Natural Numbers, Integers, and Real Numbers - Set Theory

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Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

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Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

This means that all elements in will be divisible by two and thus be an even number therefore, 3 will never be an element in .

Thus, it is concluded that

$T\neq S$

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Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

Tap to reveal answer

Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

Since four also belongs to $\mathbb{Z}$ this means that all elements in will be the same as those in .

Thus, it is concluded that

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Question

What is the correct expression of the relationships between the sets comprised of natural numbers, real numbers, and integers?

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Answer

The natural numbers are defined as $\mathbb{N}=\left \{ 0,1,2,3,...\right \}$ , the integers are defined as $\mathbb{Z}=\left \{ ...-3,-2,-1,0,1,2,3,...\right \}$ , and the real numbers ( $\mathbb{R}$ ) are defined as the set of all non-complex numbers. As such, $\mathbb{N}$ is a subset of $\mathbb{Z}$ , and $\mathbb{Z}$ is a subset of $\mathbb{R}$ .

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Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

Tap to reveal answer

Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

This means that all elements in will be divisible by two and thus be an even number therefore, 3 will never be an element in .

Thus, it is concluded that

$T\neq S$

← Didn't Know|Knew It →

Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

Tap to reveal answer

Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

Since four also belongs to $\mathbb{Z}$ this means that all elements in will be the same as those in .

Thus, it is concluded that

← Didn't Know|Knew It →

Question

What is the correct expression of the relationships between the sets comprised of natural numbers, real numbers, and integers?

Tap to reveal answer

Answer

The natural numbers are defined as $\mathbb{N}=\left \{ 0,1,2,3,...\right \}$ , the integers are defined as $\mathbb{Z}=\left \{ ...-3,-2,-1,0,1,2,3,...\right \}$ , and the real numbers ( $\mathbb{R}$ ) are defined as the set of all non-complex numbers. As such, $\mathbb{N}$ is a subset of $\mathbb{Z}$ , and $\mathbb{Z}$ is a subset of $\mathbb{R}$ .

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Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

Tap to reveal answer

Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

This means that all elements in will be divisible by two and thus be an even number therefore, 3 will never be an element in .

Thus, it is concluded that

$T\neq S$

← Didn't Know|Knew It →

Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

Tap to reveal answer

Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

Since four also belongs to $\mathbb{Z}$ this means that all elements in will be the same as those in .

Thus, it is concluded that

← Didn't Know|Knew It →

Question

What is the correct expression of the relationships between the sets comprised of natural numbers, real numbers, and integers?

Tap to reveal answer

Answer

The natural numbers are defined as $\mathbb{N}=\left \{ 0,1,2,3,...\right \}$ , the integers are defined as $\mathbb{Z}=\left \{ ...-3,-2,-1,0,1,2,3,...\right \}$ , and the real numbers ( $\mathbb{R}$ ) are defined as the set of all non-complex numbers. As such, $\mathbb{N}$ is a subset of $\mathbb{Z}$ , and $\mathbb{Z}$ is a subset of $\mathbb{R}$ .

← Didn't Know|Knew It →

Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

Tap to reveal answer

Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

This means that all elements in will be divisible by two and thus be an even number therefore, 3 will never be an element in .

Thus, it is concluded that

$T\neq S$

← Didn't Know|Knew It →

Question

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

Tap to reveal answer

Answer

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=$\text{Integers}{\begin{Bmatrix}$ ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=$\text{Reals}$ \\mathbb{N}=$\text{Naturals}$ {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 1+n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

Since four also belongs to $\mathbb{Z}$ this means that all elements in will be the same as those in .

Thus, it is concluded that

← Didn't Know|Knew It →

Question

What is the correct expression of the relationships between the sets comprised of natural numbers, real numbers, and integers?

Tap to reveal answer

Answer

The natural numbers are defined as $\mathbb{N}=\left \{ 0,1,2,3,...\right \}$ , the integers are defined as $\mathbb{Z}=\left \{ ...-3,-2,-1,0,1,2,3,...\right \}$ , and the real numbers ( $\mathbb{R}$ ) are defined as the set of all non-complex numbers. As such, $\mathbb{N}$ is a subset of $\mathbb{Z}$ , and $\mathbb{Z}$ is a subset of $\mathbb{R}$ .

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