Understanding Definition of Complex Number
$begingroup$
In the book "Linear Algebra Done Right", a complex number is defined as: "an ordered pair $(a, b)$ , where $a, b in Bbb{R}$, but we will write this as $a + bi$."
"The set of all complex numbers is denoted by $Bbb{C}$:
$$ Bbb{C}={a+bi : a, b in Bbb{R}} $$"
Where did this imaginary number come from? What is wrong just an ordered pair?
I am not understanding what the imaginary number has to do with representing an ordered pair. Please note that I am revisiting math years after taking any type of formal math course. Currently I am interesting in linear algebra, and interested in understanding all math.
complex-numbers
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
asked Jul 7 '18 at 5:25
craydencrayden
1134
$endgroup$
add a comment |
$begingroup$
In the book "Linear Algebra Done Right", a complex number is defined as: "an ordered pair $(a, b)$ , where $a, b in Bbb{R}$, but we will write this as $a + bi$."
"The set of all complex numbers is denoted by $Bbb{C}$:
$$ Bbb{C}={a+bi : a, b in Bbb{R}} $$"
Where did this imaginary number come from? What is wrong just an ordered pair?
I am not understanding what the imaginary number has to do with representing an ordered pair. Please note that I am revisiting math years after taking any type of formal math course. Currently I am interesting in linear algebra, and interested in understanding all math.
complex-numbers
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
asked Jul 7 '18 at 5:25
craydencrayden
1134
$endgroup$
$begingroup$
Where did this imaginary number come from?There is no reference to any "imaginary number" in the definition you quoted. It's just a matter of notation, like saying that the ordered pair $(a,b)$ will, for convenience, be hereafter written as $a$ ☆ + $b$ ★, where in this case ☆$=$ nothing, and ★$ = i$.
$endgroup$
– dxiv
Jul 7 '18 at 5:34
$begingroup$
$a+bi$ is just a synomym for $(a,b)$ to start with. But you will want to think of it later (when we have defined a multiplication for these pairs) as $a times 1 + b times i$, where $i$ is a shorthand for the pair $(0,1)$. We will identify all pairs $(a,0)$ with $a in mathbb{R}$ with $mathbb{R}$ itself, and extend the multiplication of pairs to extend the one on the reals.
$endgroup$
– Henno Brandsma
Jul 7 '18 at 5:46
$begingroup$
Just two different ways of writing the same idea. You are free to do this in Mathematics, so long as you clearly define what you mean, and you would only write such things if there's good justification to do so.
$endgroup$
– Antinous
Jul 7 '18 at 6:13
add a comment |
$begingroup$
In the book "Linear Algebra Done Right", a complex number is defined as: "an ordered pair $(a, b)$ , where $a, b in Bbb{R}$, but we will write this as $a + bi$."
"The set of all complex numbers is denoted by $Bbb{C}$:
$$ Bbb{C}={a+bi : a, b in Bbb{R}} $$"
Where did this imaginary number come from? What is wrong just an ordered pair?
I am not understanding what the imaginary number has to do with representing an ordered pair. Please note that I am revisiting math years after taking any type of formal math course. Currently I am interesting in linear algebra, and interested in understanding all math.
complex-numbers
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
asked Jul 7 '18 at 5:25
craydencrayden
1134
$endgroup$
In the book "Linear Algebra Done Right", a complex number is defined as: "an ordered pair $(a, b)$ , where $a, b in Bbb{R}$, but we will write this as $a + bi$."
"The set of all complex numbers is denoted by $Bbb{C}$:
$$ Bbb{C}={a+bi : a, b in Bbb{R}} $$"
Where did this imaginary number come from? What is wrong just an ordered pair?
I am not understanding what the imaginary number has to do with representing an ordered pair. Please note that I am revisiting math years after taking any type of formal math course. Currently I am interesting in linear algebra, and interested in understanding all math.
complex-numbers
complex-numbers
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
asked Jul 7 '18 at 5:25
craydencrayden
1134
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
asked Jul 7 '18 at 5:25
craydencrayden
1134
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
edited Dec 23 '18 at 3:32
Eric Wofsey
188k14216347
188k14216347
asked Jul 7 '18 at 5:25
craydencrayden
1134
asked Jul 7 '18 at 5:25
craydencrayden
1134
asked Jul 7 '18 at 5:25
craydencrayden
1134
1134
$begingroup$
Where did this imaginary number come from?There is no reference to any "imaginary number" in the definition you quoted. It's just a matter of notation, like saying that the ordered pair $(a,b)$ will, for convenience, be hereafter written as $a$ ☆ + $b$ ★, where in this case ☆$=$ nothing, and ★$ = i$.
