Answer :
Step-by-step explanation:
To show that \((-1) \times (-1) = 1\) using a pattern, let's start from the product \((-1) \times 5\) and demonstrate how the products change as we progressively decrease the multiplier. This pattern will help us understand why multiplying two negative numbers results in a positive number.
### Step-by-Step Pattern
1. **Starting point:**
\[ (-1) \times 5 = -5 \]
2. **Next step:**
\[ (-1) \times 4 = -4 \]
3. **Continue the pattern:**
\[ (-1) \times 3 = -3 \]
4. **Further steps:**
\[ (-1) \times 2 = -2 \]
5. **Almost there:**
\[ (-1) \times 1 = -1 \]
6. **Key step:**
\[ (-1) \times 0 = 0 \]
### Understanding the Transition to Negative Multipliers
Now, let's use the distributive property to understand the product of \((-1)\) with negative numbers.
Consider the expression:
\[ 0 = (-1) \times 0 \]
But we can also write:
\[ 0 = (-1) \times (1 + (-1)) \]
By the distributive property:
\[ (-1) \times (1 + (-1)) = (-1) \times 1 + (-1) \times (-1) \]
Since we know:
\[ (-1) \times 1 = -1 \]
We substitute this into our equation:
\[ 0 = -1 + (-1) \times (-1) \]
To isolate \((-1) \times (-1)\):
\[ -1 + (-1) \times (-1) = 0 \]
Adding 1 to both sides:
\[ (-1) \times (-1) = 1 \]
### Conclusion
Thus, by observing the consistent pattern of multiplying \((-1)\) by positive integers and using the distributive property at zero, we demonstrate that:
\[ (-1) \times (-1) = 1 \]
Answer:
To demonstrate that \((-1) \times (-1) = 1\), we can show a pattern of multiplication with negative numbers leading up to this result. Let's start from \((-1) \times 5\) and reduce the positive factor step by step, observing the pattern.
### Step-by-Step Pattern
1. **\((-1) \times 5\)**:
\[
(-1) \times 5 = -5
\]
2. **\((-1) \times 4\)**:
\[
(-1) \times 4 = -4
\]
3. **\((-1) \times 3\)**:
\[
(-1) \times 3 = -3
\]
4. **\((-1) \times 2\)**:
\[
(-1) \times 2 = -2
\]
5. **\((-1) \times 1\)**:
\[
(-1) \times 1 = -1
\]
Now, let's see what happens when we multiply \((-1)\) by \(0\):
6. **\((-1) \times 0\)**:
\[
(-1) \times 0 = 0
\]
Next, let's extend this to negative multipliers and observe the pattern:
7. **\((-1) \times (-1)\)**:
\[
\text{Let's derive this step by step:}
\]
To show that \((-1) \times (-1) = 1\), consider the distributive property of multiplication over addition:
\[
a \times (b + c) = (a \times b) + (a \times c)
\]
Let's use \(a = -1\), \(b = 1\), and \(c = -1\):
\[
(-1) \times (1 + (-1)) = (-1) \times 1 + (-1) \times (-1)
\]
We know that:
\[
1 + (-1) = 0
\]
Thus:
\[
(-1) \times 0 = (-1) \times 1 + (-1) \times (-1)
\]
Since \((-1) \times 0 = 0\), we get:
\[
0 = (-1) \times 1 + (-1) \times (-1)
\]
From the previous steps, we know:
\[
(-1) \times 1 = -1
\]
So:
\[
0 = -1 + (-1) \times (-1)
\]
Adding 1 to both sides:
\[
1 = (-1) \times (-1)
\]
### Conclusion
Therefore, by following the pattern and using the distributive property, we have shown that \((-1) \times (-1) = 1\).