Can you tell me some interesting facts about the numerator for math?
398790
The most interesting fact about the numerator is that it can actually be the denominator. Let me explain...\n\nIf a number in the numerator (say x) is raised to a negative power (i'll show it as x^-1) then x is moved down to the denominator. for example \n\nx^-1 = 1/x \nand\nx^-4 = 1/x^4\n\nif it is already a fraction, like (x^-1)/y, then you simply move the x down with the y, 1/xy.\n\nHope that helps
30751
The Edifier has it right, but darned if I can follow the notation. You really need to see this written out like a textbook to understand what's going on.\n\nClick the link below in "sources" to see the answer hosted at my website www.calculus-help.com. Right now, Yahoo! answers, as far as I can tell, doesn't allow graphics, which makes it tricky to answer ANY math questions.\n\nPlease don't post math answers, guys, if the poster knows the answer but can't get there, and you get something completely different.
2056
You are trying to get the smallest number on top (numerator) and bottom (denominator). You divide the numerator and denominator by the same number.\n\nFor example, 180/240 can be simplified by dividing the numerator and denominator by 10 to get 18/24. Next you can divide the numerator and denominator by 6. 18 divided by 6 is 3, and 24 divided by 6 is 4 to get a simplified fraction of 3/4. \n\nThe trick is knowing which numbers divide the numerator and denominator evenly.
478213
Properties of mathematics makes it work. I'm not sure exactly what it has you do but essentially it does this.\n\nIf your number is X. You add 5 to it then subtract 5 from it and you get your original number. Of course the nice ones hide the fact that your doing exactly this. \n\nDavid Copperfield did a TV bit once where he had a 3X3 square. You start out at any square you want and move according to his directions. After each move he took away a square. In the end you always end up on one square. That was cool. Anyone heard of it?
952627
My answers are:\na_6 = (6^2+1)/2(23) = 37/46 ~= 0.709\na_7 = (7^2+1)/2(29) = 50/58 ~= 0.860\n(even though a_6 seems to violate the monotonic increasing)\n\nWrite a_i = n_i / d_i.\nI think I found the pattern for the numerator.\nn_i seems to be (i^2+1):\na_0=1/2 = (0^2+1)/2(1)\na_1=1/2 = (1^2+1)/2(2)\na_2=5/8 = (2^2+1)/2(3)\na_3=10/14 = (3^2+1)/2(7)\na_4=17/22 = (4^2+1)/2(11)\na_5=13/16 = 26/32 = (5^2+1)/2(16)\n\nThe denominator d_i is more tricky, it doesn't have a simple difference relation:\nd_5 = 2(16) = d_4 + 2(5)\nd_4 = 2(11) = d_3 + 2(4)\nd_3 = 2(7) = d_2 + 2(4) ??\nd_2 = 2(3) = d_1 + 2(1) ??\nd_1 = 2(2) = d_0 + 2(1) ??\nd_0 = 2(1)\n\nSo I think I found this pattern:\nd_5 = 2(5*3+1)\nd_4 = 2(4*3-1)\nd_3 = 2(3*2+1)\nd_2 = 2(2*2-1)\nd_1 = 2(2*1) \nd_0 = 2(1*0+1) = 2(1)\nwhich kind of has a pattern:\nd_7 = 2(7*4+1) = 2(29)\nd_6 = 2(6*4-1) = 2(23)\n\n__________________________________________________\nMy previous tries:\n\nTricky. Couldn't spot any simple algebraic or difference relation in the terms a_i.\nHowever you can see the terms are in increasing order:\na_0=a_1= 1/2 = 8/16 = 0.5\na_2= 5/8 = 10/16 = 0.625\na_3=5/7 ~= 0.714\na_4=17/22 ~= 0.773\na_5=13/16 = 0.8125\n\nSo for a_6, a_7 , we want to construct plausible answers > 13/16\n\nLooks like the terms seem to be converging on 1, or something close to 1, with a decreasing step, it might be easier to spot the pattern this way:\na_0=a_1 = (1 - 1/2)\na_2 = (1 - 3/8)\na_3 = (1 - 2/7)\na_4 = (1 - 5/22)\na_5 = (1 - 3/16)\nor a pattern looking at the steps:\n(a_2 - a_1) = 0.125\n(a_3 - a_2) ~= 0.89\n(a_4 - a_3) ~= 0.69\n(a_5 - a_4) ~= 0.38\n\nMaybe these are the closest rational approximations to (1 - k^n) with denominators <32, for some real k.\n\nSo if we look at the differences b_i = (1 - a_i )\nb_0=b_1= 1/2 = 8/16 = 0.5\nb_2= 3/8 = 6/16 = 0.375\nb_3=2/7 ~= 0.286\nb_4=5/22 ~= 0.227\nb_5=3/16 = 0.1875\n\nThis looks maybe more useful, you can see:\nb_2 = b_1 * 3/4\nb_5 = b_2 * 1/2
952176
Presumably you mean\n(x^2)/(x + 2) - 4(/x + 2)\nSince the denominators are the same, you can add the numerators without modifying them:\n(x^2 - 4)/(x + 2)\nFactor the numerator to get\n(x - 2)(x + 2)/x+2)\n(x + 2)/x+2) = 1 so you are left with\n(x - 2)\n\nIf you post this question under philosophy, you can probably get many different answers to why this is.  Because it works is mine.
966707
u cud either go to wikipedia.com else u can try mathworld.wolfram.com u'l gt al math related info here
709978
there r lots of shorcut techniques for multi. and divison such as \n\nsquare of no.\nthe 9 rule\nthe 11 rule\nmultiply  up to 30 without a calculator \ncross multiplication technique ( u can multiply any digit no.'s fatly)\nsum of first n natural no. as n(n+1)/2.\ndivisiblity rules\nthe calender trick etc.\nsorry, as these details r very lenghty so i can't able to give u details about these on internet. but u can learn these techniques from books. there r several books in market for these techniques.\nso,  best of luck.
1090065
Multiply the numerator and denominator by (√ x + ∆x ) + ( √ x ) \n[(√ x + ∆x ) - ( √ x )][( √ x + ∆x ) + ( √ x )]\n= x + ∆x - x \n= ∆x\n∆x/[∆x[(√ x + ∆x ) + ( √ x )]]\n= 1/[(√ x + ∆x ) + ( √ x )]\nReplace ∆x with 0.\n= 1/[(√ x + 0 ) + ( √ x )]\n= 1/[(√ x ) + ( √ x )]\n= 1/(2√ x )
1073227
Zero is such a complicated number.   If you look at the way numbers are sequenced -3, -2, -1, 0, 1, 2, 3 etc. it would appear to be even.  There is a problem, however.  Zero came from the Arabic Hindu system and was considered a placeholder.\n\nZero is used as a place holder.  Numbering systems start with #1 and progress to 9.  1 starts over in the tens position and is followed by the placeholder 0 so we know one is in the tens position and not the single unit position.  The same is so for 20 and 30 and so on and so forth.  Hundreds, thousands, tens of thousands, hundreds of thousands, so on and so forth.\n\nNow back to the numbering sequence -3, -2, -1, 0, 1, 2, 3.  Zero is a placeholder and is neither odd or even.  It is a null value.  You have heard of zip, zilch, nada?  That is zero.\n\nZero was brought to the West by a mathematician named  Fibonacci some time in the 1200s.  It made numbers a lot easier to write.  Most numbers were written with a space where a zero would go today.  Made for lots of problems.  The numbers could only be understood in the context of the writing and could never be used by themselves.   1 2 3 4 5 6 7 8 9 1 11 12 13 14 15 16 17 18 19 2 21 22 23 etc.  These are writen in sequence so you know the 2nd 1 is really a ten.  But standing alone, it still looks like 1.\n\nThe computer age really stirred the pot.  Programing in the binary system put 0 before 1.   Auuuuuuuuuugh! as Charlie Brown would say.  The computer needed continuity in the old on/off programing in binary language.  When 0 followed 1 the computer wouldn't compute.\n \nWhen people count, it is assumed that zero (or nothing) comes first before we have something (1).  But we couldn't use it as a placeholder, because one is a single unit.  There is nothing to hold a place for.  10 is another situation.  Zero is holding an empty unit spot.\n\nYou can have the shovel now, I think I've dug a deep enough hole. (0)  Chuckle.
579225
Division by zero is undefined, even if you divide zero into zero.\n\nhttp://en.wikipedia.org/wiki/Division_by_zero
1030456
Multiply the numerator and denominator by  (√ x + ∆x ) + ( √ x ) \n[(√ x + ∆x ) - ( √ x )][( √ x + ∆x ) + ( √ x )]\n=   x + ∆x  -  x \n= ∆x\n∆x/[∆x[(√ x + ∆x ) + ( √ x )]]\n= 1/[(√ x + ∆x ) + ( √ x )]\nReplace ∆x with 0.\n= 1/[(√ x + 0 ) + ( √ x )]\n= 1/[(√ x ) + ( √ x )]\n= 1/(2√ x )
207651
http://mathworld.wolfram.com\nhttp://en.wikipedia.org/wiki/Mathematics
173341
Crap!\n\nSomething to do with Pi.  When my friend was in college she used to say the first one.\n\nETA:  They are both frases said to remember the numerics of Pi.
619440
7 will be after five\nand eight will be after six.\n\nReason: The upper numbers in 'numerals' contains straight lines, the lower one contains curves.\nWell...
595633
When you are dividing by something you are trying to find out how many of a certain amount (say, 3) is in another amount (say, 12). The result for my example, is of course 4, since 3 goes into the number 12 four times.\n\nNow the number 0 is nothing, literally. So if you're dividing into, say 12, you're wondering how much nothing is in 12. That's just not possible, because 12 is all "something".\n\nI hope that's clear enough...I tried.\n\nEdit: The guy above me is wrong.
616629
First of all, change this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first numerator: <input type=text name=num2 size=20></p>\n<p>Enter second denomenator: <input type=text name=num1 size=20></p>\n<p>Enter second denomenator: <input type=text name=num2 size=20></p>\n\n\nto this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first denomenator: <input type=text name=num2 size=20></p>\n<p>Enter second numerator: <input type=text name=num3 size=20></p>\n<p>Enter second denomenator: <input type=text name=num4 size=20></p>\n\nThen based on the operation selected, implement the proper logic there (its basic math) and then your program will work.\n\nGood luck!!