Course: 7.7 上岗和因素 | The Way To Learn

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上岗和因素

is common in mathematics. In this lesson we will use it to prove different kinds of hypotheses and to practice the use of the process in different situations. Specifically, we will be applying the principles of induction to prove various statements regarding factors and factorability.
::在数学中很常见。在这个教训中,我们将用它来证明不同种类的假设,并实践在不同情况下使用这个过程。具体地说,我们将运用诱导原则来证明关于各种因素和因素的各种说明。

Induction and Factors
::上岗和因素

There are two properties of integers and their factors that will be useful for the proofs in this lesson.
::整数及其因数有两个属性,对本课中的证明有用。

Property 1 : If a is a factor of b , and a is a factor of c , then a is a factor of the sum
::财产1:如果a为b的系数,而a为c的系数,则a为总和的系数。

b + c .
::b + c。

For example, the set of numbers a = 3, b = 6, and c = 9 satisfies the hypothesis because 3 is a factor of 6 and 3 is a factor of 9. Property 1 states that therefore 3 is a factor of 9 + 6 = 15, which we know is true because 3 × 5 = 15. We can prove that this property is true for all integers if we think about what the term factor means. If a is a factor of b , then there exists some integer M such that aM = b . Similarly if a is a factor of c , then there exists some integer N such that aN = c . So we can write the sum b + c as aM + aN . We know
::例如,数字a = 3, b = 6, c = 9 的一组数a = 3, c = 9 满足了假设,因为 3 是一个因数 6, 3 和 3 是一个因数 9, 财产1 表示, 因此, 3 + 6 = 15, 我们知道这是事实, 因为 3 × 5 = 15 , 我们可以证明, 对于所有整数来说, 这个属性都是真实的, 如果我们想一想该词的含义, 我们可以证明这个属性是真实的。如果是一个因数 b, 那么就存在某种整数M, 这样, aM = b。同样, 如果是一个因数, 那么也存在某种整数N, a N = c。所以我们可以把数 b + c = a M + a N。我们知道, 和 b + aN。

			b + c = aM + aN = a ( M + N ).

Because we can write the sum as a product of a and another number, a is a factor of the sum b + c .
::因为我们可以以一个和另一个数字的产物来写总和, a 是和 b + c 的一个系数。

Property 2 : If a is a factor of b and b is a factor of c , then a is a factor of c .
::财产2:如果b和b的系数是c的系数,则c的系数是c。

We can prove this property in a similar manner. If a is a factor of b , then there exists some integer M such that aM = b . If b is a factor of c , then there exists some integer N such that bN = c . We can write c in the following manner:
::我们可以以类似的方式证明这种属性。如果是 b 的系数,那么就存在某种整数Mm, 即 am = b。如果b 是 c 的系数,那么就存在某种整数N,即 bN = c。我们可以以下列方式写 c :

c = bN = ( aM ) N = a ( MN )
::c = bN = (aM)N = a(MN)

Therefore a is a factor of c because we can write c as the product of a and another integer, MN .
::因此, a 是一个 c 的因数,因为我们可以将 c 写成一个和另一个整数 MN 的产物。

As noted above, these properties of integers are useful for proving statements about integers and factors via induction.
::如上所述,整数的这些特性有助于通过上岗证明关于整数和因数的报表。

Examples
::实例

Example 1
::例1

Prove that 3 is a factor of 4 ⁿ - 1 for all positive integers n .
::证明对于所有正整数 n 而言,3是4n-1的因数。

Proof by induction:
::上岗培训证明:

1. The base case: if n = 1, then 4 ⁿ - 1 = 4 - 1 = 3. 3 is a factor of itself because 3 × 1 = 3.
::1. 基本情况:如果 n = 1,那么如果 4n - 1 = 4 - 1 = 3. 3 本身就是一个因素,因为 3 x 1 = 3。

If the base case does not convince you, you can always test out additional values of n . For example, if n = 2, 4 ² - 1 = 16 - 1 = 15 = 5 × 3.
::如果基数无法说服您, 您总是可以测试 n 的额外值。例如, 如果 n= 2, 42 - 1 = 16 - 1 = 15 = 5 × 3, 则您总是可以测试 n 的额外值。

