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If $b \ge 0$, this is trivial, as $b$ itself will be in the set (consider $k=0$). Although this result doesn't seem too profound, it is nonetheless quite handy. In Definition 1.3.10 we had defined multiplication as repeated addition. First, we need to show that $q$ and $r$ exist. }\), Using Algorithm 3.2.2 or Algorithm 3.2.10, we can compute the quotient and remainder of the division of any integer \(a\) by any natural number \(b\text{. See more ideas about math division, math classroom, teaching math. We rst prove this result under the additional assumption that b>0 is a natural number. }\) Thus we have: In Example 3.2.5 we have seen that when dividing \(a=30\) by \(b=8\) the quotient is \(q=3\) and the remainder is \(r=6\text{. \newcommand{\vect}[1]{\overrightarrow{#1}} We illustrate the process of dividing a negative number by dividing \(-33\) by \(9\text{. Let S= fa xbjx2Z;a xb 0g: If we put x= j ajthen a xb= a+ jajb jaj+ a jajj aj = 0: 3.2.2. If $a$ and $b$ are integers, with $a \gt 0$, there exist unique integers $q$ and $r$ such that $$b = qa + r \quad \quad 0 \le r \lt a$$ The integers $q$ and $r$ are called the. -15 + 9 7\amp = -6\\ Jul 26, 2018 - Explore Brenda Bishop's board "division algorithm" on Pinterest. }\), We typically use the variable \(q\) for the quotient and the variable \(r\) for the remainder. It remains to show that these values for $q$ and $r$ are unique. \(a \fdiv b\) and \(a \fmod b\) from \(a=q\cdot b+r\). (See Section 3.5, page 143.) There are many different algorithms that could be implemented, and we will focus on division by repeated subtraction. \newcommand{\Tm}{\mathtt{m}} 3. Dividend = Quotient × Divisor + Remainder In Example 3.2.3 we have seen that when dividing \(a=4\) by \(b=7\) the quotient is \(0\) and the remainder is \(4\text{. Having demonstrated that the set $S$ is a non-empty set of non-negative values, the well-ordering principle applies and guarantees that $S$ has a least element. \end{equation*}, \begin{equation*} (For some of the following, it is sufficient to choose a ring of constants; but in order for the Division Algorithm for Polynomials to hold, we need to be able to divide constants.) The algorithm by which \(q\) and \(r\) are found is just long division. }\) So we need a different algorithm for the case \(a\lt 0\text{. }\) If we try to use Algorithm 3.2.2 when \(a\) is negative, the algorithm always returns \(0,a\) which does not satisfy the condition \(0\le r\) for the output since \(r=a\lt 0\text{. \newcommand{\To}{\mathtt{o}} \newcommand{\Th}{\mathtt{h}} The division algorithm is by far the most complicated of all the written algorithms taught in primary/elementary school. Then, we need to show that $q$ and $r$ are unique. \), \begin{equation*} \end{equation*}, \begin{equation*} For example, if we divide 26 by 3, then we get a quotient of 8 and remainder or … Apply the division algorithm x= yq+ r, 0 ≤ r r_1$, it must be the case that $0 \le r_2 - r_1 \lt a$. Solution : As we have seen in problem 1, if we divide 400 by 8 using long division, we get. Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License 30=(3 \cdot 8) + 6 The theorem is frequently referred to as the division algorithm (although it is a theorem and not an algorithm), because its proof as given below lends itself to a simple division algorithm for computing q and r (see the section Proof for more). }\) Thus \(4=0\cdot 7+4\text{. \newcommand{\F}{\mathbb{F}} Here, we follow the tradition and call it the division algorithm. Without loss of generality, let us suppose that $r_2 \ge r_1$. }\) After \(s\) additions of \(b\) to \(a\) we have, If we let \(q:=-s\text{,}\) we get \(r=a-(q\cdot b)\) (compare this to what we wanted). \newcommand{\Tl}{\mathtt{l}} r=a\underbrace{+b + b+\ldots+b}_{\mbox{\(s\) times} }=a+(b\cdot s)\text{.