$endgroup$
– dxiv
Jul 7 '18 at 5:34
$begingroup$
$a+bi$ is just a synomym for $(a,b)$ to start with. But you will want to think of it later (when we have defined a multiplication for these pairs) as $a times 1 + b times i$, where $i$ is a shorthand for the pair $(0,1)$. We will identify all pairs $(a,0)$ with $a in mathbb{R}$ with $mathbb{R}$ itself, and extend the multiplication of pairs to extend the one on the reals.
$endgroup$
– Henno Brandsma
Jul 7 '18 at 5:46
$begingroup$
Just two different ways of writing the same idea. You are free to do this in Mathematics, so long as you clearly define what you mean, and you would only write such things if there's good justification to do so.
$endgroup$
– Antinous
Jul 7 '18 at 6:13
add a comment |
$begingroup$
Where did this imaginary number come from?There is no reference to any "imaginary number" in the definition you quoted. It's just a matter of notation, like saying that the ordered pair $(a,b)$ will, for convenience, be hereafter written as $a$ ☆ + $b$ ★, where in this case ☆$=$ nothing, and ★$ = i$.
$endgroup$
– dxiv
Jul 7 '18 at 5:34
$begingroup$
$a+bi$ is just a synomym for $(a,b)$ to start with. But you will want to think of it later (when we have defined a multiplication for these pairs) as $a times 1 + b times i$, where $i$ is a shorthand for the pair $(0,1)$. We will identify all pairs $(a,0)$ with $a in mathbb{R}$ with $mathbb{R}$ itself, and extend the multiplication of pairs to extend the one on the reals.
$endgroup$
– Henno Brandsma
Jul 7 '18 at 5:46
$begingroup$
Just two different ways of writing the same idea. You are free to do this in Mathematics, so long as you clearly define what you mean, and you would only write such things if there's good justification to do so.
$endgroup$
– Antinous
Jul 7 '18 at 6:13
$begingroup$
Where did this imaginary number come from? There is no reference to any "imaginary number" in the definition you quoted. It's just a matter of notation, like saying that the ordered pair $(a,b)$ will, for convenience, be hereafter written as $a$ ☆ + $b$ ★, where in this case ☆$=$ nothing, and ★$ = i$.$endgroup$
– dxiv
Jul 7 '18 at 5:34
$begingroup$
Where did this imaginary number come from? There is no reference to any "imaginary number" in the definition you quoted. It's just a matter of notation, like saying that the ordered pair $(a,b)$ will, for convenience, be hereafter written as $a$ ☆ + $b$ ★, where in this case ☆$=$ nothing, and ★$ = i$.$endgroup$
– dxiv
Jul 7 '18 at 5:34
$begingroup$
$a+bi$ is just a synomym for $(a,b)$ to start with. But you will want to think of it later (when we have defined a multiplication for these pairs) as $a times 1 + b times i$, where $i$ is a shorthand for the pair $(0,1)$. We will identify all pairs $(a,0)$ with $a in mathbb{R}$ with $mathbb{R}$ itself, and extend the multiplication of pairs to extend the one on the reals.
$endgroup$
– Henno Brandsma
Jul 7 '18 at 5:46
$begingroup$
$a+bi$ is just a synomym for $(a,b)$ to start with. But you will want to think of it later (when we have defined a multiplication for these pairs) as $a times 1 + b times i$, where $i$ is a shorthand for the pair $(0,1)$. We will identify all pairs $(a,0)$ with $a in mathbb{R}$ with $mathbb{R}$ itself, and extend the multiplication of pairs to extend the one on the reals.
$endgroup$
– Henno Brandsma
Jul 7 '18 at 5:46
$begingroup$
Just two different ways of writing the same idea. You are free to do this in Mathematics, so long as you clearly define what you mean, and you would only write such things if there's good justification to do so.
$endgroup$
– Antinous
Jul 7 '18 at 6:13
$begingroup$
Just two different ways of writing the same idea. You are free to do this in Mathematics, so long as you clearly define what you mean, and you would only write such things if there's good justification to do so.
$endgroup$
– Antinous
Jul 7 '18 at 6:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There is a confluence of notation here. The pattern "$a + b i$" could be interpreted in two different ways:
- As defined by the passage you quoted: this is just a funny way to write $(a,b)$.
- An arithmetic expression denoting the result of adding $a$ and the product of $b$ and $i$
As it turns out, both of these interpretations mean the same thing if we define $i$ to mean the ordered pair $(0,1)$ and make the complex numbers a real algebra by embedding the real numbers in the complexes by $r mapsto (r,0)$.
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
$endgroup$
add a comment |
$begingroup$
This is just a definition, and not a proof. In this context, things do not need to "come from somewhere". You can think of it as a simple matter of notation, that is, how you represent things.
The text is just saying that you can equivalently represent any complex number as an ordered pair of real numbers. In fact, the same procedure is applied to other mathematical objects as well (e.g polynomials and matrices).
