Problem based on mapping of matrix

Question

Question 1: Define a relation ∼ on R² by stating that (a, b) ∼ (c, d) if and only if a² + b² ≤ c² + d². Show that ∼ is reflexive and transitive but not symmetric.

- Reflexive

Let (a, b) ∈ R².

Then, a² + b² ≤ a² + b²

That is (a, b) ∼ (a, b)

Thus, ∼ is reflexive.

- Transitive

Let (a, b), (c, d), (e, f) ∈ R²

Suppose (a, b) ∼ (c, d) and (c, d) ∼ (e, f).

Then, a² + b² ≤ c² + d² and c² + d² ≤ e² + f²

That is a² + b² ≤ c² + d² ≤ e² + f²

In particular, a² + b² ≤ e² + f²

Then, (a, b) ∼ (e, f).

Thus, ∼ is transitive.

- Not Symmetric

Counterexample: Pick (1,1), (2,2) ∈ R²

(1,1) ∼ (2,2) is true since 1² + 1² ≤ 2² + 2²

but (2,2) ∼ (1,1) is false since 2² + 2² ? 1² + 1²

Question 2: Show that an m x n matrix give rise to a well-defined map from Rⁿ to R^m

We can define a mapping: Rⁿ → R^m

Question 3: Find the error in the following argument by providing a counterexample. "The reflexive property is redundant to the axioms for an equivalence relation. If x ∼ y, then y ∼ x by the symmetric property. Using the transitive property, we can deduce that x ∼ x."

Let R be the relation on the set A = {a, b, c} defined by R = {(a, a), (b, b), (a, b), (b, a)}

This is not an equivalence relation because it is not reflexive ((c, c) ∉ R) even though R is symmetric and transitive.

Since c is not related to some element say x ∈ A satisfying (c, x) ∈ R. Then, we are not allowed to use the argument cRx, xRc ⇒ cRc; since x does not exist.

Let σ = (1 3 5 7) and τ = (2 7 3) be elements of S₇

a) Find all elements of the subgroups <σ> and <τ>.

b) What order does σ have? What order does τ have? Can you generalize this to a statement for any cycle? σ is order 4 while τ because orders of permutations are determined by least common multiple of the lengths of the cycles.

c) Compute the product στ. What is its order?

στ = (1 2 7)(3 5), order 6.

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