Let a, b ∈ R. Suppose that for every ε > 0, we have a < b + 3ε. Prove that a ≤ b
Question
Let a, b ∈ R. Suppose that for every ε > 0, we have a < b + 3ε. Prove that a ≤ b
Solution
Step 1: Assume for contradiction that a > b.
Step 2: Then, a - b > 0. Let's denote a - b by δ. So, δ > 0.
Step 3: Now, choose ε such that ε = δ/3. Note that ε > 0 because δ > 0.
Step 4: According to the given condition, we have a < b + 3ε.
Step 5: Substituting ε = δ/3 into the inequality, we get a < b + δ, which simplifies to a < a.
Step 6: This is a contradiction because a cannot be less than itself.
Step 7: Therefore, our assumption that a > b must be false.
Step 8: So, we conclude that a ≤ b.
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