# Division and Fraction Bars (Vinculum) - Examples, Exercises and Solutions

When we study the order of mathematical operations we come across the terms division bar and fraction bar, but what do they mean and why are they so special?

First of all, we must remember that the fraction bar—or vinculumis exactly the same as a division. $10:2$ is the same as ${\ {10 \over 2}}$ and ${\ {10/ 2}}$.

Two things to remember:

• You cannot divide by $0$. To prove this, let's look at the following example: ${\ {3:0=}}$.
To solve this, we must be able to do the following: ${\ {0 \cdot ?=3}}$. However, since there is no number that can be multiplied by $0$ to give the result $3$, there is also therefore no number that can be divided by $0$.
• When we have a fraction bar, it is as if there are parentheses in the numerator. We solve the numerator first and then continue with the exercise. For example:

${\ {{10-2 \over 2}= {8 \over 2} = 4}}$

## examples with solutions for division and fraction bars (vinculum)

### Exercise #1

$(3\times5-15\times1)+3-2=$

### Step-by-Step Solution

This simple rule is the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations enclosed in parentheses precede all others,

Following the simple rule, multiplication comes before division and subtraction, therefore we calculate the values of the multiplications and then proceed with the operations of division and subtraction

$3\cdot5-15\cdot1+3-2= \\ 15-15+3-2= \\ 1$ Therefore, the correct answer is answer B.

$1$

### Exercise #2

$(5+4-3)^2:(5\times2-10\times1)=$

### Step-by-Step Solution

This simple rule is the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations enclosed in parentheses precede all others,

In the given expression, the establishment of division between two sets of parentheses, note that the parentheses on the left indicate strength, therefore, in accordance to the order of operations mentioned above, we start simplifying the expression within those parentheses, and as we proceed, we obtain the result derived from simplifying the expression within those parentheses with given strength, and in the final step, we divide the result obtained from the simplification of the expression within the parentheses on the right,

We proceed similarly with the simplification of the expression within the parentheses on the left, where we perform the operations of multiplication and division, in strength, in contrast, we simplify the expression within the parentheses on the right, which, according to the order of operations mentioned above, means multiplication precedes division, hence we first perform the operations of multiplication within those parentheses and then proceed with the operation of division:

$(5+4-3)^2:(5\cdot2-10\cdot1)= \\ (-2)^2:(10-10)= \\ 4:0\\$We conclude that the sequence of operations within the expression that is within the parentheses on the left yields a smooth result, this result we leave within the parentheses, these we raised in the next step in strength, this means we remember that every number (positive or negative) in dual strength gives a positive result,

As we proceed, note that in the last expression we received from establishing division by the number 0, this operation is known as an undefined mathematical operation (and this is the simple reason why a number should never be divided by 0 parts) therefore, the given expression yields a value that is not defined, commonly denoted as "undefined group" and use the symbol :

$\{\empty\}$In summary:

$4:0=\\ \{\empty\}$Therefore, the correct answer is answer A.

No solution

### Exercise #3

$(5\times4-10\times2)\times(3-5)=$

### Step-by-Step Solution

This simple rule is the order of operations which states that multiplication precedes addition and subtraction, and division precedes all of them,

In the given example, a multiplication occurs between two sets of parentheses, thus we simplify the expressions within each pair of parentheses separately,

We start with simplifying the expression within the parentheses on the left, this is done in accordance with the order of operations mentioned above, meaning that multiplication comes before subtraction, we perform the multiplications in this expression first and then proceed with the subtraction operations within it, in reverse we simplify the expression within the parentheses on the right and perform the subtraction operation within them:

What remains for us is to perform the last multiplication that was deferred, it is the multiplication that occurred between the expressions within the parentheses in the original expression, we perform it while remembering that multiplying any number by 0 will result in 0:

$0$

### Exercise #4

$0.18+(1-1)=$

### Step-by-Step Solution

According to the order of operations rules, we first solve the expression in parentheses:

$1-1=0$

And we get the expression:

$0.18+0=0.18$

0.18

### Exercise #5

Indicates the corresponding sign:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_—}(5^2-3+6):7\cdot\frac{1}{4}$

### Step-by-Step Solution

For a given problem, whether it involves addition or subtraction each of the terms that come into play separately,

this is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are performed before all others,

A. We will start with the terms that are on the left in the given problem:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2$First, we simplify the terms that are in the denominators in accordance with the order of operations, this is done by calculating the numerical value of the denominator in strength (this within that we remember that in defining the root as strong, the root itself is strong for everything) and then perform the operation of division and subtraction:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\ \frac{1}{\sqrt{16}}\cdot(125+3-4):2^2 =\\ \frac{1}{\sqrt{16}}\cdot124:2^2$Next, we calculate the numerical values of the part that was passed in strength (practically, if we were to represent the operation of division as broken, this part would have been in the broken position) and as such the numerical value of the part in strength that in the broken position in the terms, next we perform the operations of multiplication and division:

