To determine a counterexample to the conjecture "The sum of two numbers will always be greater than one of the numbers," we need to find an instance where the sum is not greater than one of the addends.
Let's analyze each option:
A. \( -3 + (-1) = -4 \)
- Here, \(-4\) is less than both \(-3\) and \(-1\), so it's less than both of the numbers. Hence, this does not directly provide a counterexample because the conjecture only states the sum should be greater than one of the numbers.
B. \( -3 + 2 = -1 \)
- Here, \(-1\) is greater than \(-3\) but not greater than \(2\). Though it's not clear here, let’s keep it as a potential counterexample.
C. \( 3 + 1 = 4 \)
- Here, \(4\) is greater than both \(3\) and \(1\). This supports the conjecture.
D. \( 3 + 2 = 5 \)
- Here, \(5\) is greater than both \(3\) and \(2\). This also supports the conjecture.
Upon examining:
- Option B provides a clear counterexample; \( -3 + 2 = -1 \), where \(-1\) is not greater than \(2\), but it's actually equivalent to the other number \(-1\).
Therefore, the counterexample to the conjecture is:
**B. \(-3\) + \(2\) = \( -1 \)**.
Which is a counterexample to the following conjecture?
The sum of two numbers will always be greater than one of the numbers.
A.
(–3) + (–1) = –4
B.
(–3) + 2 = –1
C.
3 + 1 = 4
D.
3 + 2 = 5
1 answer
3 answers