Explain partition(integer partition)with an example?
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A partition of a positive integer n is a set of k strictly positive integers whose sum is n.\n\nHere are the eleven partitions of 6. Observe that they are given in reverse lexicographic order.\n\n{\n {6},\n {5,1},\n {4, 2},\n {4,1,1},\n {3,3},\n {3,2,1},\n {3,1,1,1},\n {2,2,2},\n {2,2,1,1},\n {2,1,1,1,1},\n {1, 1, 1, 1, 1, 1}\n}\n\n\nA set partition of n elements is a grouping of all the elements into nonempty, nonintersecting subsets.\n\nSet partitions are different than integer partitions, representing the ways we can partition distinct elements into subsets. They are useful for representing colorings and clusterings. \n\nSet Partitition of {1,2,3}:\n\n{\n {{1,2,3}},\n {{1}, {2,3}},\n {{1,2}, {3}},\n {{1,3}, {2}},\n {{1}, {2}, {3}}\n}
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It's a little tricky, but here goes:\n\nOk, assume that rt2 can be written as a rational fraction, then there must be some integer, "a", such that a*rt2 is an integer.  For example, if rt2 were 3/4, then "a" would be 4.  4*3/4 = 3, an integer.\n\nAlso assume that, whatever number "a" is, it is the smallest possible integer.  As in the example, 8,12,16 etc would also yield integers, but 4 is the smallest one possible.\n\nSo to recap, we have a*rt2 is some integer.\n\nNow we also know that 1<rt2<2.  Subtract 1 all the way across 0<rt2 - 1<1\n\nso now we know that rt2-1<1, and, logically, a*(rt2 - 1) < a*rt2\n\nTake a moment, let all of that sink in...and moving on:\n\na*(rt2-1) = a*rt2 - a  --- THIS WILL ALSO BE AN INTEGER, we'll assign it a new variable, b, and remember that above we proved that b is less than a.\n\nIf b = a*rt2 - a \n\nThen by multiplying both sides by rt2 you get\n\nb*rt2 = 2a - a*rt2\n\nNow, if a is an integer, and a*rt2 is an integer, then 2a-a*rt2 must also be an integer.  And since it is equal to b*rt2, then b must also be an integer.\n\nBut wait - we said that in a*rt2, a was the smallest possible integer, and we know that b is smaller than a, so there is your contradiction.\n\n\nPersonally, I like the other proof better I think.  This one is quite nice, but it was rather difficult to explain.  Once it sinks in, though, it is actually simpler.  I hope this helps.
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you either have one hard drive with two partitions or two hard drives..  when you boot your system it will tell you but you can (assuming you have XP) click on start -> all programs -> accesories -> system tools --> system info.\n\nHere go to components -> storage -> drives\n\nThis will break out your drive info - you are interested in Local fixed disks.  It details size, space used/unused of each fixed partition.\n\nYou can also go to the control panel -> admin tools -> computer management -> storage -> disk management .  Here it will show you the disk layout and partitions.  If disk 0 has morethan one drive letter defined to the right of it then you have those partitions on a single hard drive (disk 0 is a single physical drive which can have 1, 2, 3... partions defined as C, D, E).
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I can do all that, but except for small numbers I wold never bother.\n\nThe "easy" way, such as it is, is to always go through binary on your way from decimal to hex. \nFor the others, list the bits and add/partition.
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look at a few examples and it will make sense\n\nfor example \n\n02 (where the 2 is small)   -> _ O    (the _ is where a number goes)  (2)
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Yah what butchell said.\n\nHello Hello\nEarth to Alireza G\nEarth to Alireza G\n\nYou can only have one (1) (uno)\nPRIMARY partition.