$\frac{1}{\sqrt{16}}\cdot124:2^2 =\\ \frac{1}{4}\cdot124:4 =\\ \frac{1\cdot124}{4}:4=\\ \frac{\not{124}}{\not{4}}:4=\\ 31:4=\\ \frac{31}{4}=\\ 7\frac{3}{4}$In the final stages, we performed the multiplication of the number 124 in break, this we did within that we remember that multiplication in break means multiplication in the broken position, next we performed the operation of division of the break (by condensing the break) and in the final stage we performed the operation of division in the number 4, this operation resulted in a complete answer, and therefore we marked it as break (a break from understanding, an assumption that the position is greater than the position) and in the continuation we converted the break from understanding to a mixed break, by extracting the wholes (the answer to the question: "How many times does the division enter the divisor?") and adding the remaining division to the divisor,

We finished simplifying the terms that are on the left in the given problem, we will summarize the simplification stages:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2 =\\ \frac{1}{\sqrt{16}}\cdot124:2^2 =\\ \frac{1\cdot124}{4}:4=\\ 7\frac{3}{4}$

B. We will continue and simplify the terms that are on the right in the given problem:

$(5^2-3+6):7\cdot\frac{1}{4}$In this part, to simplify the terms within the framework of the order of operations,

In this term, the operation of division of the beginning on terms that are in the denominators, therefore we will first simplify this term,

Let's note that multiplication precedes addition and subtraction, which precede division and subtraction, therefore we will start by calculating the numerical value of the part in strength that in this term, next we perform the operations of division and subtraction:

$(5^2-3+6):7\cdot\frac{1}{4} =\\ (25-3+6):7\cdot\frac{1}{4} =\\ 28:7\cdot\frac{1}{4}$We will continue and simplify the received term, noting that between multiplication and division there is no precedence defined in the order of operations, we perform the operations in this term one after the other according to the order from left to right, which is the natural order of operations:

$28:7\cdot\frac{1}{4} =\\ 4\cdot\frac{1}{4} =\\ \frac{4\cdot1}{4}=\\ \frac{\not{4}}{\not{4}}=\\ 1$ In the second stage, we performed the multiplication in break, this within that we remember (again) that multiplication in break means multiplication in the broken position, in the next stage we performed the operation of division of the break (by condensing the break).

We finished simplifying the terms that are on the right in the given problem, we will summarize the simplification stages:

$(5^2-3+6):7\cdot\frac{1}{4} =\\ 28:7\cdot\frac{1}{4} =\\ \frac{\not{4}}{\not{4}}=\\ 1$We return to the original problem, and we will present the results of simplifying the terms that were reported in A and B:

$\frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_—}(5^2-3+6):7\cdot\frac{1}{4} \\ \downarrow\\ 7\frac{3}{4} \text{ }\textcolor{red}{_—}1$As a result, we receive that:

$7\frac{3}{4} \text{ }\textcolor{red}{\neq}1$Therefore, the correct answer here is answer B.

$\ne$

## examples with solutions for division and fraction bars (vinculum)

### Exercise #1

Indicates the corresponding sign:

$-5+(5-3\cdot2)+6\textcolor{red}{☐}((3+2)\cdot2):2\cdot0$

### Step-by-Step Solution

In order to solve the given problem, whether it involves addition or subtraction each of the terms that appear in the equation must be dealt with separately,

This is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are for all,

A. We will start with the terms that are on the left in the given problem:

$-5+5-3\cdot2+6$ Simplify the terms that are in the parentheses in accordance with the order of operations, start with the multiplication that is in the terms and continue to perform the operations of addition and subtraction:

$-5+5-3\cdot2+6=\\ -5+5-6+6=\\ 0$

We finish simplifying the terms that are on the left in the given problem.

B. We will continue and simplify the terms that are on the right in the given problem:

$\big((3+2)\cdot2\big):2\cdot0$Note that in this term there is a multiplication between the term and the number 0, in addition note that in this term it is defined (since it does not include division by 0), we remember that multiplying any number by 0 will yield the result 0, therefore:

$\big((3+2)\cdot2\big):2\cdot0 =\\ 0$

We return now to the original problem, and we will present the results of simplifying the terms that were reported in A and B:

$-5+5-3\cdot2+6\textcolor{red}{☐}\big((3+2)\cdot2\big):2\cdot0 \\ \downarrow\\ 0\text{ }\textcolor{red}{_—}0$As a result that we find that:

$0 \text{ }\textcolor{red}{=}0$Therefore the correct answer here is answer A.

$=$

### Exercise #2

$[(5-2):3-1]\times4=$

### Step-by-Step Solution

In the order of operations, parentheses come before everything else.

