Writaban Sarkar
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Yes, since making equivalence relations means that the elements are in the same set, we can say that stating elements equivalent means to break the set and form some new set (partition).
Let 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.
1. We first check if gluing points is reflexive. If we have a point A , then A glued to A itself, since it is the point itself. Now we check if gluing points is symetric. If we have A is glued to B then we should have that B is glued to A. Now atlast we check if it is transitive. If A is glued to B and B is glued to C, then we have A is glued to C, since both are glued to B. So we find that gluing points is an example of equivalence, since it is true for all 3 cases.
2. We see that if we have a triangle ABC then it ABC is congruent, since it approves SSS Congruence. Now if we have 2 triangles ABC and DEF such that ABC is congruent is DEF then we also see that DEF is congruent is ABC. Now if we have three triangles ABC, DEF and IJK such that ABC is congruent DEF and DEF is congruent IJK, then we have ABC is congruent IJK, since they are both congruent to DEF. So we find that congruence is an example of equivalence, since it is true for all 3 cases.
Therefore, we see that both gluing and congruence are examples of equivalence, since they are true for all 3 cases.
Let the number \(n=\) \(d_k \cdot 10^k+d_{k-1}\cdot 10^{k-1}+ . . .+d_1\cdot 10^1+d_0\cdot 10^0 \) . If we divide \(n\) by 3 , we get \(d_k+d_{k-1}+ . . .+d_1+d_0\) as remainder. Therefore , for \(n\) to be divisble by 3 the sum of it digits i.e. \(d_k+d_{k-1}+ . . .+d_1+d_0\) must divisible by 3.
Hence proven.
