Let T be the set of all triangles in a plane with R a relation in T given by R={(T1, T2):T1 is congruent to T2}. Show that R is an equivalence relation.
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Step-by-Step Solution
Step 1: Reflexivity
For a relation to be reflexive, every element must be related to itself. In this case, we need to show that any triangle T1 is congruent to itself. This is true by definition, as a triangle is always congruent to itself.
Step 2: Symmetry
For a relation to be symmetric, if T1 is related to T2, then T2 must be related to T1. If triangle T1 is congruent to triangle T2, it logically follows that triangle T2 is also congruent to triangle T1. Congruence is a symmetric property.
Step 3: Transitivity
For a relation to be transitive, if T1 is related to T2 and T2 is related to T3, then T1 must be related to T3. If triangle T1 is congruent to triangle T2, and triangle T2 is congruent to triangle T3, then triangle T1 must be congruent to triangle T3. This is a fundamental property of congruence.
Step 4: Conclusion
Since the relation R satisfies all three properties: reflexivity, symmetry, and transitivity, it is an equivalence relation.