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| − | '''Bezout's Lemma''' states that if <math>x</math> and <math>y</math> are nonzero [[Integer|integers]] and <math>g = \gcd(x,y)</math>, then there exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. In other words, there exists a linear combination of <math>x</math> and <math>y</math> equal to <math>g</math>.
| + | #REDIRECT[[Bézout's Identity]] |
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| − | Furthermore, <math>g</math> is the smallest positive integer that can be expressed in this form, i.e. <math>g = \min\{x\alpha+y\beta|\alpha,\beta\in\mathbb Z, x\alpha+y\beta > 0\}</math>.
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| − | In particular, if <math>x</math> and <math>y</math> are [[relatively prime]] then there are integers <math>\alpha</math> and <math>\beta</math> for which <math>x\alpha+y\beta=1</math>.
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| − | ==Proof==
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| − | Let <math>x = gx_1</math>, <math>y = gy_1</math>, and notice that <math>\gcd(x_1,y_1) = 1</math>.
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| − | Since <math>\gcd(x_1,y_1)=1</math>, <math>\text{lcm}(x_1,y_1)=x_1y_1</math>. So <math>\alpha=y_1</math> is smallest positive <math>\alpha</math> for which <math>x_1\alpha\equiv 0\pmod{y}</math>. Now if for all integers <math>0\le a,b<y_1</math>, we have that <math>x_1a\not\equiv x_1b\pmod{y_1}</math>, then one of those <math>y_1</math> integers must be 1 from the [[Pigeonhole Principle]]. Assume for contradiction that <math>x_1a\equiv x_1b\pmod{y_1}</math>, and WLOG let <math>b>a</math>. Then, <math>x_1(b-a)\equiv 0\pmod {y_1}</math>, and so as we saw above this means <math>b-a\ge y_1</math> but this is impossible since <math>0\le a,b<y_1</math>. Thus there exists an <math>\alpha</math> such that <math>x_1\alpha\equiv 1\pmod{y_1}</math>.
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| − | Therefore <math>y_1|(x_1\alpha-1)</math>, and so there exists an integer <math>\beta</math> such that <math>x_1\alpha - 1 = y_1\beta</math>, and so <math>x_1\alpha + y_1\beta = 1</math>. Now multiplying through by <math>g</math> gives, <math>gx_1\alpha + gy_1\beta = g</math>, or <math>x\alpha+y\beta = g</math>.
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| − | Thus there does exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>.
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| − | Now to prove <math>g</math> is minimum, consider any positive integer <math>g' = x\alpha'+y\beta'</math>. As <math>g|x,y</math> we get <math>g|x\alpha'+y\beta' = g'</math>, and as <math>g</math> and <math>g'</math> are both positive integers this gives <math>g\le g'</math>. So <math>g</math> is indeed the minimum.
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| − | ==Generalization/Extension of Bezout's Lemma==
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| − | Let <math>a_1, a_2,..., a_m</math> be non-zero integers. Then there exists integers <math>x_1, x_2, ..., x_m</math> such that
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| − | <cmath>\sum_{i=1}^{m} a_ix_i = \gcd(a_1, a_2, ..., a_m)</cmath> Also, <math>\gcd(a_1, a_2, ..., a_m)</math> is the least positive integer satisfying this property.
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| − | ===Proof===
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| − | Consider the set <math>P = \{n \in \mathbb{Z}^{+}|n= \sum_{i=1}^{m} a_iu_i: u_1, \dots, u_m \in \mathbb{Z}\}</math>. Obviously, <math>P \neq \emptyset</math>. Thus, because all the elements of <math>P</math> are positive, by the [[Well-Ordering Principle|Well Ordering Principle]], there exists a minimal element <math>d \in P</math>. So
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| − | <cmath>d=a_1x_1 +a_2x_2 + \dots +a_mx_m</cmath>
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| − | if <math>n >d</math> and <math>n \in S</math> then
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| − | <cmath>n=a_1u_1 +a_2u_2 + \dots +a_mu_m</cmath>
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| − | But by the [[Division Algorithm|Division Algorithm]]:
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| − | <cmath>n=qd +r \Longrightarrow r=n-qd</cmath>
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| − | <cmath>= \sum_{i=1}^m a_i(u_i-qx_i)</cmath>
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| − | <cmath> \Longrightarrow r\in P</cmath>
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| − | But <math>0 \le r<d</math> so this would imply that <math>r \in P</math> which contradicts the assumption that <math>d</math> is the minimal element in <math>P</math>. Thus, <math>r=0</math> hence, <math>d|n</math>. But this would imply that <math>d|a_i</math> for <math>i \in \{1, 2,\dots,m\}</math> because <math>a_i = a_i \cdot1 + \sum_{k=1; k \neq i}^{m}(a_k\cdot0) \Longrightarrow \{a_1, a_2, \dots, a_m \} \subset P</math>.
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| − | Now, because <math>d|a_i</math> for <math>i \in \{1, 2,\dots,m\}</math> we have that <math>d|\gcd(a_1, a_2,\dots, a_m)|a_i</math>. But then we also have that <math>\gcd(a_1, a_2,\dots, a_m)|\sum_{i=1}^m a_iu_i =d</math>. Thus, we have that <math>\boxed{d=\gcd(a_1, a_2,\dots, a_m)}</math> <math>\Box</math>
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| − | ~qwertysri987
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| − | ==Generalization to Principal Ideal Domains==
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| − | Bezout's Lemma can be generalized to [[Principal ideal domain|principal ideal domains]].
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| − | Let <math>R</math> be a principal ideal domain, and consider any <math>x,y\in R</math>. Let <math>g = \gcd(x,y)</math>. Then there exist elements <math>r_1,r_2\in R</math> for which <math>xr_1+yr_2 = g</math>. Furthermore, <math>g</math> is the minimal such element (under divisibility), i.e. if <math>g' = xr_1'+yr_2'</math> then <math>g|g'</math>.
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| − | Note that this statement is indeed a generalization of the previous statement, as the [[ring]] of integers, <math>\mathbb Z</math> is a principal ideal domain.
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| − | ===Proof===
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| − | Consider the [[ideal]] <math>I = (x,y) = \{xr_1+yr_2|r_1,r_2\in R\}</math>. As <math>R</math> is a principal ideal domain, <math>I</math> must be principle, that is it must be generated by a single element, say <math>I = (g)</math>. Now from the definition of <math>I</math>, there must exist <math>r_1,r_2\in R</math> such that <math>g = xr_1+yr_2</math>. We now claim that <math>g = \gcd(x,y)</math>.
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| − | First we prove the following simple fact: if <math>z\in I</math>, then <math>g|z</math>. To see this, note that if <math>z\in I = (g)</math>, then there must be some <math>r\in R</math> such that <math>z = rg</math>. But now by definition we have <math>g|z</math>.
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| − | Now from this, as <math>x,y\in I</math>, we get that <math>g|x,y</math>. Furthermore, consider any <math>s\in R</math> with <math>s|x,y</math>. We clearly have that <math>s|xr_1+yr_2 = g</math>. So indeed <math>g = \gcd(x,y)</math>.
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| − | Now we shall prove minimality. Let <math>g' = xr_1'+yr_2'</math>. Then as <math>g|x,y</math>, we have <math>g|xr_1'+yr_2' = g'</math>, as desired.
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| − | ==See also==
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| − | [[Category:Number theory]]
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| − | [[Category:Abstract algebra]]
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| − | [[Category:Ring theory]]
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| − | {{stub}}
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| − | [[Category:Theorems]]
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