a system of notation by means of the binary digits, its algebra?
144487
the answer is exponents.
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4, 0, -2, -2, 0\n\nFigure it out! Don't be lazy!
973911
It might be a little to high of a level but thesymbol for integers Z (in set theoy) comes from the german word for number.  Also, in cardinality of sets uses some hebrew letters.  Algebra is an Arabic word.  Our number system is called the Hindu-Arabic.  \n\nWikipedia has some good stuff on math history....\n\nOf course, lots of things are named for the developer.  Cartesian coordinate for Descarte.  Gaussian, or normal, distibution: the so-called bell curve.  Pythagoras...Fibonacci numbers...
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I only need four 0.\nbinary:0!+0!+0!+0!=100
38023
In binary zero means off
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Brackets, Orders, Division, Multiplication, Addition, Subtraction.\nIn some countries its BEDMAS, where E is exponents.
121493
We are used to ten digits in our numbering system, but binary only has two (zero and one, or on and off) In our system when we get to nine, our ones spot is at the top and we have to use two columns for higher numbers. Same applies in binary. Column starts at zero and is full at one.\nOn our regular system our first two columns fill at nine nine - in binary the first columns fill at one one. nine nine in our system equals 99. one one in binary equals 3.\n00000000 = 0\n00000001 = 1\nnow that column is full, so to count to 2 you have to use the next column.\n00000010 = 2\n00000011 = 3\nnow that column is full, so to count to 4 you have to use the next column.\n00000100 = 4\nand so on.\n\nInsted of having the "ones" "tens" "hundreds" etc. columns as our regular system does - \nbinary has the "ones" "twos" "fours" "eights" "sixteens" thirty-twos" etc colmns\n\nA byte is 8 columns.
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Each row coresponds to coeficients of the binomial expansion.  The first row is the n=0 row with the entry 1, the second row n=1 with the entries 1,1 and so on.  The entries correspond to the coefficients of (a+b)^n.  \n\nAlso you can get relationships between the entries of pascal's triangle and the fibonaci numbers.  There are many other applications and patterns to be found.
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we use binary numbers is boolean algebra\n1's and 0's\n\nthe common operators in boolean algebra are\nand (x)\nor (+)\nnot (/)\n\nthink of it as a series of true/false questions\n1 being true and 0 being false\n\nif we use the AND (x) operator\n1 x 1 = 1   in the AND operator, both the first term AND the second term must be true for the answer to be true\n\nthe truth table for the AND (x) will look like\n0 x 0 = 0 \n0 x 1 = 0\n1 x 0 = 0\n1 x 1 = 1  only when both are true we get a 1\n\nfor the OR (+) operator, as long as one of the terms is true, the answer is true. (one OR the other must be true)\n0 + 0 = 0 \n0 + 1 = 1\n1 + 0 = 1 \n1 + 1 = 1\n\nthe not negates the value. /1 = 0 & /0 = 1
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You don't have to actually take the exponents to do this.  Instead, let's look at the units digit of exponents of 8:\n\n8^1 = 8\n8^2 = 4 (looking only at the units digit)\n8^3 = 2\n8^4 = 6\n8^5 = 8 (notice we are at 8 again, so we will just repeat our pattern from here on out)\n\nSo the digits unit of 8^n repeats every 4 digits, so we look at the remainder of n/4 to determine our units digit.\n\nSince n = 1999 in our case, and 1999/4 = 499 remainder 3, we can know therefore that 1998^1999 has the same units digit as 8^3, namely 2.\n\nSimilarly for 1999^1998, let's look at the units digits for exponents of 9:\n\n9^1 = 9\n9^2 = 1 (looking only at units digit)\n9^3 = 9 (notice we are at 9 again, so we will just repeat our pattern from here on out)\n\nSo here our pattern is pretty easy -- for 9^n, the units digit is 1 if n is odd, 2 if n is even.\n\nSo, 1999^1998 will have a 1 as the units digit (since 1998 is even)\n\nTherefore the units digit of (1998^1999) - (1999^1998) = \n2-1 = 1\n\nPretty neat way to solve it, huh?  :-)\n\n.
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- Its called Binary.\n- The binary consists of 0's and 1's\n- At basic level of hardware, whenever you pass a command to the computer, the operating system translates this to machine language.\n- The machine language interprets as 0's and 1's which means create some electronic signals.\n- Computer hardware is consists of thousands of Transsistors.\n- A '0' generation means the transistor produces the 'Negative' signal and '1' genaration means a 'positive' signal.\n- This how a computer's base system works and so the binary called 0 and 1.\n- Its a broad topic and so i can suggest you to search in Googe for more elaborated results.\n- all the best.\n--------------
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Binary digits stand for powers of 2.  For example, 2^0 is 1, and 1 in binary is 1.  2 is equal to 2^1, so 2 is 10 in binary.  3 is 2^1 + 2^0, so it's 11 in binary.  \n\nSo the joke is really saying there are 2 types of people.  Now you fall into the former.  :-)\n\n*** There are two types of people, those who know binary and those who don't, now you fall into the former category of those who know.
725157
assuming the numbers after the parenthesis and variables are exponents...\n(x-1)^(3/2) [6x(x-1)^2-x^2(x-1)^3]\njust add the exponents together from the outside to the inside- (3/2) + (2) = 7/2\n\n(x-1)^(3/2) [6x(x^2-2x+1)-x^2(x^3-3x^2+3x-1)]\n(x-1)^(3/2) [6x^3-12x^2+6x-x^5-3x^4+3x^3-x^2]\n(x-1)^(3/2) [-x^5-3x^4+9x^3-13x^2+6x]\nx(x-1)^(3/2)(-x^4-3x^3+9x^2-13x+6)\nokay, that took me for friggin ever to type on a regular keyboard. Please, seriously consider me for the best answer!
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The binary digit. That's what everything else boils down to.
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So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
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3 in the binary number system is written 11.  Stands for 2^1 + 2^0.
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Binary numbers are just normal (base 10) numbers written in a different way.\n\nTo convert a binary number to a number you're used to seeing, start at the right end. Each place in the number represents a power of 2, starting with 2^0, or 1. Each place to the left increases by a power of 2. To determine what number you have, simply add the powers of 2 which correspond to the 1's.\n\nFor example, take the binary number 110101.\n\nStarting from the left, we have\n\n1 * 2^0 = 1*1  = 1\n0 * 2^1 = 0*2  = 0\n1 * 2^2 = 1*4  = 4\n0 * 2^3 = 0*8  = 0\n1 * 2^4 = 1*16 = 16\n1 * 2^5 = 1*32 = 32\n\nSo we add 1 + 4 + 16 + 32 = 53\n\nSo 53 expressed in binary is 110101\n\nHope this helps.