} Divisor = 8. }\) Thus we have: For the given values of \(a\) and \(b\text{,}\) determine the quotient \(q\) and remainder \(r\) of the division of \(a\) by \(b\text{,}\) and write the equality \(a=(q \cdot b) + r\text{.}\). Follow the instructions to find quotient and remainder. There are many different algorithms that could be implemented, and we will focus on division by repeated subtraction. Set r = a – qb. The Division Algorithm. The division algorithm, therefore, is more or less an approach that guarantees that the long division process is actually foolproof. }\) Thus \(-20=(-3)\cdot 7+1\text{. }\) We repeatedly add \(b\) to negative numbers until \(0\le r\lt b\) is true. The algorithm by which \(q\) and \(r\) are found is just long division. If \(a\lt b\) then we cannot subtract \(b\) from \(a\) and end up with a number greater than or equal to \(b\text{. 1.4. When we speak of the quotient and the remainder when we “divide an integer \(a\) by the positive integer \(b\),” we will always mean the quotient \(q\) and the remainder \(r\) guaranteed by the Division Algorithm. We will see that in fact there is sometimes a choice of remainders. \newcommand{\Tv}{\mathtt{v}} As \(r=-20\) the statement \(r\ge 0\) is false. Dividing \(30\) by \(8\) with Algorithm 3.2.2 shorter. With this in mind, suppose $b = 27$ and $a = 5$, and let us consider the set of all integers that results from either adding to 27, or subtracting from 27, some multiple of 5: $$\{ \ldots, -8, -3, 2, 7, 12, 17, 22, 27, 32, 37, \ldots \}$$ Now let us reduce this set down to just the positive values present, to produce a set $$S = \{2, 7, 12, 22, 27, 32, 37, \ldots\}$$ We can see that the smallest element of this set (i.e., $2$) is the remainder we seek -- but in the more general case, how can we be guaranteed that this smallest element exists? \end{equation*}, \begin{equation*} Problem 3 : Divide 400 by 8, list out dividend, divisor, quotient, remainder and write division algorithm. A negative integer \(a\) and a natural number \(b\), We find the output values of the Division Algorithm (Algorithm 3.2.10) for the input values \(a=-20\) and \(b=7\text{.}\). \newcommand{\lcm}{\mathrm{lcm}} \newcommand{\degre}{^\circ} q_1 a - q_2 a &=& r_2 - r_1 \quad \textrm{and thus,}\\ 1.5 The Division Algorithm We begin this section with a statement of the Division Algorithm, which you saw at the end of the Prelab section of this chapter: Theorem 1.2 (Division Algorithm) Let a be an integer and b be a positive integer. }\) We write: In Example 3.2.5 we have seen that when dividing \(a=30\) by \(b=8\) the quotient is \(q=3\) and the remainder is \(r=6\text{. As \(r=14\) and \(q=8\) the statement \(r \lt q\) is false. For instance, it is used in proving the Fundamental Theorem of Arithmetic, and will also appear in the next chapter. As \(a=4\) and \(b=7\) statement \(a \lt b\) is true. It is not actually an algorithm, but this is this theorem’s traditional name. The division algorithm for polynomials has several important consequences. Then there are unique polynomials q(x) and r(x) 2F[x] such that f(x) = q(x)g(x) + r(x) and either r(x) = 0 or the degree of r(x) is less than the degree mof g(x). Hint: First show that the algorithm terminates. Answer: First we show that the algorithm terminates. Prove the “uniqueness” part of the Division Algorithm. The statement of the division algorithm as given in the theorem describes very explicitly and formally what long division is. If $b$ is negative, the remainder is found in a slightly different way -- by adding as many $a$'s as necessary until the result becomes non-negative. -20=(7 \cdot (-3)) + 1 Since a negative number plus \(b\) is always less than \(b\) and we check the value of \(r\) after every addition, it is sufficient to check whether \(0\le r\text{.