Great question. Let's let r be a rational number and s be irrational. Note r has to be nonzero for this to work. In other words, it's not true that when we multiply zero, a rational number, by an irrational number like π we get an irrational number. We of course get zero.
The question is: why is the product
irrational?
In math "why" questions are usually answered with an illuminating proof. Here the indirect proof is enlightening.
Suppose p was rational. Then
would be rational as well, being the ratio of two rational numbers, so ultimately the ratio of two integers.
But we're given that s is irrational so we have our contradiction and must conclude our assumption that p is rational is false, that is, we conclude p is irrational.
A proof by contradiction.
Let assume that the product of a rational number and an irrational number is rational.
Let and be rational numbers, where and an irrational number.
Then
Integers are closed under multiplication, therefore and are integers, making the number rational, which is contradictory with the earlier statement that is an irrational number.
-16 = 8x/5
x=-25 3/5
x=10
x=-10
Answer:
x = -10
Step-by-step explanation:
-16 = 8x/5
-80 = 8x
x = -10
Answer:
7
Step-by-step explanation: