Home › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Math Circle › Proof of equivalence

Tagged: Congruence is an equivalence

- This topic has 13 replies, 9 voices, and was last updated 1 year, 8 months ago by Writaban Sarkar.

- AuthorPosts
- November 18, 2018 at 9:14 pm #24370Ramanjaneyulu KakumaniMember
2) Congruance is Equivalance Numbers

- for Any numbers N,M — |N-N| is divisible by M –> Reflexive ( N~ N)
- if two numbers N, P — |N-P| is divisible by M, then |P-N| also divisible by M –> Symmetric (N~P)
- If three NumberN,P,Q — |N-P| is divisible by M and |P-Q| is divisible by M then|N-Q| is also divisible by M — Transitive ( N ~ P ~ Q )

Hence, Congruance of Numbers is Equivalance

November 18, 2018 at 9:16 pm #24372Gouri BasakParticipantTo prove – congruence of numbers is an equivalence relation

Let us take a line segment AB of length n and a scale of 3 m to measure it

In this way we can easily prove it when it is reflexive,symmetric and transitive.

i) It is reflexive if we take the point A itself,it measures 0 m and 0 is multiple of every number

ii) It is symmetric because if we take the length between point A and B and if it is divisible by 3, then it is obviously true that the length between point B and A is also divisible by 3

iii) It is also transitive because if we make an extra point C outside AB and the length AB is divisible by 3 and the length BC is divisible by 3 then it is also true that AC is divisible by 3

Thus it proves that congruency of numbers is an equivalence relation

November 18, 2018 at 9:16 pm #24373haimanti royParticipantCongruence of glued points:

Proof: A must be glued to A since it is itself.(reflexive)

If A is glued to B, then B is glued to A(symmetric)

If A is glued to B, and B is glued to C, then A must be glued to C.(transitive)

Thus the statement is proved

*faciliment.*🙂November 18, 2018 at 9:17 pm #24374Writaban SarkarParticipantLet us have a scale with length m.We see that if we have a line AB, such that m divides the length AB, then AB is congruent to itself, since the length is 0. We also see that AB~BA since they are of same length. Now if we have two lines AB and BC, such that m divides the length AB and CD, then we have that the length of AC is divisible by m, since they are AC=AB+BD. So, congruence is an equivalence. Hence, proven.

- This reply was modified 1 year, 8 months ago by Writaban Sarkar.

- AuthorPosts

- You must be logged in to reply to this topic.