}\). The division algorithm is often employed to verify the correctness of a division problem. The division algorithm is an algorithm in which given 2 integers N N N and D D D, it computes their quotient Q Q Q and remainder R R R, where 0 ≤ R < ∣ D ∣ 0 \leq R < |D| 0 ≤ R < ∣ D ∣. \newcommand{\gt}{>} \newcommand{\todo}[1]{{\color{purple}TO DO: #1}} The following result is known as The Division Algorithm:1 If a,b ∈ Z, b > 0, then there exist unique q,r ∈ Z such that a = qb+r, 0 ≤ r < b.Here q is called quotient of the integer division of a by b, and r is called remainder. \newcommand{\gro}[1]{{\color{gray}#1}} }\), \(\newcommand{\longdivision}[2]{#1\big)\!\!\overline{\;#2}} The key part here is that you can use the fact that naturals are well ordered by looking at the degree of your remainder. The division algorithm is an algorithm in which given 2 integers N N N and D D D, it computes their quotient Q Q Q and remainder R R R, where 0 ≤ R < ∣ D ∣ 0 \leq R < |D| 0 ≤ R < ∣ D ∣. Since its proof is very similar to the corresponding proof for integers, it is worthwhile to review Theorem 2.9 at this point. Then they use this in the proof of the division algorithm by constructing non-negative integers and applying WOP to this construction. Proof of Division Algorithm Date: 11/13/97 at 10:02:51 From: James Lester Subject: Proof of division algorithm We are looking for a proof for the division algorithm: a,b are positive integers, b does not equal 0; there are unique integers q and r such that a = qb + r; 0 is less than or equal to r, r is less than modulus value of b. r_2 - r_1 &=& a(q_1 - q_2) We now formalize this procedure in an algorithm. The process of division often relies on the long division method. Then they use this in the proof of the division algorithm by constructing non-negative integers and applying WOP to this construction. By its construction, S is only allowed to contain non-negative integers, so it remains to show that it is non-empty. Some mathematicians prefer to call it the division theorem. We can use the division algorithm to prove The Euclidean algorithm. \newcommand{\Sni}{\Tj} Let S= fa xbjx2Z;a xb 0g: If we put x= j ajthen a xb= a+ jajb jaj+ a jajj aj = 0: \newcommand{\fmod}{\bmod} Its handiness draws from the fact that it not only makes the process of division easier, but also in its use in finding the proof … \newcommand{\RR}{\R} }\), When \(a\lt 0\text{,}\) we still want find \(q\) and \(r\) such that \(a=(q\cdot b)+r\) with \(0\le r\lt b\text{. \newcommand{\Tr}{\mathtt{r}} \newcommand{\N}{\mathbb{N}} \newcommand{\lt}{<} The division algorithm for polynomials has several important consequences. a natural number \(a\) and a natural number \(b\), Two integers \(q\) and \(r\) such that \(a=(q\cdot b)+r\) and \(0 \leq r\lt b\). \end{equation*}, \begin{equation*} To show that $q$ and $r$ exist, let us play around with a specific example first to get an idea of what might be involved, and then attempt to argue the general case. ˛ ˚ !$ 1" Title: 3613-l07.dvi Author: binegar Created Date: 9/9/2005 8:51:21 AM \newcommand{\abs}[1]{|#1|} In Checkpoint 3.2.13 we unroll the loop in Algorithm 3.2.2. Division is not defined in the case where b … We find the output values of the Algorithm 3.2.2 for the input values \(a=30\) and \(b=8\text{.}\). So we continue with step 4. The importance of the division algorithm is demonstrated through examples. \newcommand{\fdiv}{\,\mathrm{div}\,} Effectively, we need to show that if $b = q_1 a + r_1$ and $b = q_2 a + r_2$, then it can only be the case that $q_1 = q_2$ and $r_1 = r_2$. The Euclidean algorithm can be proven to work in vast generality. We find the output values of Algorithm 3.2.2 for the input values \(a=30\) and \(b=8\text{.