"Proofs involving the addition of natural numbers This article contains mathematical proofs for some properties of addition of the natural numbers: the additive identity, commutativity, and associativity. These proofs are used in the article Addition of natural numbers. This article will use the Peano axioms for the definitions of addition of the natural numbers, and the successor function S(a). In particular: For the proof of commutativity, it is useful to define another natural number closely related to the successor function, namely ""1"". We define 1 to be the successor of 0, in other words, Note that for all natural numbers"
"10 values are valid (0000 through 1001). So adding 6 to the invalid entries results in the following: Thus the result of the subtraction is 1001 1001 0010 0101 (−925). To confirm the result, note that the first digit is 9, which means negative. This seems to be correct, since 357 − 432 should result in a negative number. The remaining nibbles are BCD, so 1001 0010 0101 is 925. The ten's complement of 925 is 1000 − 925 = 75, so the calculated answer is −75. If there are a different number of nibbles being added together (such as"
"between 56.25 and 72.25. So 6 × 10 is the best one can give, as other possible answers give a false sense of accuracy. Further, the 6 × 10 is itself confusing (as it might be considered to imply 60 ±5, which is over-optimistic; more accurate would be 64 ±8). When adding or subtracting using significant figures rules, results are rounded to the ""position"" of the least significant digit in the most uncertain of the numbers being summed (or subtracted). That is, the result is rounded to the last digit that is significant in ""each"" of the numbers being summed."
"an exhaustive search of [1,2916] one finds that for summing the cubes of digits base 10 there are happy numbers and eight different types of unhappy number: those that eventually reach formula_30 or formula_31, which perpetually produce themselves. those that eventually reach the loops: formula_32 formula_33, as well as those that alternate between formula_34 and formula_35 or between formula_36 and formula_37. All multiples of three end in 153. This fact can be proved by the exhaustive search up and noting that a number is a multiple of three if and only if the sum of digits is a multiple of"
"is equal to its own missing-digit-sum in base ten. Now consider the set of seven-digit numbers ending in ...04. For these numbers, the final digit of the sum is equal to (0 x 1 + 4 x 6) modulo 10 = (0 + 24) mod 10 = 24 mod 10 = 4. This set may therefore contain one or more missing-digit sums. Next consider seven-digit numbers ending ...404. The penultimate (last-but-one) digit of the sum is equal to (2 + 4 x 2 + 0 x 4) modulo 10 = (2 + 8 + 0) mod 10 = 10 mod"
"the case of ""9"" (), ""20"" (), and ""1000"" (), the final vowel is assimilated into the suffix. Most multiples of ten end in a vowel in their cardinal form and form their ordinal form by adding the suffix to their genitive singular form, which ends in ; this is not reflected in writing. Exceptions are ""20"" () and ""40"" (), both of whom form their ordinals by adding the suffix directly to the cardinal ( and ). When counting objects (2) becomes and (4) becomes . As in French, the vigesimal system is widely used, particularly in people's ages."
"the last number is either 0 or 5, the entire number is divisible by 5. If the last digit in the number is 0, then the result will be the remaining digits multiplied by 2. For example, the number 40 ends in a zero (0), so take the remaining digits (4) and multiply that by two (4 × 2 = 8). The result is the same as the result of 40 divided by 5(40/5 = 8). If the last digit in the number is 5, then the result will be the remaining digits multiplied by two (2), plus one (1)."
