Exercise Section 2.1 & 2.2.docx

Exercise Section2.1 18. Let a,b ∈ ℝ, and suppose that for every ℰ > 0 we have a ≤ b + ℰ. Show that a ≤ b. Answer : Suppose that b < a then we have that 0 < a –b and then taking r = a≤b+ℰ=b+

𝑎−𝑏 2

. Since b +

𝑎−𝑏 2

=

𝑎+𝑏 2

we have that a ≤

𝑎+𝑏 2

𝑎−𝑏 2

we have that

which gives us that 2a ≤ a + b and

then therefore a ≤ b. Thus, we have that b < a and a ≤ b which contradicts the Trichotomy Property. Hence, a ≤ b. Exercise Section 2.2 9. Find all values of x that satisfy the following inequalities. Sketch graphs. Answer : Lets first consider the left- hand side of the inequality |𝑥 − 2| = {

𝑥 − 2 𝑖𝑓 𝑥 ≥ 2 −𝑥 + 2 𝑖𝑓 𝑥 < 2

Therefore, we have two cases : Case 1 : x ≥ 2 Then |𝑥 − 2| = 𝑥 − 2 so we have 𝑥 − 2 ≤ 𝑥 + 1 −2 ≥ 1 That is correct for all x ∈ ℝ. Since this case is for all x ≥ 2 the final solution for this case is the intersection of thee solutions, that is x ∈ [2, + ∞ ⟩. Case 2 : x < 2 Then |𝑥 − 2|= -x +2 so we have -x +2 ≤ x +1 -2x ≤ -1 1

x≥ 2 1

From there, x ∈ [2 , + ∞⟩. Since this case is for all x< 2 the final solution for this case is for all 1

x < 2 the final solution for this case intersection of these solutions, that is x ∈ [2 , 2⟩. This solution to the given equation is the union of the solutions of both case, that is, the final 1

solution is x ∈ [2 , + ∞⟩.