}\). Suppose f = a 0 + a 1x+ + a nxn g = b 0 + b 1x+ + b mxm are polynomials of degree n and m, respectively. Putting the left side of this last inequality back into its original form, we end up with $b - ka \ge 0$. \newcommand{\C}{\mathbb{C}} Feb 14, 2020 - Explore Heather Kraus's board "division algorithm" on Pinterest. Let f(x) = a nxn+ a n 1xn 1 + + a 1x+ a 0 = X a ix i g(x) = b mx m+ b m 1x 1 + + b 1x+ b 0 = X b ix i be two polynomials over a eld F of degrees nand m>0. We already know $r$ is non-negative as it was in $S$, so all we need to do now is assure ourselves that it is less than $a$. Jul 16, 2019 #2 H. HallsofIvy Well-known member. Hence, the HCF of 675 and 81 is 27. Recall the well-ordering principle tells us that any non-empty set of non-negative integers contains a least element. -33 + 9 \amp = -24\\ THE EUCLIDEAN ALGORITHM 53 3.2. Dividing \(4\) by \(7\) with Algorithm 3.2.2. The number qis called the quotientand ris called the remainder. \newcommand{\Tq}{\mathtt{q}} \renewcommand{\emptyset}{\{\}} a \fmod 42 = 7. \newcommand{\PP}{\mathbb{P}} \newcommand{\fixme}[1]{{\color{red}FIX ME: #1}} In our first version of the division algorithm we start with a non-negative integer \(a\) and keep subtracting a natural number \(b\) until we end up with a number that is less than \(b\) and greater than or equal to \(0\text{. \newcommand{\Ty}{\mathtt{y}} \newcommand{\Tk}{\mathtt{k}} Perhaps this could prove helpful... Now, with a hint of what tool might prove useful, let us turn our attention to the general case, for some arbitrary $a$ and $b$, with $a > 0$. The construction of \(q\) and \(r\) in Algorithm 3.2.2 and Algorithm 3.2.10 yields a proof of the following theorem. \newcommand{\sol}[1]{{\color{blue}\textit{#1}}} Then, there exist unique integers \(q\) and \(r\) with \(0 \leq r \lt b\) such that. So we continue with step 2. \(-20\fdiv 7\) and \(-20\fmod 7\) with Algorithm 3.2.10. It involves processes of division with remainders, multiplication, subtraction and regrouping, making lots of potential chances to make a mistake. -6 + 9 \amp = 3 Find \(a\) from \(a \fdiv b\) and \(a \fmod b\). Interesting applications of this principle, as well as the pigeonhole principle are discussed. \newcommand{\tox}[1]{\texttt{\##1} \amp \cox{#1}} 5. The Division Algorithm for Polynomials Handout Monday March 5, 2012 Let F be a field (such as R, Q, C, or Fp for some prime p). Quotient = 3x 2 + 4x + 5 Remainder = 0. There is a lot more left to learn in real numbers. This will allow us to divide by any nonzero scalar. 1. Proof. So regardless of the value of $b$, $S$ will have at least one element, and is thus non-empty. 5. With the division algorithm we find a quotient and remainder. We rst prove this result under the additional assumption that b>0 is a natural number. This may appear to be confusing at rst, but it is literally just saying that you can divide an in- teger a by a nonzero integer b and get a remainder which is less than the number we are dividing by. \newcommand{\blanksp}{\underline{\hspace{.25in}}} \newcommand{\Tc}{\mathtt{c}} Then there exist unique integers q and r such that. In Z[i] we measure "size" by the norm. As \(r=6\) and \(q=8\) the statement \(r\lt q\) is true. We have. Remark. To find the very first term of the quotient, divide the first term of the dividend by the highest degree term in the divisor. Establish the uniqueness of \(q\) and \(r\). Proof. In this example we go through the repeat_until loop several times. Since a is an integer, it must lie in some interval [qb,(q+1)b). \newcommand{\Tz}{\mathtt{z}} If \(a>0\text{,}\) then Algorithm 3.2.2 returns the quotient and remainder of the division of \(a\) by \(b\text{. [DivisionAlgorithm] Suppose a>0 and bare integers. r := x \fdiv 42 = 7 \end{equation*}, \begin{equation*} \newcommand{\gexp}[3]{#1^{#2 #3}} }\) We have added \(9\) four times, so the quotient is \(-4\text{. 