Question 684518: Multiplicative inverse of 2
Answer by RedemptiveMath(80)   (Show Source):
You can put this solution on YOUR website! The multiplicative inverse of 2 is simply a term we call a "reciprocal." A reciprocal of a number n is that number n under 1. In other words, the reciprocal of n is 1/n. Be mindful of signs on n (positive and negative). When you multiply a number and its reciprocal, you will get the product of 1: n * 1/n = (1n)/n = (1n)/(1n) = 1/1 = 1.
If n would equal 2, then your answer would be 1/2. 2 * 1/2 = [2(1)]/2 = 2/2 = 1. 1/2 is the multiplicative inverse of 2.
If you were meaning to submit a problem concerning multiplicative inverses of complex numbers, then I'll briefly explain below.
Multiplicative inverses of complex numbers are a little more complex than multiplicative inverses of most real numbers. There is a significant similarity between the two, however. If you remember dealing with rational expressions, then you may recall the instances of having a radical in the denominator. These radicals were oftentimes accompanied by constants being added or subtracted to or from them. An example is 2/{[sqrt](3)+2}. It was expressed in algebra that radicals in the denominator of a fraction makes the fraction unsimplified. Thus, we needed to manipulate the fraction in order to move the radical out of the denominator. To simplify the example above, we multiply the entire fraction by what we call a conjugate. Simply put, conjugates are like their counterparts, but the signs between the parts are different. In fractions, we would multiply by the conjugate of the denominator in these cases. Simplifying 2/{[sqrt](3)+2}:
2/{[sqrt](3)+2} * {[sqrt](3)-2}/{[sqrt](3)-2}
{2[sqrt](3)-4}/({[sqrt](3)+2}{[sqrt](3)-2})
{2[sqrt](3)-4}/{[sqrt](3)^2-2[sqrt](3)+2[sqrt](3)-4}
{2[sqrt](3)-4}/(3-4)
{2[sqrt](3)-4}/-1
-{2[sqrt](3)-4}.
As you can see, the conjugate of the original denominator {[sqrt](3)+2} is {[sqrt](3)-2}. You notice that the only difference is the sign (positive to negative). We can multiply the fraction by the conjugate over the conjugate because it technically does not change the output of the expression. That is, 2/{[sqrt](3)+2} and -{2[sqrt](3)-4} are equivalent fractions. We arrived at the latter from the former by basically multiplying the former by 1. If we multiply anything by 1, then it really doesn't change that number. Similarly, multiplying an expression by 1 doesn't change the output of that expression. {[sqrt](3)-2}/{[sqrt](3)-2} = 1, so 2/{[sqrt](3)+2} multiplied by that doesn't change the output of 2/{[sqrt](3)+2}.
Now onto complex numbers, we use the same notion of conjugates with complex numbers. Complex numbers have the form of a+bi, where a and b are real numbers and i is the imaginary number (sqrt)(-1). When we have that form in a fraction, it is unconventional to have the imaginary in the denominator. Just like radicals in the denominator, we must manipulate the fraction to remove the imaginary from the denominator. Let's use the fraction 2/(8-3i). Simplifying, we have:
2/(8-3i) * (8+3i)/(8+3i)
[2(8+3i)]/[(8-3i)(8+3i)]
16+6i/[64+24i-24i-9(i^2)]
16+6i/[64-9(-1)]
16+6i/(64+9)
16+6i/73.
When dealing with the squares, cubes, and so forth of imaginaries in complex numbers, we must remember what those imaginaries equal. I equals (sqrt)(-1), so the square of that would remove the square root notation and leave us with -1. The expression "rationalizing the denominator" is what we have used when removing an unwanted term from a denominator. The general form of any complex fraction, its multiplicative inverse, and the operations to get from the former to the latter is explained shown below (let x equal the numerator).
x/(a+bi)
x/(a+bi) * (a-bi)/(a-bi)
[x(a-bi)]/[(a+bi)(a-bi)]
(xa-xbi)/[a^2-b(i^2)]
(xa-xbi)/[a^2-b(-1)]
(xa-xbi)/(a^2+b).
Be mindful of signs on a and b.
|
|
|