"a number between 1 and 10 using a two-out-of-five code with the following numeric equivalence: 1, 2, 4, 7, 0. The numbers are summed together. For example, the number 6 is encoded NWWNN, with wide bars occupying the positions for 2 and 4 (2+4=6). In the case of NNWWN which is it is assigned to 0 for digits (+0), and 10 for the letter columns (+10 – +30). When encoding the (+10 to +30) Letters the equation needs a ""−1"" added so 'A' is WNNNW → 1 + 10 − 1 → 10 as shown in the table. The last"
"10. At this point the process is repeated enough times to reach a stopping point: The largest number by which the divisor 4 can be multiplied without exceeding 10 is 2, so 2 is written above the 0 that is next to the 5 – that is, directly above the last digit in the 10. Then the latest entry to the quotient, 2, is multiplied by the divisor 4 to get 8, which is the largest multiple of 4 that does not exceed 10; so 8 is written below 10, and the subtraction 10 minus 8 is performed to get"
"the multiply-by-10 proof above in the opposite direction: if ""x"" = ...999 then 10""x"" = ...990, so 10""x"" = ""x"" − 9, hence ""x"" = −1 again. As a final extension, since 0.999... = 1 (in the reals) and ...999 = −1 (in the 10-adics), then by ""blind faith and unabashed juggling of symbols"" one may add the two equations and arrive at ...999.999... = 0. This equation does not make sense either as a 10-adic expansion or an ordinary decimal expansion, but it turns out to be meaningful and true if one develops a theory of ""double-decimals"" with eventually"
"(mod 8), (mod 12), (mod 5) and (mod 28), then by the law of quadratic reciprocity 2, 3, 5, and 7 will all be residues modulo ""p"", and thus all numbers 1–10 will be. The CRT says that this is the same as ""p"" ≡ 1 (mod 840), and Dirichlet's theorem says there are an infinite number of primes of this form. 2521 is the smallest, and indeed 1 ≡ 1, 1046 ≡ 2, 123 ≡ 3, 2 ≡ 4, 643 ≡ 5, 87 ≡ 6, 668 ≡ 7, 429 ≡ 8, 3 ≡ 9, and 529 ≡ 10 (mod"
"The following example is decimal, which simply means the base is 10. This is nothing other than converting to scientific notation. In detail: This is the true result, the exact sum of the operands. It will be rounded to 7 digits and then normalized if necessary. The final result is: Note that the low 3 digits of the second operand (654) are essentially lost. This is round-off error. In extreme cases, the sum of two non-zero numbers may be equal to one of them: Another problem of loss of significance occurs when two close numbers are subtracted. e=5; s=1.234571 and"
"1, 0), (0, 8, 3, 8, 3, 0) Multiply the group containing the rightmost character (which is the FIRST group) by 2: Add up the individual digits: (6 + 4 + 0 + (1 + 4) + 6 + 2 + 0) + (0 + 8 + 3 + 8 + 3 + 0) = 45 Take the 10s modulus of the sum: 45 mod 10 = 5 Subtract from 10: 10 - 5 = 5 Take the 10s modulus of the result (this final step is important in the instance where the modulus of the sum is 0, as"
"one-digit mistakes and digit swaps. Checking validity of PESEL: 44051401358 (number 8, the last digit, is the check digit for this PESEL): 1 * 4 + 3 * 4 + 7 * 0 + 9 * 5 + 1 * 1 + 3 * 4 + 7 * 0 + 9 * 1 + 1 * 3 + 3 * 5 = 101 Getting the last digit of the result (101 modulo 10): 101/10 = 10 remainder 1 In order to get the check digit one needs to take the 10s complement of the number. It is 0 if the"
"of ""m"" to the right end are They are the period of ""n""/(10""n"" - 1), also the period of the decadic integer -""n""/(10""n"" - 1). Number of digits of them are Parasitic number An ""n""-parasitic number (in base 10) is a positive natural number which can be multiplied by ""n"" by moving the rightmost digit of its decimal representation to the front. Here ""n"" is itself a single-digit positive natural number. In other words, the decimal representation undergoes a right circular shift by one place. For example, 4•128205=512820, so 128205 is 4-parasitic. Most authors do not allow leading zeros to"
"""n"" ""x""s) mod ""p"" are all different. That is, one is 0, another is 1, another is 2, ..., the other is ""p"" − 1. Thus, since the first ""p"" − 1 numbers are all primes, the last number (the repdigit with ""n"" ""x""s) must be divisible by ""p"". Since ""p"" does not divide ""x"", so ""p"" must divide the repunit with ""n"" 1s. Since ""b"" is a primitive root mod ""p"", the multiplicative order of ""n"" mod ""p"" is ""p"" − 1. Thus, ""n"" must be divisible by ""p"" − 1) Thus, if ""b"" = 10, the digits coprime"
"another example, if ""n"" = 2, then 10""n"" − 1 = 19 and the repeating decimal for 1/19 is So that for 2/19 is double that: The length ""m"" of this period is 18, the same as the order of 10 modulo 19, so = 105263157894736842. 105263157894736842 × 2 = 210526315789473684, which is the result of moving the last digit of 105263157894736842 to the front. The smallest ""n""-parasitic numbers are also known as Dyson numbers, after a puzzle concerning these numbers posed by Freeman Dyson. They are: (leading zeros are not allowed) In general, if we relax the rules to"