4. Sometimes one is not interested in both the quotient and the remainder. The division algorithm is a theorem in mathematics which precisely expresses the outcome of the usual process of division of integers.Its name is a partial misnomer: it is not a true algorithm (a well-defined procedure for achieving a specific task), but it can be used to find the greatest common divisor of two integers.. Theorem 17.6. Proof of Division Algorithm. Find quotient and remainder with division algorithm. \newcommand{\Ta}{\mathtt{a}} For \(a=7\) and \(b=3\text{,}\) we have \(q=2\) and \(r=1\text{,}\) and write \(7=(3\cdot 2)+1\text{. As \(a=30\) and \(b=8\) the statement \(a \lt b\) is false. \newcommand{\R}{\mathbb{R}} In Checkpoint 3.2.7 we unroll the loop in Algorithm 3.2.2 in a similar way. We will use the well-ordering principle to obtain the quotient qand remainder r. Since we can take q= aif d= 1, we shall assume that d>1. \end{equation*}, \begin{equation*} The only multiple of $a$ in this range is zero, so $r_2 - r_1 = 0$, or equivalently, $r_1 = r_2$. \end{equation*}, MAT 112 Ancient and Contemporary Mathematics. \newcommand{\Tx}{\mathtt{x}} In Example 3.2.3 we have seen that when dividing \(a=4\) by \(b=7\) the quotient is \(q=0\) and the remainder is \(r=4\text{. We aim to show that $b - ka \ge 0$ for this value of $k$ and must then necessarily be in the set S. To this end, note $$b - ka = b - ba = b (1-a)$$ and the factor $(1-a)$ must be either zero or negative as $a \gt 0$. In grade school you Theorem 16.1 (Division Algorithm). Jan 29, 2012 1,151. It's rather obvious isn't it? }\) For \(a=0\) and any natural number \(b\) we have \(a=(q\cdot b)+r\) and \(0\le r\lt b\) when \(q=0\) and \(r=0\text{.}\). \newcommand{\Tg}{\mathtt{g}} Further, as seen before, $$q_2 = \frac{b - r_2}{a} \quad \textrm{and} \quad q_1 = \frac{b - r_1}{a}$$ Since $r_1$ and $r_2$ agree in value, this forces $q_1$ and $q_2$ to agree in value as well. But consider the alternative: if $r \ge a$, then $b - k_r a \ge a$, which implies (after subtracting $a$ from both sides) that $$b - (k_r + 1)a \ge 0$$ This contradicts the fact that $r$ was the smallest element of the set $S$. A similar theorem exists for polynomials. In many books on number theory they define the well ordering principle (WOP) as: Every non- empty subset of positive integers has a least element. In Example 3.2.6 you can observe how the values of the variables change as you click your way through the steps of the division algorithm. As \(0\le 3\lt 9\) we are done. Note that one can write r 1 in terms of a and b. Let Sbe the set of all natural numbers of the form a kd, where kis an integer. Proof. \newcommand{\Te}{\mathtt{e}} \newcommand{\W}{\mathbb{W}} That is, prove that the integers qand rare unique, which means that if (q1,r1) satisfies b= q1a+r1, 0 ≤r10, then there exist unique q,r ∈Z such thata=qb+r, 0≤ r < b. Hereqis calledquotientof theinteger divisionofabyb, andris calledremainder. The division algorithm for Z[i] If u, v ∈ Z[i] with v ≠ 0 then ∃ q, r ∈ Z[i] such that u = vq + r with N(r) < N(v). 4. \newcommand{\Tu}{\mathtt{u}} \newcommand{\xx}{\mathtt{\#}} \end{equation*}, \begin{equation*} Step 2:In case of division we seek to find the quotient. Division Algorithm. \newcommand{\Ts}{\mathtt{s}} }\) We write: \(-20 \fdiv 7 = -3\) and \(-20\fmod 7=1\), In the Checkpoint 3.2.19 write the quotient and remainder with the new operations \(\fdiv\) and \(\fmod\text{. \newcommand{\Tn}{\mathtt{n}} a=(q\cdot 42)+ r. a = (10\cdot 42)+7=427. Is it possible to apply the WOP to a subset of non-negative integers? The next theorem shows a connection between the division algorithm and congruences. Multiplication Algorithm & Division Algorithm The multiplier and multiplicand bits are loaded into two registers Q and M. A third register A is initially set to zero. (PDF) A polynomial-based division algorithm - In addition, through the well-ordering principle, the chapter illustrates with an additional proof technique, the