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a declaration\n\nexample:\nint count; defines count as an integer
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you can use partition Logic,Partition Logic is a free hard disk partitioning and data management tool.  It can create, delete, format, defragment, resize, and move partitions,there is a floppy and a Cd version \n\nhttp://partitionlogic.org.uk/download/index.html
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Let’s call a positive integer ambiguous if there are two different partitions of equal\nsize of it into summands whose products are the same. For example, 14 is ambiguous because\n14 = 3+3+8 = 2+6+6 and 3 · 3 · 8 = 2 · 6 · 6 = 72. A number is called doubly ambiguous if\nthere are two pairs of partitions each of which has the same product, and these two products\nare different. For example, 13 is doubly ambiguous since 13 = 1 + 6 + 6 = 2 + 2 + 9 and\n1 · 6 · 6 = 2 · 2 · 9 = 36 and at the same time 13 = 1 + 1 + 3 + 4 + 4 = 1 + 2 + 2 + 2 + 6 and\n1 · 1 · 3 · 4 · 4 = 1 · 2 · 2 · 2 · 6 = 48. Now the first mathematicians response to the second mathematicianss\ncomment is equivalent to ‘the bus number is ambiguous’. Note that if n is ambiguous, then so\nis n + 1 and if n is doubly ambiguous, then n + 1 is also doubly ambiguous. An examination\nof all partitions of 11 shows that 11 is not ambiguous. Thus every integer n at least 12 is\nambiguous, and every integer bigger than or equal to 13 is doubly ambiguous. The second\nmathematician’s comment that he knew the age of the first mathematician implies that the bus number is not\ndoubly ambiguous. Thus the bus number must be 12.
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again really easy:\n\nint divisibility = (m % 10) == 0) ? 1 : 0;\n\nBut I think it is better you learn it yourself. Try a book that would be better. Just search C language tutorials.
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An integer is a whole number - 1, 2, 100, etc.  Any number with a fraction or decimal is not an integer.  A positive integer is greater than 0, and a negative integer is less than 0.
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This is sampling without replacement and without regard to order.\n\nIt's given by the binomial coefficient:\n\nfor n items, choosing k at a time\n\nnCk means n choose k \n\nor \n\n    n!\n-----------  (this is a division)\n(n-k)!k!\nNote that you can get these from Pascal's triangle, nth row, kth column.\n\nSo, 4C2 is 6:\n\nAB\nBC\nCD\nDA\nAC\nBD
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format command creates a file system in the disk partition. In case of windows it is either FAT32 or NTFS depending on your choice.
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Partitioning is separating or dividing something into parts.  So, I would imagine adding or subtracting using partitioning would be using physical or graphic blocks to show the problem.  Remember when children first learn to add, worksheets often contain a tens unit block?  Partitioning gives the problem a visual aid.  It's easier to learn when you can see that two groups make more...or taking away equals subtraction.
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Partition - This is how you organise your harddisk space. Say you have a 100GB harddisk. You can use all of it as C: drive or you can "partition" 50GB for C: drive and 50GB for D: drive. Some people put different Operating Systems in different partitions, while others put Programs in one partition and Data in another partition. There are innovative ways to partition your harddisk to get the optimal performance for your particular needs.\n\nFAT - File Allocation Table. This is the place where the locations of your files are registered so that it can be saved, retrieved, deleted, moved, etc.\n\nFormat - This is a process of organising your harddisk drive's space so that the operating system and applications can use it.\n\nIDE - This is harddisk hardware interface to the PC system's board. It is normally connected via a ribbon cable.
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If you are thinking of f : ? -> { 0, 1, 2, 3, ... } (i.e., the domain is mapped into the set of non-negative integers) then clearly the domain "?" contains [at least] all finite bit strings.  If you keep the same restriction on the range, you can even extend the domain to include those infinite bit strings (however you wish to define that) which contain only finitely many ones.  If you extend the range to also include infinite cardinals and/or ordinals then you can include even more infinite bit strings into the domain.
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Because your C: drive contains Windows and all the programs you have installed (if you didn't choose to install them on the D: drive).  I am also assuming that these are two hard drives and not just separate partitions because Windows will see a partitioned drive as two drives also.