is my proof correct?
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
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.
No, your proof is not correct. Let's go step by step to identify the mistake and correct it.
We start with the assumption that x + 2y ≤ 4 is true, and we want to prove that if x > 2, then y < 1.
1. Start by assuming x > 2 and prove by contradiction.
2. Assume y ≥ 1 (instead of 1 ≤ y, as stated in your proof).
Now let's continue with the proof:
Assume x + 2y ≤ 4 is true, and x > 2.
1. Start by negating the conclusion: Assume y ≥ 1.
2. Substitute this assumption into the original inequality:
x + 2(1) ≤ 4
x + 2 ≤ 4
x ≤ 2
3. Notice that our assumption y ≥ 1 led to the conclusion x ≤ 2, which contradicts our initial assumption x > 2.
Therefore, our assumption that y ≥ 1 is incorrect, and y must be less than 1 when x > 2.
Hence, if x > 2, then y < 1.
To determine if your proof is correct, let's break it down step by step:
1. You correctly stated the given inequality: x + 2y ≤ 4.
2. Your goal is to show that if x > 2, then y < 1.
3. You started your proof by assuming the given inequality is true and using proof by contradiction to prove the desired result.
4. You assumed that y ≥ 1 and proceeded to substitute y = 1 into the inequality.
5. By substituting y = 1, you obtained x + 2 ≤ 4, which simplifies to x ≤ 2.
6. At this point, you correctly identified that this result contradicts the assumption x > 2.
Based on these steps, your proof is correct. However, to make it clearer, we can rephrase the proof as follows:
Proof:
We assume that x + 2y ≤ 4 and x > 2.
Suppose, for the sake of contradiction, that y ≥ 1. Then, substituting y = 1 into the inequality, we have:
x + 2(1) ≤ 4
x + 2 ≤ 4
x ≤ 2
However, this contradicts our assumption that x > 2. Therefore, the assumption y ≥ 1 must be false, which means y < 1.
Thus, we have shown that if x > 2, then y < 1.