Math / Numbers & Operations

QuestionМатематикт $а$ нь $A$ олонлогийн элемент мөн бол $a \in A$ , элемент биш бол $a \notin A$ гэж тэмдэглэх ба үүнийг $а$ элемент А олонлогт харьяалагдана ( харьяалагдахгүй ) гэж уншина. $A$ олонлог нь натурал тоон олонлог бол $2 \sqrt{3}$ гэсэн тоо $A$ олонлогт харьяалагдах уу? харьяалагдахгүй юу?

Studdy Solution

STEP 1

Assumptions
1. The set $A$ is the set of natural numbers.
2. Natural numbers are positive integers starting from 1, i.e., $\{1, 2, 3, \ldots\}$ .
3. We need to determine if the number $2 \sqrt{3}$ is an element of the set $A$ .

STEP 2

Calculate the value of $2 \sqrt{3}$ to determine its nature.
$2 \sqrt{3} = 2 \times \sqrt{3}$

STEP 3

Approximate the value of $\sqrt{3}$ .
$\sqrt{3} \approx 1.732$

STEP 4

Substitute the approximate value of $\sqrt{3}$ into the expression for $2 \sqrt{3}$ .
$2 \sqrt{3} \approx 2 \times 1.732$

STEP 5

Calculate the product to find the approximate value of $2 \sqrt{3}$ .
$2 \times 1.732 = 3.464$

STEP 6

Determine if $3.464$ is a natural number.
- A natural number is a positive integer. - $3.464$ is not an integer; it is a decimal number.

STEP 7

Conclude whether $2 \sqrt{3}$ is an element of the set $A$ .
Since $2 \sqrt{3} \approx 3.464$ is not an integer, it is not a natural number. Therefore, $2 \sqrt{3} \notin A$ .

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Studdy Solution

STEP 1

Assumptions
1. The set $A$ is the set of natural numbers.
2. Natural numbers are positive integers starting from 1, i.e., $\{1, 2, 3, \ldots\}$ .
3. We need to determine if the number $2 \sqrt{3}$ is an element of the set $A$ .

STEP 2

Calculate the value of $2 \sqrt{3}$ to determine its nature.
$2 \sqrt{3} = 2 \times \sqrt{3}$

STEP 3

Approximate the value of $\sqrt{3}$ .
$\sqrt{3} \approx 1.732$

STEP 4

Substitute the approximate value of $\sqrt{3}$ into the expression for $2 \sqrt{3}$ .
$2 \sqrt{3} \approx 2 \times 1.732$

STEP 5

Calculate the product to find the approximate value of $2 \sqrt{3}$ .
$2 \times 1.732 = 3.464$

STEP 6

Determine if $3.464$ is a natural number.
- A natural number is a positive integer. - $3.464$ is not an integer; it is a decimal number.

STEP 7

Conclude whether $2 \sqrt{3}$ is an element of the set $A$ .
Since $2 \sqrt{3} \approx 3.464$ is not an integer, it is not a natural number. Therefore, $2 \sqrt{3} \notin A$ .

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Studdy Solution

STEP 1

Assumptions1. The set AAA is the set of natural numbers.2. Natural numbers are positive integers starting from 1, i.e., {1,2,3,…}\{1, 2, 3, \ldots\}{1,2,3,…}.3. We need to determine if the number 232 \sqrt{3}23​ is an element of the set AAA.

STEP 2

Calculate the value of 232 \sqrt{3}23​ to determine its nature.23=2×32 \sqrt{3} = 2 \times \sqrt{3}23​=2×3​

STEP 3

Approximate the value of 3\sqrt{3}3​.3≈1.732\sqrt{3} \approx 1.7323​≈1.732

STEP 4

Substitute the approximate value of 3\sqrt{3}3​ into the expression for 232 \sqrt{3}23​.23≈2×1.7322 \sqrt{3} \approx 2 \times 1.73223​≈2×1.732

STEP 5

Calculate the product to find the approximate value of 232 \sqrt{3}23​.2×1.732=3.4642 \times 1.732 = 3.4642×1.732=3.464

STEP 6

Determine if 3.4643.4643.464 is a natural number.- A natural number is a positive integer. - 3.4643.4643.464 is not an integer; it is a decimal number.

STEP 7

Conclude whether 232 \sqrt{3}23​ is an element of the set AAA.Since 23≈3.4642 \sqrt{3} \approx 3.46423​≈3.464 is not an integer, it is not a natural number. Therefore, 23∉A2 \sqrt{3} \notin A23​∈/A.

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Assumptions
1. The set $A$ is the set of natural numbers.
2. Natural numbers are positive integers starting from 1, i.e., $\{1, 2, 3, \ldots\}$ .
3. We need to determine if the number $2 \sqrt{3}$ is an element of the set $A$ .

Calculate the value of $2 \sqrt{3}$ to determine its nature.
$2 \sqrt{3} = 2 \times \sqrt{3}$

Approximate the value of $\sqrt{3}$ .
$\sqrt{3} \approx 1.732$

Substitute the approximate value of $\sqrt{3}$ into the expression for $2 \sqrt{3}$ .
$2 \sqrt{3} \approx 2 \times 1.732$

Calculate the product to find the approximate value of $2 \sqrt{3}$ .
$2 \times 1.732 = 3.464$

Determine if $3.464$ is a natural number.
- A natural number is a positive integer. - $3.464$ is not an integer; it is a decimal number.

Conclude whether $2 \sqrt{3}$ is an element of the set $A$ .
Since $2 \sqrt{3} \approx 3.464$ is not an integer, it is not a natural number. Therefore, $2 \sqrt{3} \notin A$ .