Finally, I would suggest how to look at the first 10 or 11 Chapters of 'Naive Set Theory' from Halmos if you want to get started with understanding mathematical formalism.
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
$endgroup$
add a comment |
$begingroup$
You can interpret the equivalente between $(a,b)$ and $a+ib$ in the following sense:
$(a,b)+(c,d) = (a+c,b+d)$
$(a+ib)+(c+id)=(a+c) + i(b+d)$ (as adition of polinomials!)
Do you see whats going on?
As we have a product in $mathbb{C}$, we have by definition of the product that
$(a,b)(c,d) = (ac-bd, ad+bc)$
but we don't have to remember that formula if we remember that $i^2=-1$ and we write
$(a+ib)(c+id)=ac+aid+ibc+i^2bd = (ac-bd) +i(ad+bc) $ which is the same as $(ac-bd,ad+bc)$ above.
So we prefer to write $a+ib$ because it allow us to extrapolate usual notions of algebra and remember less stuff.
Hope it helps.
answered Jul 7 '18 at 6:05
HeManHeMan
858832
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a confluence of notation here. The pattern "$a + b i$" could be interpreted in two different ways:
- As defined by the passage you quoted: this is just a funny way to write $(a,b)$.
- An arithmetic expression denoting the result of adding $a$ and the product of $b$ and $i$
As it turns out, both of these interpretations mean the same thing if we define $i$ to mean the ordered pair $(0,1)$ and make the complex numbers a real algebra by embedding the real numbers in the complexes by $r mapsto (r,0)$.
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
$endgroup$
add a comment |
$begingroup$
There is a confluence of notation here. The pattern "$a + b i$" could be interpreted in two different ways:
- As defined by the passage you quoted: this is just a funny way to write $(a,b)$.
- An arithmetic expression denoting the result of adding $a$ and the product of $b$ and $i$
As it turns out, both of these interpretations mean the same thing if we define $i$ to mean the ordered pair $(0,1)$ and make the complex numbers a real algebra by embedding the real numbers in the complexes by $r mapsto (r,0)$.
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
$endgroup$
add a comment |
$begingroup$
There is a confluence of notation here. The pattern "$a + b i$" could be interpreted in two different ways:
- As defined by the passage you quoted: this is just a funny way to write $(a,b)$.
- An arithmetic expression denoting the result of adding $a$ and the product of $b$ and $i$
As it turns out, both of these interpretations mean the same thing if we define $i$ to mean the ordered pair $(0,1)$ and make the complex numbers a real algebra by embedding the real numbers in the complexes by $r mapsto (r,0)$.
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
$endgroup$
There is a confluence of notation here. The pattern "$a + b i$" could be interpreted in two different ways:
- As defined by the passage you quoted: this is just a funny way to write $(a,b)$.
- An arithmetic expression denoting the result of adding $a$ and the product of $b$ and $i$
As it turns out, both of these interpretations mean the same thing if we define $i$ to mean the ordered pair $(0,1)$ and make the complex numbers a real algebra by embedding the real numbers in the complexes by $r mapsto (r,0)$.
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
answered Jul 7 '18 at 5:58
HurkylHurkyl
112k9120262
112k9120262
add a comment |
add a comment |
$begingroup$
This is just a definition, and not a proof. In this context, things do not need to "come from somewhere". You can think of it as a simple matter of notation, that is, how you represent things.
The text is just saying that you can equivalently represent any complex number as an ordered pair of real numbers. In fact, the same procedure is applied to other mathematical objects as well (e.g polynomials and matrices).
Finally, I would suggest how to look at the first 10 or 11 Chapters of 'Naive Set Theory' from Halmos if you want to get started with understanding mathematical formalism.
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
$endgroup$
add a comment |
$begingroup$
This is just a definition, and not a proof. In this context, things do not need to "come from somewhere". You can think of it as a simple matter of notation, that is, how you represent things.
The text is just saying that you can equivalently represent any complex number as an ordered pair of real numbers. In fact, the same procedure is applied to other mathematical objects as well (e.g polynomials and matrices).
Finally, I would suggest how to look at the first 10 or 11 Chapters of 'Naive Set Theory' from Halmos if you want to get started with understanding mathematical formalism.
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
$endgroup$
add a comment |
$begingroup$
This is just a definition, and not a proof. In this context, things do not need to "come from somewhere". You can think of it as a simple matter of notation, that is, how you represent things.
The text is just saying that you can equivalently represent any complex number as an ordered pair of real numbers. In fact, the same procedure is applied to other mathematical objects as well (e.g polynomials and matrices).
Finally, I would suggest how to look at the first 10 or 11 Chapters of 'Naive Set Theory' from Halmos if you want to get started with understanding mathematical formalism.
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
$endgroup$
This is just a definition, and not a proof. In this context, things do not need to "come from somewhere". You can think of it as a simple matter of notation, that is, how you represent things.
