltDestroy wrote:
Sorry if they've already been posted.
1.
Disprove this: 64 = 65
If you look at the slopes of the bottom two pieces, you can see (looking at the second picture) that the green piece has a slope of 8/3, while the orange piece has a slope of 5/2. The slopes are close enough that it's hard to tell from the picture that the pieces don't fit together evenly.
Edit: Explanatory picture:
Edit again:
CtrlAltDestroy wrote:
2.
Find a 10-digit number such that the number created by the first two digits is divisible by 2, the number created by the first 3 digits is divisible by 3, et cetera, all the way up to the entire number being divisible by 10.
0000000000! Seriously, though, it's relatively easy to use an iterative-type method for solving this; the first digit has to be 0 (since the overall number must be divisible by 10), and you can just go through the possibilities for each digit, checking if the n-digit number conforms to the defined rules. One example number is: 1615111120.
Edit again again: What I mean by iterative is, start with the second digit, try a "1" there, then test if the resulting number (10) is divisible by 2. Since it is, move on to digit three, try a "1" there, see that "110" is not divisble by 3, try a "2" there instead, etc. etc. If you use this method, you only have to "backtrack" once (since there are no solutions for numbers ending in -10, you have to start with -20 as the last two digits).
2.
Find a 10-digit number such that the number created by the first two digits is divisible by 2, the number created by the first 3 digits is divisible by 3, et cetera, all the way up to the entire number being divisible by 10.
1236543210 is one of extremely many solutions. (edit: 1020005640 is the smallest solution and 9876545640 is the largest)
CtrlAltDestroy wrote:
3.
A state lottery drawing involves picking X numbers out of a pool of Y numbers. Each number can only be picked once per drawing. A lottery ticket contains X numbers. A contestant wins if all X of the numbers on his ticket were drawn in any order. How many tickets must be purchased in order to guarantee winning the lottery?
All of them, after removing duplicate tickets, since the number of draws and the number of symbols on the ticket is the same number (so there is only one winning ticket).
someone is out there who will like you. take off your mask so they can find you faster.
I support the new Nekketsu Kouha Kunio-kun.
I guess I left out the part where you can only use each digit once. Oops.
Edit:
Boco wrote:
All of them, after removing duplicate tickets, since the number of draws and the number of symbols on the ticket is the same number (so there is only one winning ticket).
Say X = 4 and Y = 6. There are 6 possible numbers in the drawing. Lottery tickets contain 4 numbers, and 4 numbers will be drawn in the drawing. You need to buy 15 tickets to guarantee a win. The answer should be an equation.
This works :P
For my solution now:
No apparent mistake :)
Sorry if this was posted before.
I can't exactly say why, but it has to do with the properties of the exponential function. It's just... wrong. :P Perhaps someone can explain better why this is an illegal operation?
Edit again:
