jim1174
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What is the easiest method for adding and subtracting rational numbers
The discussion revolves around the easiest method for adding and subtracting rational numbers, specifically focusing on those represented as fractions. Participants explore various approaches and clarify steps involved in the process.
Participants generally agree on the necessity of finding a common denominator for adding and subtracting fractions, but there are variations in the details of the methods presented. No consensus is reached on a singular "easiest" method, as different approaches are discussed.
Some assumptions about the familiarity with fractions and the concept of least common denominators are present, but not all participants clarify these assumptions explicitly. The discussion does not resolve the potential complexities involved in different scenarios of adding and subtracting rational numbers.
You need to be more specific. Do you mean rational numbers represented as fractions, like ##\frac{1}{2} + \frac{3}{5}## or rational numbers represented as decimal fractions, like .5 + .6?jim1174 said:What is the easiest method for adding and subtracting rational numbers
jim1174 said:What is the easiest method for adding and subtracting rational numbers
To elaborate on what AMenendez is saying, we are multiplying each fraction by 1 in some form so as to get the denominator we want.AMenendez said:So, the standard way of adding / subtracting numbers is to first identify a ``least common denominator". So, say I had:
##\frac{1}{2} + \frac{3}{7}##
Well, what is the smallest number that is a multiple of both 2 and 7? It's 14. So now, how do we write the original fractions with denominators of 14? Well, we multiply 2 by 7 to get 14, so we do the same to the numberator, 1. Likewise, we multiply 7 by 2 to get 14, so we multiply 3 by 2 to get 6. So we may write
##\frac{1}{2} + \frac{3}{7} = \frac{7}{14} + \frac{6}{14} = \frac{13}{14}##
AMenendez said:Which doesn't reduce. So, very roughly speaking,
##\frac{a}{bc} + \frac{d}{ef} = \frac{(aef)+(bcd)}{bcef}##