Question

1. prove (4 + i)(4 - i) is contained in the reals. r⊄ℂ, thus no i’s

Explanation:

Step1: Expand the product

Use the formula $(a + b)(a - b)=a^{2}-b^{2}$. Here $a = 4$ and $b = i$, so $(4 + i)(4 - i)=4^{2}-i^{2}$.

Step2: Simplify the expression

We know that $i^{2}=-1$, so $4^{2}-i^{2}=16-(-1)$.

Step3: Calculate the result

$16-(-1)=16 + 1=17$. Since 17 is a real - number, $(4 + i)(4 - i)$ is in the set of real numbers.

Answer:

The product $(4 + i)(4 - i)$ is 17, which is a real number, thus $(4 + i)(4 - i)$ is contained in the reals.