2. The inductive hypothesis: assume that 3 is a factor of 4 ^k - 1.
::2. 感想假设:假设3是4k-1系数。

3. The inductive step: show that 3 is a factor of 4 ^k+1 - 1.
::3. 感应步骤:显示3是4k+1-1-1的因数。

If 3 is a factor of 4 ^k - 1, then there exists some integer M such that 3 M = 4 ^k - 1. We can write 4 ^k+1 - 1 in a manner that allows us to use the inductive hypothesis:
::如果3是4k-1系数,那么就存在某种整数Mm,3M=4k-1,我们可以写4k+1-1,这样我们就可以使用感应假设:

	4 ^k+1 - 1
	4(4 ^k -1) + 3	Factor out 4, but add 3 in order to keep the value of 4 ^k+1 -1
	4(3 M ) + 3	Substitute: 3 M = 4 ^k - 1

Note that the substitution is the same as using property 2 above: if 3 is a factor of 4 ^k - 1 , then 3 is a factor of 4(4 ^k - 1). Using the substitution simply makes the fact a bit more obvious. This last step proves that 3 is a factor of 4 ^k+1 - 1 , by property 1 above.
::请注意,替换与使用上述财产2相同:如果3是4k-1的因数,那么3是4(4k-1)的因数,则3是4(4k-1)的因数。使用替代只是使事实更加明显。最后一步证明,按以上财产1计算,3是4k+1-1的因数。

The technique of rewriting the k + 1 term can also be used to prove statements about polynomials and factors.
::重写 k + 1 术语的技术也可以用来证明关于多语种和因素的陈述。

Example 2
::例2

Prove that x - y is a factor of x ⁿ - y ⁿ for all positive integers n .
::证明 x - y 是所有正整数 n 的 xn - yn 系数。

Note: Since we are talking about polynomials that are factorable now, not integers, then we say that if x - y is a factor of x ⁿ - y ⁿ , then there exists a polynomial P such that P ( x - y ) = x ⁿ - y ⁿ .
::注意: 既然我们谈论的是现在可以考虑的多数值,而不是整数, 那么我们说,如果 x - y 是 xn - yn 的系数, 那么就有一个多数值P, 即 P( x - y) = xn - yn 。

Proof by induction:
::上岗培训证明:

1. The base case: If n = 1, we have x ⁿ - y ⁿ = x - y , and x - y is a factor of itself, as x - y = 1( x - y ).
::1. 基本情况:如果 n = 1,我们有 xn - y = x - y 和 x - y 是其本身的一个系数,如 x - y = 1,x - y 。

As we did above, we can also check n = 2 in order to convince ourselves. If n = 2, we have x ² - y ² = ( x - y )( x + y ), so x - y is clearly a factor.
::正如我们在上文中所做的那样,我们也可以检查n=2,以便说服自己。如果 n=2,我们有x2 - y2 = (x - y(x - y)(x) + y),那么x - y显然是一个因素。

2. The inductive hypothesis: assume that x - y is a factor of x ^k - y ^k .
::2. 感想假设:假设x - y是xk - yk的一个系数。

3. Show that x - y is a factor of x ^k+1 - y ^k+1 .
::3. 显示 x - y 是 xk+1 - yk+1 的系数。

From the inductive step, we know that there is some polynomial P such that P ( x - y ) = x ^k - y ^k . We can rewrite x ^k+1 - y ^k+1 in a manner that allows us to use the inductive hypothesis:
::从感应步骤中,我们知道有一些多式P,例如P(x-y) = xk - yk。我们可以重写 xk+1 - yk+1, 从而使我们能够使用感想假设:

	x ^k+1 - y ^k+1
	= x ^k+1 - xy ^k + xy ^k - y ^k+1
	= x ( x ^k - y ^k ) + y ^k ( x - y )
	= x ( P ( x - y )) + y ^k ( x - y )
	= Px ( x - y ) + y' ^k-1 ( x - y )

Again, by property 1 above, this shows that x - y is a factor of x ^k+1 - y ^k+1 . Therefore we have shown that x - y is a factor of x ⁿ - y ⁿ for all positive integers n .
::以上属性1再次显示,x-yis是xk+1-yk+1的系数。因此,我们已经显示,x-yis是所有正整数的xn-yn系数。

Example 3
::例3

Without adding, determine if 7 a factor of 49 + 70.
::不增加,确定7个因数是否为49+70。

Use Property 1 from the lesson: If a is a factor of b, and a is a factor of c, then a is a factor of the sum b + c.
::使用从教训中得出的财产1:如果a为b系数,而a为c系数,则a为和 b + c的系数。

