Neutral Element (Identiy Element)

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

What is a Neutral Element?

In mathematics, a neutral element is an element that does not alter the rest of the numbers when we perform an operation with it.

Neutral Element - Addition

With addition, 0 0 is considered a neutral element because it does not modify the number to which it is added.

0+3=3 0+3=3

Neutral Element - Multiplication

In multiplication, 1 1 is considered a neutral element because it does not affect the result.

4Γ—1=4 4\times1=4

Neutral Element - Subtraction and Division

The neutral element in subtraction is 0 0 , while in division it is 1 1 .

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

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\( 0+0.2+0.6= \)

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Addition and Subtraction Exercises with a Neutral Number

  • 18βˆ’0=18 18-0=18
  • 22+0=22 22+0=22
  • 1000000βˆ’0=1000000 1000000-0=1000000
  • 0+12βˆ’0=12 0+12-0=12

Multiplication Exercises with a Neutral Number

  • 13Γ—1=13 13\times 1=13
  • 220Γ—1=220 220\times 1=220
  • 12Γ—1=12 \frac{1}{2}\times 1=\frac{1}{2}
  • XΓ—1=X X\times 1=X
  • 2000000Γ—1=2000000 2000000\times 1=2000000

Note that in a multiplication operation, βˆ’1 -1 is not considered a neutral number because it affects the the number it is multiplied with by changing its sign.


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Review Questions

What is a neutral element?

A neutral element is a number that does not alter the other numbers when an operation is applied to it.


Why is there no neutral element for addition?

The only neutral element for addition is 0 0 and zero is not part of the set of natural numbers. Therefore, there is no neutral of addition in this set.


Do you know what the answer is?

How do you determine the neutral element for a product?

If there is a non-zero number a and we must find the value of x x when ax=a ax=a , then by solving the equation we find that the neutral of the product is 1.


What is the neutral element for division?

The neutral element for division is 1 1 .


Check your understanding

What does absorbing element mean?

The absorbing element of multiplication is 0 because any number multiplied by 00 is 00.


What defines the additive neutral of real numbers?

The additive neutral of real numbers does not modify the value of any number that it is added to.

Example: 00 is the additive neutral of real numbers since a+0=a a+0=a .


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examples with solutions for neutral element (identiy element)

Exercise #1

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

Video Solution

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

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.

Answer

1 1

Exercise #2

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

Video Solution

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+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.

Answer

No solution

Exercise #3

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

Video Solution

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:

Therefore, the correct answer is answer d.

Answer

0 0

Exercise #4

0.18+(1βˆ’1)= 0.18+(1-1)=

Video Solution

Step-by-Step Solution

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

1βˆ’1=0 1-1=0

And we get the expression:

0.18+0=0.18 0.18+0=0.18

Answer

0.18

Exercise #5

Indicates the corresponding sign:

116β‹…(125+3βˆ’16):22Β β€”(52βˆ’3+6):7β‹…14 \frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_β€”}(5^2-3+6):7\cdot\frac{1}{4}

Video Solution

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:

116β‹…(125+3βˆ’16):22 \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:

116β‹…(125+3βˆ’16):22=116β‹…(125+3βˆ’4):22=116β‹…124:22 \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:

116β‹…124:22=14β‹…124:4=1β‹…1244:4=1ΜΈ244ΜΈ:4=31:4=314=734 \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:

We received that:

116β‹…(125+3βˆ’16):22=116β‹…124:22=1β‹…1244:4=734 \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:

(52βˆ’3+6):7β‹…14 (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:

(52βˆ’3+6):7β‹…14=(25βˆ’3+6):7β‹…14=28:7β‹…14 (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:

28:7β‹…14=4β‹…14=4β‹…14=4ΜΈ4ΜΈ=1 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:

We received that:

(52βˆ’3+6):7β‹…14=28:7β‹…14=4ΜΈ4ΜΈ=1 (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:

116β‹…(125+3βˆ’16):22Β β€”(52βˆ’3+6):7β‹…14↓734Β β€”1 \frac{1}{\sqrt{16}}\cdot(125+3-\sqrt{16}):2^2\text{ }\textcolor{red}{_β€”}(5^2-3+6):7\cdot\frac{1}{4} \\ \downarrow\\ 7\frac{3}{4} \text{ }\textcolor{red}{_β€”}1 As a result, we receive that:

734Β β‰ 1 7\frac{3}{4} \text{ }\textcolor{red}{\neq}1 Therefore, the correct answer here is answer B.

Answer

β‰  \ne

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