"10 10 (ten) is an even natural number following 9 and preceding 11. Ten is the base of the decimal numeral system, by far the most common system of denoting numbers in both spoken and written language. The reason for the choice of ten is assumed to be that humans have ten fingers (digits). As is the case for any base in its system, ten is the first two-digit number in decimal and thus the lowest number where the position of a numeral affects its value. Any integer written in the decimal system can be multiplied by ten by adding"
"10 (and also in base 2). The decimal expansion of 1/131 repeats the digits 007633587786259541984732824427480916030534351145038167938931 297709923664122137404580152671755725190839694656488549618320 6106870229 indefinitely. 131 is also: 131 (number) 131 (one hundred [and] thirty-one) is the natural number following 130 and preceding 132. 131 is a Sophie Germain prime, an irregular prime, the second 3-digit palindromic prime, and also a permutable prime with 113 and 311. It can be expressed as the sum of three consecutive primes, 131 = 41 + 43 + 47. 131 is an Eisenstein prime with no imaginary part and real part of the form formula_1. Because the next odd number, 133,"
"+ 1φ) = 5 − 3φ ≈ 0.145898034..., the result so far being 1000.10000... Highest power of φ ≤ 5 − 3φ ≈ 0.145898034... is φ = 5 − 3φ ≈ 0.145898034... Subtracting this from 5 − 3φ ≈ 0.145898034..., we have 5 − 3φ − (5 − 3φ) = 0 + 0φ = 0, with the final result being 1000.1001. Just as with any base-n system, numbers with a terminating representation have an alternative recurring representation. In base-10, this relies on the observation that 0.999...=1. In base-φ, the numeral 0.1010101... can be seen to be equal to 1 in"
"10,000 10,000 (ten thousand) is the natural number following 9,999 and preceding 10,001. Many languages have a specific word for this number: in Ancient Greek it is (the etymological root of the word myriad in English), in Aramaic , in Hebrew [], in Chinese (Mandarin , Cantonese , Hokkien ""bān""), in Japanese [], in Khmer [], in Korean [], in Vietnamese , in Thai [], and in Malayalam []. It is often used to mean an indefinite very large number. The Greek root was used in early versions of the metric system in the form of the decimal prefix myria-."
"last significant figure in the ""tenths"" place. The first term has its last significant figure in the tenths place and the second term has its last significant figure in the thousandths place. The leftmost of the decimal places of the last significant figure out of all the terms of the sum is the tenths place from the first term, so the calculated result should also have its last significant figure in the tenths place. The rules for calculating significant figures for multiplication and division are opposite to the rules for addition and subtraction. For multiplication and division, only the total"
"10=(,) and results from either crossing of the thumbs, or holding both hands up in a cross. Another possibility is that each represents a finger and represents the thumb of one hand. This way the numbers between 1–10 can be counted on one hand using the order: =P, =PR, =PRM, =IT, =T, =TP, =TPR, =TPRM, =IN, =N (P=pinky, R=ring, M=middle, I=index, T=thumb N=no fingers/other hand). This pattern can also be continued using the other hand with the fingers representing and the thumb . A third hypothesis about the origins states that the basic ciphers were , , and (or )"
"The number 10000 can also be written 10,000 (UK and US), 10.000 (Europe mainland), 10 000 (transition metric), or 10•000 (with the dot raised to the middle of the zeroes; metric). 10,000 10,000 (ten thousand) is the natural number following 9,999 and preceding 10,001. Many languages have a specific word for this number: in Ancient Greek it is (the etymological root of the word myriad in English), in Aramaic , in Hebrew [], in Chinese (Mandarin , Cantonese , Hokkien ""bān""), in Japanese [], in Khmer [], in Korean [], in Vietnamese , in Thai [], and in Malayalam []."
"by another numeral: ""tám mươi"" (""8"" x ""10"" = ""80""). Consonantal and tonal alternations occur in some compound numerals. The numeral ""mười"" ""10"" in multiplicative compounds has a tonal change (""huyền"" tone > ""ngang"" tone) to ""-mươi"" ""times 10"", as in: The numeral ""một"" ""1"" undergoes a tonal alternation (""nặng"" tone > ""sắc"" tone) to ""mốt"" when it occurs after ""mươi"" (with ""ngang"" tone) in multiples of 10, as in: The numeral ""năm"" ""5"" undergoes an initial consonant alternation (""n"" > ""l"") to ""lăm"" as the final element in additive compounds, as in: The Vietnamese word for 100 is ""trăm""."
