Multiplicative Inverse

πŸ†Practice special cases (0 and 1, inverse, fraction line)

Two numbers are multiplicative inverses when their product results in 1 1 .

For example:

12{\Large {1 \over 2}} and 2 2 are multiplicative inverses because 2β‹…12=1{\Large 2 \cdot {1 \over 2}=1}

Formulation of the Rule for Multiplicative Inverse Numbers:

Whenever a is different from 00, it follows that aβ‹…1a=1{\Large a\cdot{1 \over a} = 1}

Multiplication and Division of Multiplicative Inverses

Division is equivalent to multiplication by its multiplicative inverse,

That is: Β 213=2β‹…3=6{\Large {{2 \over {1 \over 3}} = 2 \cdot 3 = 6}}

Because 3 3 is the multiplicative inverse of Β 13{\Large {1 \over 3}}

Generally: a1b=aβ‹…b \frac{a}{\frac{1}{b}}=aβ‹…b

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Test yourself on special cases (0 and 1, inverse, fraction line)!

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\( 12+3\times0= \)

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Two numbers are multiplicative inverses when their multiplication results in 1 1 .

For example:

12{\Large {1 \over 2}} and 2 2 are multiplicative inverses because 2β‹…12=1{\Large 2 \cdot {1 \over 2}=1}

More examples:

The multiplicative inverse of 5 5 is 15{\Large {1 \over 5}}
5β‹…15=1{\Large 5 \cdot {1 \over 5}=1}


The multiplicative inverse of 3 3 is 13{\Large {1 \over 3}}
3β‹…13=1{\Large 3 \cdot {1 \over 3}=1}


The multiplicative inverse of 57{\Large {5 \over 7}} is 75{\Large {7 \over 5}}
75β‹…57=1{\Large {7 \over 5} \cdot {5 \over 7}=1}


The multiplicative inverse of 923{\Large {9 \over 23}} is 239{\Large {23 \over 9}}
239β‹…923=1{\Large {23 \over 9} \cdot {9 \over 23}=1}


The multiplicative inverse of 0.5 0.5 is 2 2

2β‹…0.5=1{\Large 2 \cdot 0.5=1}


The multiplicative inverse of 0.25 0.25 is 4 4

4β‹…0.25=1{\Large 4 \cdot 0.25=1}


Formulation of the Rule for Multiplicative Inverse Numbers:

Whenever a is different from 0 0 , it happens that aβ‹…1a=1{\Large a\cdot{1 \over a} = 1}


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Multiplication and Division of Multiplicative Inverses

Division is equivalent to multiplication by the multiplicative inverse,

that is: Β 213=2β‹…3=6{\Large {{2 \over {1 \over 3}} = 2 \cdot 3 = 6}}

This is because 3 3 is the multiplicative inverse of Β 13{\Large {1 \over 3}}

In general: Β a/1b=aβ‹…b{\Large a /{1 \over b} = a \cdot b}


Exercises on Multiplicative Inverses

Solve the following exercises

  • 5+472={\Large {5+{{4 \over 7} \over 2} =}}
  • 60.75βˆ’2β‹…3={\Large {{6 \over 0.75} - 2 \cdot 3 =}}
  • 312βˆ’1316={\Large {{3{1 \over 2}-{{1 \over 3} \over {1 \over 6}}}=}}
  • 1072+278={\Large {{{{10 \over 7} \over 2 } + {2 \over {7 \over 8} }}=}}
  • 35910+7913={\Large {{{3 \over 5} \over {9 \over 10}} + {{7 \over 9} \over {1 \over 3}}=}}
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If you are interested in this article, you might also be interested in the following articles:

Positive, Negative Numbers and Zero

The Real Number Line

Opposite Numbers

Absolute Value

Elimination of Parentheses in Real Numbers

Multiplication and Division of Real Numbers

Abbreviated Multiplication Formulas

On the Tutorela blog, you will find a variety of articles about mathematics.


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Examples and Exercises with Solutions for Multiplicative Inverse

examples.example_title

(3Γ—5βˆ’15Γ—1)+3βˆ’2= (3\times5-15\times1)+3-2=

examples.explanation_title

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

3β‹…5βˆ’15β‹…1+3βˆ’2=15βˆ’15+3βˆ’2=1 3\cdot5-15\cdot1+3-2= \\ 15-15+3-2= \\ 1 Therefore, the correct answer is answer B.

examples.solution_title

1 1

examples.example_title

(5Γ—4βˆ’10Γ—2)Γ—(3βˆ’5)= (5\times4-10\times2)\times(3-5)=

examples.explanation_title

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:

Therefore, the correct answer is answer d.

examples.solution_title

0 0

examples.example_title

(5+4βˆ’3)2:(5Γ—2βˆ’10Γ—1)= (5+4-3)^2:(5\times2-10\times1)=

examples.explanation_title

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+4βˆ’3)2:(5β‹…2βˆ’10β‹…1)=(βˆ’2)2:(10βˆ’10)=4:0 (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 :

{βˆ…} \{\empty\} In summary:

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

examples.solution_title

No solution

examples.example_title

[(4+3):7+2:2βˆ’2]:5= \lbrack(4+3):7+2:2-2\rbrack:5=

examples.explanation_title

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:

[(4+3):7+2:2βˆ’2]:5=[7:7+2:2βˆ’2]:5 \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:

[7:7+2:2βˆ’2]:5[1+1βˆ’2]:5=0:5=0 \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,

And the correct answer here is answer A.

examples.solution_title

0 0

examples.example_title

(3+2βˆ’1):(1+3)βˆ’1+5= (3+2-1):(1+3)-1+5=

examples.explanation_title

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+2βˆ’1):(1+3)βˆ’1+5=4:4βˆ’1+5 (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:

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

examples.solution_title

5 5

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