Art of Problem Solving

Cauchy-davenport: Difference between revisions

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The Cauchy-Davenport theorem states that for all nonempty sets <math>A,B \subseteq \mathbb{Z}/p\mathbb{Z}</math> , we have that  
The Cauchy-Davenport Theorem states that for all nonempty sets <math>A,B \subseteq \mathbb{Z}/p\mathbb{Z}</math> , we have that  
<cmath>|A+B| \geq \min\{|A|+|B|-1,p\},</cmath>
<cmath>|A+B| \geq \min\{|A|+|B|-1,p\},</cmath>
where <math>A+B</math> is defined as the set of all <math>c \in \mathbb{Z}/p\mathbb{Z}</math> that can be expressed as <math>a+b</math> for <math>a \in A</math> and <math>b \in B</math>.
where <math>A+B</math> is defined as the set of all <math>c \in \mathbb{Z}/p\mathbb{Z}</math> that can be expressed as <math>a+b</math> for <math>a \in A</math> and <math>b \in B</math>.
== Proof of the Cauchy-Davenport Theorem by the Combinatorial Nullstellensatz ==
== Proof by Induction ==
== Applications of the Cauchy-Davenport Theorem ==

Latest revision as of 12:39, 20 September 2019

The Cauchy-Davenport Theorem states that for all nonempty sets $A,B \subseteq \mathbb{Z}/p\mathbb{Z}$ , we have that \[|A+B| \geq \min\{|A|+|B|-1,p\},\] where $A+B$ is defined as the set of all $c \in \mathbb{Z}/p\mathbb{Z}$ that can be expressed as $a+b$ for $a \in A$ and $b \in B$.

Proof of the Cauchy-Davenport Theorem by the Combinatorial Nullstellensatz

Proof by Induction

Applications of the Cauchy-Davenport Theorem