is my proof correct?

Question

Suppose that x and y are real numbers and x + 2y ≤ 4. Show
that if x > 2 then y < 1

suppose x + 2y ≤ 4 is true, we prove by contradiction and assume 1≤y

x + 2y ≤ 4
x + 2(1) ≤ 4
x + 2 ≤ 4
x ≤ 2
Therefore, we obtain a contradiction when we assume 1≤y and obtain x ≤ 2 which contradicts x>2

Answer 1

Your proof is not correct. Here's a corrected proof:

Suppose x and y are real numbers and x + 2y ≤ 4. We want to show that if x > 2, then y < 1.

Assume x > 2.

If y ≥ 1, then x + 2y ≥ 2 + 2(1) = 4, which contradicts the given condition x + 2y ≤ 4. Therefore, it must be the case that y < 1.