"1 or 4, and progresses in the sequence 1, 4, 1. The last two digits of a star number in base 10 are always 01, 13, 21, 33, 37, 41, 53, 61, 73, 81, or 93. Unique among the star numbers is 35113, since its prime factors (i.e. 13, 37 and 73) are also consecutive star numbers. Geometrically, the ""n""th star number is made up of a central point and 12 copies of the (""n""−1)th triangular number — making it numerically equal to the ""n""th centered dodecagonal number, but differently arranged. Infinitely many star numbers are also triangular numbers, the"
"9 9 (nine) is the natural number following and preceding . 9 is a composite number, its proper divisors being and . It is 3 times 3 and hence the third square number. Nine is a Motzkin number. It is the first composite lucky number, along with the first composite odd number and only single-digit composite odd number. 9 is the only positive perfect power that is one more than another positive perfect power, by Mihăilescu's Theorem. 9 is the highest single-digit number in the decimal system. It is the second non-unitary square prime of the form (""p"") and the"
"at most formula_8, or formula_9. For formula_10 and above, so any number over 1000 gets smaller under this process and in particular becomes a number with strictly fewer digits. Once we are under 1000, the number for which the sum of squares of digits is largest is 999, and the result is 3 times 81, that is, 243. Considering more precisely the intervals [244, 999], [164, 243], [108, 163] and [100, 107], we see that every number above 99 gets strictly smaller under this process. Thus, no matter what number we start with, we eventually drop below 100. An exhaustive"
"2, 13, 24, −15, 26, 37, 8, 9} where each integer is in a different congruence class modulo 10. The unique least residue system modulo 10 is {0, 1, 2, ..., 9}. A reduced residue system modulo 10 could be {1, 3, 7, 9}. The product of any two congruence classes represented by these numbers is again one of these four congruence classes. This implies that these four congruence classes form a group, in this case the cyclic group of order four, having either 3 or 7 as a (multiplicative) generator. The represented congruence classes form the group of units"
"""F"" must be coprime to 10. Most often it is convenient to choose the smallest that fits the above. The following summarizes some of the results obtained in this manner, where the white spaces between the digits divide the digits into 10-digit groups: In duodecimal system, the transposable integers are: (using inverted two and three for ten and eleven, respectively) Note that the “Shifting left cyclically by single position” problem has no solution for the multiplier less than 12 except 2 and 5, the same problem in decimal system has no solution for the multiplier less than 10 except 3."
"10 = 0 (where the 2 is the tens digit of 24 from the sum for the final digit). This set of numbers ending ...404 may therefore contain one or more missing-digit sums. Similar reasoning can be applied to sums in which two, three and more digits are deleted from the original number. 14358846, 14400000, 15000000, 28758846, 28800000, 29358846, 29400000, 1107488889, 1107489042, 1111088889, 1111089042, 3277800000, 3281400000, 4388888889, 4388889042, 4392488889, 4392489042, 4500000000, 5607488889, 5607489042, 5611088889, 5611089042, 7777800000, 7781400000, 8888888889, 8888889042, 8892488889, 8892489042, 10000000000, 20000000000, 30000000000, 40000000000, 50000000000, 60000000000, 70000000000, 80000000000, 90000000000 Missing-digit sum Missing-digit sums are integer numbers that are equal"
"number of significant figures in each of the factors matter; the decimal place of the last significant figure in each factor is irrelevant. For addition and subtraction, only the decimal place of the last significant figure in each of the terms matters; the total number of significant figures in each term is irrelevant. In a base 10 logarithm of a normalized number, the result should be rounded to the number of significant figures in the normalized number. For example, log(3.000×10) = log(10) + log(3.000) ≈ 4 + 0.47712125472, should be rounded to 4.4771. When taking antilogarithms, the resulting number should"
"formed using 2, 3, and 4 (respectively), linked to the word for 20 by ""m"" (a vestigial multiplicative): Any other number between 21 and 99 is formed using 20, 40, 60, or 80, dropping the final ""i,"" then adding ""da"" (= and) followed by the appropriate number from 1 to 19; e.g.: The hundreds are formed by linking 2, 3, . . ., 10 directly to the word for 100 (without the multiplicative ""m"" used for 40, 60, and 80). 1000 is expressed as ""atasi"" (10 x 100), and multiples of 1000 are expressed using ""atasi"" — so, for example,"