principle of mathematical induction. }\) That number is the remainder. Prove that gcd(a,b) = r n, the last nonzero remainder. }\), Replacing \(q\) with \(10\) and \(r\) with \(7\) we get, In Checkpoint 3.2.21 use conduct the computation from the example above to find the integer \(a\) when given \(a \fdiv b\) and \(a \fmod b\text{. Division is not defined in the case where b = 0; see division … Consequently $b(1-a) \ge 0$. As \(r=1\) the statement \(r\ge 0\) is false. \newcommand{\ttx}[1]{\texttt{\##1}} Thus for all integers \(a\) and all natural numbers \(b\) we can find integers \(q\) and \(r\) such that \(a=(q\cdot b)+r\) and \(0\le r\lt b\text{. \newcommand{\amp}{&} Proof of the Divison Algorithm The Division Algorithm If $a$ and $b$ are integers, with $a \gt 0$, there exist unique integers $q$ and $r$ such that $$b = qa + r \quad \quad 0 \le r \lt a$$ The integers $q$ and $r$ are called the quotient and remainder , respectively, of the division of $b$ by $a$. }\) We write: In Example 3.2.5 we have seen that when dividing \(a=-20\) by \(b=7\) the quotient is \(-3\) and the remainder is \(1\text{. Let \(a\) be an integer and \(b\) be a natural number, and let \(q\) and \(r\) be the unique integers such that \(0 \leq r \lt b\) and \(a=(q\cdot b)+r\text{. a=(q\cdot b)+ r\text{.} a \fdiv 42 = 10 \newcommand{\A}{\mathbb{A}} A similar theorem exists for polynomials. }\), We call \(q\) the quotient of the division of \(a\) by \(b\) and denote it by \(a\fdiv b\text{. Given two numbers, for instance, 1052 }\), For \(a=20\) and \(b=4\text{,}\) we have \(q=5\) and \(r=0\text{,}\) and write \(20=(4\cdot 5)+0\text{. \newcommand{\Z}{\mathbb{Z}} Note that r is an integer with 0 ≤ r < b and a = qb + r as required. The simplest division algorithm, historically incorporated into a greatest common divisor algorithm presented in Euclid's Elements, Book VII, Proposition 1, finds the remainder given two positive integers using only subtractions and comparisons: Then use mathematical induction and Question 2. Theorem 0.1 Division Algorithm Let a and b be integers with b > 0. }\) This is the same as repeatedly adding \(-b\text{. \newcommand{\nr}[1]{\##1} The Division Algorithm by Matt Farmer and Stephen Steward Subsection 3.2.1 Division Algorithm for positive integers In our first version of the division algorithm we start with a non-negative integer \(a\) and keep subtracting a natural number \(b\) until we end up with a number that is less than \(b\) and greater than or equal to \(0\text{. It is indeed the case that given integers $a$ and $b$, with $a > 0$, there exist unique integers $q$ and $r$ such that $b = qa + r$ and $0 \le r \lt a$. \newcommand{\id}{\mathrm{id}} }\) We get a positive remainder when \(a\) is negative by repeated addition of \(b\text{. If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = q(x) × g(x) + r(x) where r(x) = 0 or degree of r(x) < degree of g(x). Since its proof is very similar to the corresponding proof for integers, it is worthwhile to review Theorem 2.9 at this point. The multiplicative groups \((\Z_p^\otimes,\otimes)\). We return the quotient \(q=-3\) and the remainder \(r=1\), Thus the quotient or the division of \(-20\) by \(7\) is \(-3\) and the remainder is \(1\text{.}\). \newcommand{\So}{\Tf} So it must be the case that $r \lt a$. a = bq + r and 0 r < b. Quotient = 50. Remainder = 0 \newcommand{\Sno}{\Tg} So we continue with step 5. We return the quotient \(q=3\) and the remainder \(r=6\), Thus the quotient of the division of \(30\) by \(8\) is \(3\) and the remainder is \(6\text{.}\). \newcommand{\Tw}{\mathtt{w}} Preview Activity \(\PageIndex{1}\): The GCD and the Division Algorithm. See more ideas about Math division, Fourth grade math, 4th grade math. Then there is a unique pair of integers qand rsuch that b= aq+r where 0 ≤r

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