suppose x,y e R. Prove ff: if x+y≥2, then x≥1 or y≥1

We are given the implication to prove: 

**Statement**: If \( x + y \geq 2 \), then \( x \geq 1 \) or \( y \geq 1 \), where \( x, y \in \mathbb{R} \).

This is a **disjunction** proof (an "or" statement), so we need to show that whenever \( x + y \geq 2 \), at least one of the conditions \( x \geq 1 \) or \( y \geq 1 \) must hold.

### Proof by Contrapositive:

We will prove this by proving the **contrapositive** of the statement. The contrapositive of the statement is logically equivalent to the original statement. The contrapositive of "If \( x + y \geq 2 \), then \( x \geq 1 \) or \( y \geq 1 \)" is:

**Contrapositive Statement**: If \( x < 1 \) and \( y < 1 \), then \( x + y < 2 \).

If we can prove this contrapositive, the original statement will also be proven.

#### Step 1: Assume \( x < 1 \) and \( y < 1 \).

This means both \( x \) and \( y \) are less than 1.

#### Step 2: Add these inequalities together.

\[
x + y < 1 + 1 = 2
\]

Therefore, \( x + y < 2 \).

#### Conclusion:
We have shown that if \( x < 1 \) and \( y < 1 \), then \( x + y < 2 \). This proves the contrapositive of the original statement. Hence, the original statement "If \( x + y \geq 2 \), then \( x \geq 1 \) or \( y \geq 1 \)" is true.

### Summary:
We have proved that if \( x + y \geq 2 \), then at least one of \( x \geq 1 \) or \( y \geq 1 \) must hold, by showing that if neither condition holds (i.e., if both \( x < 1 \) and \( y < 1 \)), then \( x + y < 2 \), which contradicts the hypothesis \( x + y \geq 2 \).

Would you like further details on this proof?

Here are some related questions for practice:
1. Prove that if \( x + y = 2 \), then \( x = 1 \) or \( y = 1 \).
2. Show that if \( x + y > 2 \), then \( x > 1 \) or \( y > 1 \).
3. What happens if \( x + y \geq 0 \)? Can we still conclude something similar?
4. Prove that if \( x \geq 0 \) and \( y \geq 0 \), then \( x + y \geq 2 \) implies \( x \geq 1 \) or \( y \geq 1 \).
5. Prove that if \( x + y < 2 \), then \( x < 1 \) and \( y < 1 \).

**Tip**: When proving statements involving "or" (disjunctions), considering the contrapositive can often simplify the proof process.

Suppose x, y ∈ R. Prove that if x + y ≥ 2, then x ≥ 1 or y ≥ 1.

Explore a proof of the logical implication that if x + y ≥ 2, then x ≥ 1 or y ≥ 1. This problem involves key mathematical concepts such as inequalities and the contrapositive proof technique, making it suitable for students in Grades 10-12.

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