Euclidean algorithm
Supplementary material for thelecture of Monday, July
The Euclidean algorithm
The Euclidean algorithm is a way to find the greatest common
divisor of two positive integers, a and b. First let me show the
computations for a=210 and b=45.
Divide 210 by 45, and get the result 4 with remainder 30, so
210=4&45+30.
Divide 45 by 30, and get the result 1 with remainder 15, so
45=1&30+15.
Divide 30 by 15, and get the result 2 with remainder 0, so
30=2&15+0.
The greatest common divisor of 210 and 45 is 15.
Let's see. Several questions occur immediately..
What's going on here?
Why does the algorithm stop?
Why is the answer correct?
O.k. I will answer first question by giving a formal description of
the algorithm, which supposedly finds the greatest common divisor
(GCD) of two integers. This "greatest common divisor" must exist,
since positive integer divisors of integers can't be any larger than
the integers:
Formal description of the Euclidean algorithm
Two positive integers, a and b.
The greatest common divisor, g, of a and b.
Internal computation
If a&b, exchange a and b.
Divide a by b and get the remainder, r. If r=0, report b as the
GCD of a and b.
Replace a by b and replace b by r. Return to the previous step.
which describes the algorithm. Here's another link,
with more information. If you want to "do" some further examples, Paul
Garrett of the University of Minnesota has a Visible Euclidean
algorithm page where you can specify further examples and see the
algorithm work..
Why does the algorithm stop?
At each step, the remainder, r, decreases by at least 1. Therefore r
must eventually be 0. A formal proof would use mathematical
induction.
Why is the answer actually equal to the GCD?
In the final stage, when r=0, we see that the "final b" must divide
the final a. Go backwards along the string of equations
(a=q&b+r), and you will see that at each step, the "final b"
divides each of the parts of the equation. Therefore the "final b"
must be a divisor of both of the initial a and initial b.
Now look at the first equation. In the equation a=q&b+r,
the GCD divides both a and b and therefore must divide r as well
(because a-q&r is a multiple of the GCD). Now carry this
observation through all of the equations, forward. The GCD must divide
two of the three pieces in all of the equations, and thus must divide
the third. Therefore, the GCD also divides the "final b". So the
"final b" divides both a and b, and is itself a multiple of the
GCD. Well, the GCD is the greatest such divisor, and therefore
the "final b" must be equal to the GCD of the initial values.
The extended Euclidean algorithm
Here's a true statement:
If a and b are positive integers, then there are always
integers m and n so that the GCD of a and b equals m&a+n&b.
extended Euclidean algorithm (described, for example, here,
allows the computation of multiplicative inverses mod P. First let's see
an example. Since the GCD of 210 and 45 is 15, we should be able to
write 15 as a sum of multiples of 210 and 45. Here's how
to do it. We look carefully at the steps above and change them each.
Divide 210 by 45, and get the result 4 with remainder 30, so
210=4&45+30.
30=1&210-4&45.
Divide 45 by 30, and get the result 1 with remainder 15, so
45=1&30+15.
15=45-1&30=45-1&(1&210-4&45)=-1&210+5&45
The greatest common divisor of 210 and 45 is 15, and we have written
15 as a sum of integer multiples of 210 and 45.
The extended Euclidean algorithm has a very important use: finding
multiplicative inverses mod P. Choose a prime, P: how about
97. I know 97 is prime, because 2 and 3 and 5 and 7 and even 11 aren't
factors of 97, and I only need to check division by primes up to the
square root of 97.
Now let me take a fairly random integer, say 20. Since 20 is less than
97, and 97 is prime, the GCD of 20 and 97 should be 1. (Remember,
since 97 is prime, its only divisors are itself and 1.) I will verify
this by "running" the Euclidean algorithm:
97=4&20+17
The extended Euclidean algorithm allows us to write 1 as a sum of 97
and 20. Here we go:
17=1&97-4&20
20-1&17=3 so
3=1&20-1&17=1&20-(1&97-4&20)=-1&97+5&20
17=5&3+2 so
2=17-5&3=(1&97-4&20)-5(-1&97+5&20)=6&97-29&20
1=3-2=(-1&97+5&20)-(6&97-29&20)=-7&97+34&20
The final equation tells us that
1=-7&97+34&20, which means that the
product of 34 and 20 is equal to 1 plus a multiple of 97. But in mod
97, we ignore multiples of 97. Therefore 34
is the multiplicative inverse of 20 mod 97.
