find the additive and multiplicative inverses for 8?
The answer depends on which ring you are considering. Over the real numbers, the additive inverse is -8, and the multiplicative inverse is 1/8. If this problem arises from mathematics at any level below that of Abstract Algebra, then that would be the correct answer. There would be different answers over any of the other rings which have an element that might be written as 8. The ring of integers modulo an odd number k would be an example of this. Here, the additive inverse would be k-8. For the multiplicative inverse, choose (rk-1)/8 if the remainder of k on division by 8 is r (here, r must be one of the numbers 1, 3, 5, or 7).
They are correct; the answers are -8 and 1/8. Jaden's wrong though about getting 0 when you multiply a number by its multiplicative inverse. You get 1. Here's a little more of an explanation: An identity is a number such that, when you {add, multiply, etc} another number to it, you get that number back. The second number keeps its identity. So the additive identity is 0, because 8 + 0 0 + 8 , and the multiplicative identity is 1, because 8 x 1 1 x 8 8. The inverse of a number is a number such that, when you {add, multiply} by the original number, you get the identity number. So the additive inverse of 8 is the number that you would add to 8 in order to get 0 (the identity); that number is -8. The multiplicative inverse of 8 is the number that you would multiply by 8 in order to get 1 (the identity); that number is 1/8.
The additive inverse of 8 is ?8. To find the additive inverse, simply take the opposite of what you are given. When you add a number and its inverse you will get zero. The multiplicative inverse of 8 is 1/8. To find the multiplicative inverse, simply make the given number the denominator of a fraction with a numerator of one. When you multiply a number and its inverse you will get one. UPDATE: uhh, yeah.you get one. Sorry! And thanks for catching my error.