We start by solving the inner parentheses in the subtraction operation:

$((3):3-1)\times4=$ We continue with the inner parentheses in the division operation and then subtraction:

$(1-1)\times4=$

We continue solving the subtraction exercise within parentheses and then multiply:

$0\times4=0$

$0$

### Exercise #3

$\lbrack(4+3):7+2:2-2\rbrack:5=$

### Step-by-Step Solution

Simplifying this expression emphasizes the order of operations, which states that multiplication precedes addition and subtraction, and that division precedes all of them,

In the given expression, the establishment of division operations between the parentheses (the outermost) to a number, therefore according to the order of operations as mentioned, is handled by simplifying the expression in these parentheses, this expression includes division operations that begin on the expression within the parentheses (the innermost), therefore according to the order of operations as mentioned is handled by simplifying the expression in these parentheses and performing the subtraction operations in it, there is no hindrance to calculate the outcome of the division operations in the expression in the outermost parentheses, but for the sake of good order this is done afterwards:

$\lbrack(4+3):7+2:2-2\rbrack:5= \\ \lbrack7:7+2:2-2\rbrack:5$Continuing and simplifying the expression in the parentheses we noted, since division precedes addition and subtraction, start with the division operations in the expression and only then calculate the outcome of the addition and subtraction, ultimately perform the division operations on this expression in the parentheses:

$\lbrack7:7+2:2-2\rbrack:5 \\ \lbrack1+1-2\rbrack:5=\\ 0:5=\\0$In the last stage we mentioned that multiplying a number by 0 gives the result 0,

Therefore, this simplifying expression is short so there is no need to elaborate,

$0$

### Exercise #4

$(3+2-1):(1+3)-1+5=$

### Step-by-Step Solution

This simple rule is the order of operations which states that multiplication and division come before addition and subtraction, and operations enclosed in parentheses come first,

In the given example of division between two given numbers in parentheses, therefore according to the order of operations mentioned above, we start by calculating the values of each of the numbers within the parentheses, there is no prohibition against calculating the result of the addition operation in the given number, for the sake of proper order, this operation is performed later:

$(3+2-1):(1+3)-1+5= \\ 4:4-1+5$In continuation of the principle that division comes before addition and subtraction the division operation is performed first and then the operations of subtraction and addition which were received in the given number and in the last stage:

$4:4-1+5= \\ 1-1+5=\\ 5$Therefore the correct answer here is answer B.

$5$

### Exercise #5

Indicates the corresponding sign:

$\frac{1}{25}\cdot(5^2-3+\sqrt{9})\textcolor{red}{☐}\sqrt{25}\cdot5\cdot\frac{1}{5}$

### Step-by-Step Solution

We solve the left side and start from the parentheses:

$5^2=5\times5=25$

We will solve the root exercise using the equation:$\sqrt{a^2}=a$

$\sqrt{9}=\sqrt{3^2}=3$

We arrange the exercise accordingly:

$\frac{1}{25}\times(25-3+3)=$

We solve the exercise in parentheses from left to right:

$\frac{1}{25}\times(22+3)=\frac{1}{25}\times25$

We convert the 25 into a simple fraction, multiply and divide:

$\frac{1}{25}\times\frac{25}{1}=\frac{25}{25}=\frac{1}{1}=1$

We solve the right side:

$\sqrt{25}=\sqrt{5^2}$

We arrange the exercise:

$\sqrt{5^2}\times5\times\frac{1}{5}$

We convert the 5 into a simple fraction and note that it is possible to reduce by 5:

$\sqrt{5^2}\times\frac{5}{1}\times\frac{1}{5}=\sqrt{5^2}\times1$

We solve the root according to the formula:$\sqrt{a^2}=a$

$5\times1=5$

Now we are going to compare the left side with the right side, and it seems that we obtained two different results and therefore the two sides are not equal.

$\ne$

## examples with solutions for division and fraction bars (vinculum)

### Exercise #1

Complete the following exercise:

$\frac{27-5\cdot3}{6\cdot2}+\frac{15\cdot4}{3}=$

### Step-by-Step Solution

According to the order of arithmetic operations, first we place the multiplication exercises within parentheses:

$\frac{27-(5\cdot3)}{(6\cdot2)}+\frac{(15\cdot4)}{3}=$

We solve the exercises within parentheses:

$5\times3=15$

$6\times2=12$

$15\times4=60$

Now we obtain the exercise:

$\frac{27-15}{12}+\frac{60}{3}=$

We solve the numerator of the fraction:

$27-15=12$

We obtain:

$\frac{12}{12}+\frac{60}{3}=$

We solve the fractions:

$\frac{12}{12}=1$

$60:3=20$

Now we obtain the exercise:

$1+20=21$

21

### Exercise #2

Complete the following exercise:

$\frac{25+3-2}{13}+5\cdot4=$

### Step-by-Step Solution

According to the order of arithmetic operations, we first place the multiplication exercise in parentheses:

$\frac{25+3-2}{13}+(5\cdot4)=$

We solve the multiplication exercise:

$5\times4=20$

We obtain the exercise:

$\frac{25+3-2}{13}+20=$

We solve the exercise in the numerator of the fraction:

$25+3-2=28-2=26$

We obtain the fraction:

$\frac{26}{13}=2$

Now we obtain the exercise:

$2+20=22$

22

### Exercise #3

$\lbrack(\sqrt{81}-3\times3):4+5\times5\rbrack=$

### Step-by-Step Solution

According to the rules of order of arithmetic operations, parentheses are resolved first.