Exercise Find the multiplicative inverse of 60 mod 97 by
hand. As I mentioned in class, doing just one of these
computations "by hand" is good enough.
The extended Euclidean algorithm is easy to implement on a computer
and the amount of memory needed is not large.
The algorithm runs
very fast. For example, I am currently running a copy of
Maple in another window on a PC which is neither very fast
nor has much memory. I just selected a 100 decimal digit prime, P, and
found the multiplicative inverse of a 50 decimal digit number mod
P. The system returned 0.00 as the amount of time used. That is, the
computation used less than .01 seconds!
Maintained by
and last modified 7/13/2004.sat2数学_百度知道
关于高次多项式Rational zero (root ) theorem 是什么意思呀 还有 if p+qi is a zero ,then p-qi is aslo a zero是为什么呀谢谢啦
提问者采纳
则P-Qi自然为0了. a，则Q=0，b可以使式子，则a=0或者b=0。（2）P+Qi=0,P=0（1）就是ab=0，也可以是数
提问者评价
其他类似问题
为您推荐：
sat2的相关知识
其他2条回答
Rational root theorem
From Wikipedia, the free encyclopedia
Jump to: navigation, search
In algebra, the rational root theorem (or rational root test) states a constraint on rational solutions (or roots) of the polynomial equation
a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\!
with integer coefficients.
If a0 and an are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1) satisfies
* p is an integer factor of the constant term a0, and
* q is an integer factor of the leading coefficient an.
Thus, a list of possible rational roots of the eq...
参考资料：
晕 楼上的英语应该给人家翻译一下啊 呵呵 楼上可能一时没注意看吧
这是带有实部虚部的表达式表达式 p+qi
p代表实数 而qi代表复数
若 p+qi=0 ，即实部虚部都等于零
即 p q 都等于零 所以 p-qi is aslo a zero
等待您来回答
下载知道APP
随时随地咨询
出门在外也不愁If x and y are integers whose least common multiple is a perfect square and whose greatest common divisor is a prime number,which could represent the ordered pair (x,y)?The answer is (18,4) ; but I just dont know why (9,1) couldn't be the right answe_百度作业帮
If x and y are integers whose least common multiple is a perfect square and whose greatest common divisor is a prime number,which could represent the ordered pair (x,y)?The answer is (18,4) ; but I just dont know why (9,1) couldn't be the right answe
If x and y are integers whose least common multiple is a perfect square and whose greatest common divisor is a prime number,which could represent the ordered pair (x,y)?The answer is (18,4) ; but I just dont know why (9,1) couldn't be the right answer,I figure it is right,too..Thanks
Because that 1 is not a prime number.
because of 1 is not a prime numbergmat数学DS部分 prep上的一道题,gmat数学DS部分 prep上的一道题,也看不到答案.177.If x and y are positive integers,what is the value of xy (1) The greatest common factor of x and y is 10.(2) The least common multiple of x and y is _百度作业帮
gmat数学DS部分 prep上的一道题,gmat数学DS部分 prep上的一道题,也看不到答案.177.If x and y are positive integers,what is the value of xy (1) The greatest common factor of x and y is 10.(2) The least common multiple of x and y is
gmat数学DS部分 prep上的一道题,gmat数学DS部分 prep上的一道题,也看不到答案.177.If x and y are positive integers,what is the value of xy (1) The greatest common factor of x and y is 10.(2) The least common multiple of x and y is 180.答案是1）+2）后,充分
1800,最大公约数是10.最小公倍数是180,故X和Y一个是10,一个是180.故XY是1800.仅供参考}