Question

prove: the difference of the reciprocals of two successive integers equals the product of their reciprocals.

\\(\frac{1}{n} - \frac{1}{n+1} = \frac{?}{n(n+1)} = \frac{\box}{n} \cdot \frac{\box}{n+1}\\)

Explanation:

Define successive integers and their reciprocals

Let the two successive integers be \(n\) and \(n+1\). Their reciprocals are \(\frac{1}{n}\) and \(\frac{1}{n+1}\).

Find a common denominator

Using the Algebraic Proof knowledge point:

$$ \frac{1}{n} - \frac{1}{n+1} = \frac{n+1}{n(n+1)} - \frac{n}{n(n+1)} $$

Simplify the numerator

Using the Algebraic Proof knowledge point:

$$ \frac{(n+1) - n}{n(n+1)} = \frac{1}{n(n+1)} $$

Thus, the missing numerator in the first blank is \(1\).

Express as a product of reciprocals

Using the Algebraic Proof knowledge point:

$$ \frac{1}{n(n+1)} = \frac{1}{n} \cdot \frac{1}{n+1} $$

Thus, the numerators for the second and third blanks are both \(1\).

Answer:

Prove: The difference of the reciprocals of two successive integers equals the product of their reciprocals.

$$\frac{1}{n} - \frac{1}{n+1} = \frac{\text{1}}{n(n+1)} = \frac{\text{1}}{n} \cdot \frac{\text{1}}{n+1}$$