We start by solving the inner parentheses, first we will solve the root using the formula:

$\sqrt{a}=\sqrt{a^2}=a$

$\sqrt{81}=\sqrt{9^2}=9$

The exercise obtained within parentheses is:

$(9-3\times3)$

First we solve the multiplication exercise and then we subtract:

$(9-9)=0$

After solving the inner parentheses, the resulting exercise is:

$0:4+5\times5$

According to the rules of the order of arithmetic operations, we first solve the exercises of multiplication and division, and then subtraction.

We place the two exercises within parentheses to avoid confusion:

$(0:4)+(5\times5)=0+25=25$

$25$

### Exercise #4

$\frac{14+8-2}{4\cdot5}\cdot2+3=$

### Step-by-Step Solution

First, we will solve the multiplication exercise that was broken:

$4\times5=20$

Now, we will solve the exercise that was broken:

$14+8-2=22-2=20$

$\frac{20}{20}=1$

$1\times2+3=$

According to the order of operations, we will first solve the multiplication exercise and then proceed:

$1\times2=2$

$2+3=5$

5

### Exercise #5

Indicates the corresponding sign:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)\textcolor{red}{☐}(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13}$

### Step-by-Step Solution

For solving a problem involving addition or subtraction each of the terms that come into play is treated separately,

This is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are performed before all others,

A. We will start with the term that appears on the left in the given problem:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)$In this term, a division operation is established between two terms that in their sum give a specific result, we take the term that in their sum gives the left, remembering that multiplication precedes subtraction, therefore the multiplication in these terms is performed first and then the subtraction operation, in contrast- the term that in their sum gives the right (the denominators) is considered as a multiplication of a number by the term that in their sum (the numerators) therefore the multiplication in this term, this is within that we remember that multiplication precedes division and that the root cause (the definition of the root as strong) is strong for everything, therefore we will calculate its numerical value and then perform the division operation that in this term:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ (36-12):\big((4+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big) \\$We will continue and note that in the term that was received in the last stage the multiplication operation is found in the terms and accordingly precedence is given to the division operation that precedes them,the multiplication is performed within that we remember that the multiplication in the break means the multiplication by the break, in continuation the division operation of the break is performed, this by summarizing:

$24:\frac{6\cdot1}{6}=\\ 24:\frac{\not{6}}{\not{6}}=\\ 24:1=\\ 24$In the last stage we performed the remaining division operation, this within that we remember that dividing any number by the number 1 will yield the number itself,

We will conclude the simplification of the term that appears on the left in the given problem, we will summarize the simplification stages:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big)= \\ 24$

B. We will continue and simplify the term that appears on the right in the given problem:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13}$In this part, in the first section, we simplify the term within the framework of the order of operations,

In this term, a division operation is established between two terms that in their sum, in this part, in the first section, we simplify that two terms in contrast, the term that in their sum gives the left is simplified within performing the division and subtraction operations, in contrast we simplify the term that in their sum gives the right, given that multiplication and division precede subtraction we start from the second simplification stage in these terms , and given that the order of operations does not define precedence to one of the multiplication or division operations is performed one after the other according to the order from left to right (which is the natural order of operations) , in continuation we will calculate the result of the subtraction operation that in these terms:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:(17-4)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\$We will continue and simplify the term that was received in the last stage, in this part, first the division and multiplication operations are performed one after the other from left to right:

$17:13\cdot\frac{1}{13} =\\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17\cdot1}{13\cdot13}=\\ \frac{17}{13^2}$In the first stage, given that the result of the division operation is a result that is not whole we mark it as a break (a break from above- given that the numerator is larger than the denominator) in continuation we perform the multiplication of the breaks within that we remember that when we multiply two breaks we multiply numerator by numerator and denominator by denominator and keep the essence of the original break.

We will conclude the simplification of the term that appears on the right in the given problem, we will summarize the simplification stages:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17}{13^2}$We will return to the original problem, and we will present the results of the simplification of the terms that were reported in A and B:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)\textcolor{red}{☐}(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} \downarrow\\ 24 \textcolor{red}{☐}\frac{17}{13^2}$As a result, the correct answer here is answer B.

$\ne$