Operations on Rational Numbers

 

Operations on Rational Numbers

 

Order of Operations:

 

In mathematics, certain operations have priority over others. Therefore, when different operations need to be performed together, they must be done sequentially according to the order of operations.

 

The order of operations is as follows:

1) Expressions inside parentheses are evaluated first. Additionally, parentheses and fraction bars determine the direction of the operation.

2) If there are exponential expressions, the exponentiation is performed.

3) Multiplication and Division operations are performed.

4) Addition and Subtraction operations are performed.

 

 

Warning:

 

Multiplication and division do not have priority over one another, nor do addition and subtraction have priority over each other. For this reason, when multiplication and division, or addition and subtraction, appear consecutively, the operation order is typically evaluated from left to right unless determined by parentheses.
*(Note: Clarified the textbook’s note regarding consecutive operations of the same priority level, which traditionally follow a strict left-to-right rule).*

 

Examples:

 

  • \[\begin{array}{l l } = 3 \cdot 2+ \frac{1}{3} : \frac{2}{3}-2\div \frac{1}{3}\\=6+ \frac{1}{3} \cdot \frac{3}{2} – 2 \cdot 3\\=6+\frac{1}{2}-6=\frac{1}{2} \end{array}\]

 

  • \[\begin{array}{l l }=2^3-[(2:3):\frac{1}{3} ] \cdot (1-3)-(2-5)^2\\= 8-[\frac{2}{3} : \frac{1}{3}] \cdot (-2)-(-3)^2\\= 8-(\frac{2}{3} \cdot 3)\cdot (-2)-9\\= 8-2 \cdot (-2)-9\\=8+4-9\\=3\end{array}\]
  • \[\begin{array}{l l }
    =\Large{ \frac{\frac{3}{4} }{2} } – \Large{ \frac{2}{\frac{3}{4} } }\\
    =\frac{3}{4} \cdot \frac{1}{2}-2 \cdot \frac{4}{3} \\
    =\frac{3}{8} -\frac{8}{3} \\
    =\frac{9-64}{24} \\
    =-\frac{55}{24}
    \end{array}\]\]

 

  • \[\begin{array}{l l }
    [(2:3) \cdot (1:2)] :[4 \cdot (1:3)]\\
    =[\frac{2}{3} \cdot \frac{1}{2} ]:[4 \cdot \frac{1}{3} ]\\
    =\frac{1}{3}: \frac{4}{3} =\frac{1}{3} \cdot \frac{3}{4} =\frac{1}{4}
    \end{array}\]\]

 

  • \[\begin{array}{l l } x-[x+y \cdot (x+1)]+ x \cdot y +x \\
    =2x-[x+x \cdot y+y]+x \cdot y\\
    =2x-x-x \cdot y-y+x \cdot y\\
    =x-y \quad \text{holds.}
    \end{array}\]\]

 

 

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