7 is a factor of 49, since $\begin{align*}7 \times 7 = 49\end{align*}$
::7 是49的因数, 因为 7x7=49

7 is a factor of 70, since $\begin{align*}7 \times 10 = 70\end{align*}$
::7 是70的系数, 因为 7x10=70

Therefore 7 is a factor of 119, since $\begin{align*}49 + 70 = 119\end{align*}$
::因此,自49+70=119以来,7是119系数。

Consider the sum 23 + 54 = 77. Is 7 a factor of 77? What does this tell you about the first factor property in the lesson?
::将23+54=77乘以7是77吗?这告诉您关于课中第一个要素财产的内容是什么?

This is a test of the converse of Property 1, which would be "If a number is a factor of the sum, then it is a factor of the factors of the sum"
::这是对财产1的对等性的一种检验,即“如果一个数字是总和的一个系数,那么它是总和的一个系数”。

7 is a factor of the sum: 77
::7是总和的一个系数:77

7 is not a factor of 23 or 54
::7 不是一个系数23或54

This tells us that the converse of the property is not necessarily true.
::这告诉我们,财产的反比并不一定是真实的。

Example 4
::例4

Prove that x ⁿ - 1 is divisible by x - 1 for all positive integers n .
::证明 xn - 1 对所有正数整数 n 的 x - 1 可除以 x - 1 。

1. Base case:	If n = 1, x ^k - 1 = x - 1 = (x - 1)(1)
2. Inductive hypothesis:	Assume that x ^k - 1 is divisible by x - 1
3. Inductive step:	Show that x ^k+1 - 1 is divisible by x - 1.

x ^k - 1 divisible by x - 1 $\begin{align*}\Rightarrow\end{align*}$ P ( x - 1) = ( x ^k - 1) for some polynomial P x ^{k + 1} - 1 = x ( x ^k - 1) + ( x - 1) = Px ( x - 1) + ( x - 1), which is divisible by x - 1
::x- 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 x (x - 1), + (x - 1) , 1 , 1 , 1 , 1 , 1 4 , 1 , 1 4 , 1 4 , 6 , 1 4 , 1 4 , 1 4 , 1 6 , 1 4 , 1 6 , 6 4 , 6 4 , 1 6 , 6 6 , 6 6 , 6 6 6 , 6 6 , 6 6 , 6 6 , 6 6 , 6 6 6 6 , 6- 1 6 6 6 6 6 , 6 6 6 6 6 , 6 6 , 6 6- 6 6- 6 6 6 , 6 6 6 6 6 , 6 6 6 6 6 6 6 6 6 6 , 6xxxxxx- 6- 6- 6xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 1 =xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 1 = 1 = 1 = 1 = 1 = 1 =xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Example 5
::例5

Prove that n ² - n is even for all positive integers n .
::证明 n2 - n 对所有正整数 n 均相等。

1. Base case:	If n = 1, 1 ² - 1 = 1 - 1 = 0 = 2 × 0
2. Inductive hypothesis:	Assume that k ² - k is even
3. Inductive step:	Show that ( k + 1) ² - ( k + 1) is even.

If k ² - k is even, then k ² - k = 2 M for some integer M ( k + 1) ² - ( k + 1) = k ² + 2 k + 1 - k - 1 = k ² - k + 2 k = 2 M + 2 k = 2( M + k ) which is even because M + K is an integer.
::偶数的 Iftk2-kis, 然后k2-k= 2M(k+1) 2M(k+1) =k2+2k+1 -k- 1=k2-k+2k=2M+2k=2(M+k), 这甚至是因为 M+Kis 是一个整数。

Example 6
::例6

Prove that 5 ^2n-1 + 1 is divisible by 6 for all positive integers n .
::证明所有正整数 n 的52n-1+1除以 6。

1. Base case:	If n = 1, 5 ¹ + 1 = 5 + 1 = 6 = 6(1)
2. Inductive hypothesis:	Assume that 5 ^2k-1 + 1 is divisible by 6.
3. Inductive step:	Show that 5 ^{2(k + 1) - 1} + 1 is divisible by 6.

If 5 ^{2k - 1} + 1 is divisible by 6, then 5 ^{2k - 1} + 1 = 6M for some integer M . 5 ^{2(k + 1) - 1} + 1 = 5 ^{2k + 1} + 1 = 5 ² (5 ^{2k - 1} + 1) - 24 = 5 ² (6M) - 24 which is divisible by 6.
::如果52k-1+1除以6,则52k-1+1=6M,对于某些整数M. 52(k+1) - 1+1=52k+1+1=52(52k-1+1) - 24=52(6M) - 24可除以6。

Review
::回顾

Without adding, determine if 7 is a factor of 49 + 70.
::不增加,确定7是49+70的因数。

Consider the sum $\begin{align*}23 + 54 = 77\end{align*}$ Is 7 a factor of 77? What does this tell you about the first factor property in the lesson?
::考虑23+54=77的总额7是77的系数吗?这告诉您关于课中的第一个要素属性是什么?

Prove that any positive integer n > 1 a) is prime or, b) can be represented as a product of prime factors.
::证明任何正整数n > 1 a)为正数或b)可作为主要因素的产物表示。

Found within set "J" are all positive integers, from the number 1 to 2n. Prove that there are two numbers, one that is a factor of another, from any (n = 1) numbers chosen from set "J"
::在设置“ J” 中发现的所有正整数都是正数, 从数字 1 到 2n 。证明从设定“ J” 中选择的任何数字( n= 1) 中有两个数字, 一个是另一个数字的一个系数。

Prove that $\begin{align*}(x^n + \frac{1}{x^n})\end{align*}$ is also an integer for any positive integer n if the following is an integer: $\begin{align*}(x + \frac{1}{x})\end{align*}$
::证明 (xn+1xn) 也是任何正整 n 的整数,如果以下为整数x+1x)

Prove the formula $\begin{align*}n_{k + m} = n_{k-1} u_m + n_k n_{m+1}\end{align*}$ for the sequence of Fibonacci numbers: $\begin{align*}n_1 = 1, n_2 = 1, u_{k+1}= n_k + n_{k-1}, k = 2, 3...\end{align*}$
::验证公式 nk+m=nk- 1um+nknm+1 用于 Fibonacci 编号序列: n1=1,n2=1,uk+1=nk+nk+nk- 1,k=2,3...

Prove the following identities.
::证明以下身份。

$\begin{align*}1^2 + 2^2 + 3^2 + ... +n^2 =\frac{ n(n + 1)(2n + 1)}{6}\end{align*}$
::12+22+32+...+n2=n(n+1)(2n+1)6

$\begin{align*}1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n + 1)^2}{4}\end{align*}$
::13+23+33+...+n3=n2(n+1)24

$\begin{align*}1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n(n + 1)(n + 2) = \end{align*}$ $\begin{align*}\frac{n(n + 1)(n + 2) (n + 3)}{4}\end{align*}$
::+...+n(n+1)(n+2)=n(n+1)(n+2(n+3)4

$\begin{align*}1 x \cdot 1! + 2 x \cdot 2! + ... + n \cdot n! = (n + 1)! - 1\end{align*}$
::1x% 1! +2x% 2! +...+nnn!=(n+1)!

$\begin{align*}n^2 \frac{(n + 1)^2}{4} = 1^3 + 2^3 + 3^3 + ... + n^3\end{align*}$
::n2(n+1)24=13+23+33+...+n3

$\begin{align*}n (n + 1) \frac{(2n + 1)}{6} = 1^2 + 2^2 + 3^2 + ... + n^2\end{align*}$
::n(n+1)(2n+1)6=12+22+32+...+n2

Prove the following divisibilities.
::证明以下可能性。

Prove that $\begin{align*}n \frac{(n+1)}{2}\end{align*}$ is a factor of $\begin{align*}1 + 2 + 3 +... + n\end{align*}$ for all positive integers $\begin{align*}n\end{align*}$
::证明 n(n+1) 2 是所有正整数 n 的 1+2+3+...+n 系数

Prove that 3 is a factor of $\begin{align*}n^3 + 2n\end{align*}$ for all positive integers n .
::证明对于所有正整数 n 而言, 3 是 n3+2n 的因数。

Prove that $\begin{align*}n^2 \frac{(n + 1)^2}{4}\end{align*}$ is a factor of $\begin{align*}1^3 + 2^3 + 3^3 + ... + n^3\end{align*}$
::证明 n2(n+1) 24 是 13+23+33+...+n3 的乘数

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

上岗和因素

Induction and Factors ::上岗和因素

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Review ::回顾

Review (Answers) ::回顾(答复)