The text is just saying that you can equivalently represent any complex number as an ordered pair of real numbers. In fact, the same procedure is applied to other mathematical objects as well (e.g polynomials and matrices).
Finally, I would suggest how to look at the first 10 or 11 Chapters of 'Naive Set Theory' from Halmos if you want to get started with understanding mathematical formalism.
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
answered Jul 7 '18 at 5:40
Ariel SerranoniAriel Serranoni
3915
3915
add a comment |
add a comment |
$begingroup$
You can interpret the equivalente between $(a,b)$ and $a+ib$ in the following sense:
$(a,b)+(c,d) = (a+c,b+d)$
$(a+ib)+(c+id)=(a+c) + i(b+d)$ (as adition of polinomials!)
Do you see whats going on?
As we have a product in $mathbb{C}$, we have by definition of the product that
$(a,b)(c,d) = (ac-bd, ad+bc)$
but we don't have to remember that formula if we remember that $i^2=-1$ and we write
$(a+ib)(c+id)=ac+aid+ibc+i^2bd = (ac-bd) +i(ad+bc) $ which is the same as $(ac-bd,ad+bc)$ above.
So we prefer to write $a+ib$ because it allow us to extrapolate usual notions of algebra and remember less stuff.
Hope it helps.
answered Jul 7 '18 at 6:05
HeManHeMan
858832
$endgroup$
add a comment |
$begingroup$
You can interpret the equivalente between $(a,b)$ and $a+ib$ in the following sense:
$(a,b)+(c,d) = (a+c,b+d)$
$(a+ib)+(c+id)=(a+c) + i(b+d)$ (as adition of polinomials!)
Do you see whats going on?
As we have a product in $mathbb{C}$, we have by definition of the product that
$(a,b)(c,d) = (ac-bd, ad+bc)$
but we don't have to remember that formula if we remember that $i^2=-1$ and we write
$(a+ib)(c+id)=ac+aid+ibc+i^2bd = (ac-bd) +i(ad+bc) $ which is the same as $(ac-bd,ad+bc)$ above.
So we prefer to write $a+ib$ because it allow us to extrapolate usual notions of algebra and remember less stuff.
Hope it helps.
answered Jul 7 '18 at 6:05
HeManHeMan
858832
$endgroup$
add a comment |
$begingroup$
You can interpret the equivalente between $(a,b)$ and $a+ib$ in the following sense:
$(a,b)+(c,d) = (a+c,b+d)$
$(a+ib)+(c+id)=(a+c) + i(b+d)$ (as adition of polinomials!)
Do you see whats going on?
As we have a product in $mathbb{C}$, we have by definition of the product that
$(a,b)(c,d) = (ac-bd, ad+bc)$
but we don't have to remember that formula if we remember that $i^2=-1$ and we write
$(a+ib)(c+id)=ac+aid+ibc+i^2bd = (ac-bd) +i(ad+bc) $ which is the same as $(ac-bd,ad+bc)$ above.
So we prefer to write $a+ib$ because it allow us to extrapolate usual notions of algebra and remember less stuff.
Hope it helps.
answered Jul 7 '18 at 6:05
HeManHeMan
858832
$endgroup$
You can interpret the equivalente between $(a,b)$ and $a+ib$ in the following sense:
$(a,b)+(c,d) = (a+c,b+d)$
$(a+ib)+(c+id)=(a+c) + i(b+d)$ (as adition of polinomials!)
Do you see whats going on?
As we have a product in $mathbb{C}$, we have by definition of the product that
$(a,b)(c,d) = (ac-bd, ad+bc)$
but we don't have to remember that formula if we remember that $i^2=-1$ and we write
$(a+ib)(c+id)=ac+aid+ibc+i^2bd = (ac-bd) +i(ad+bc) $ which is the same as $(ac-bd,ad+bc)$ above.
So we prefer to write $a+ib$ because it allow us to extrapolate usual notions of algebra and remember less stuff.
Hope it helps.
answered Jul 7 '18 at 6:05
HeManHeMan
858832
answered Jul 7 '18 at 6:05
HeManHeMan
858832
answered Jul 7 '18 at 6:05
HeManHeMan
858832
answered Jul 7 '18 at 6:05
HeManHeMan
858832
858832
add a comment |
add a comment |
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$begingroup$
Where did this imaginary number come from?There is no reference to any "imaginary number" in the definition you quoted. It's just a matter of notation, like saying that the ordered pair $(a,b)$ will, for convenience, be hereafter written as $a$ ☆ + $b$ ★, where in this case ☆$=$ nothing, and ★$ = i$.$endgroup$
– dxiv
Jul 7 '18 at 5:34
$begingroup$
$a+bi$ is just a synomym for $(a,b)$ to start with. But you will want to think of it later (when we have defined a multiplication for these pairs) as $a times 1 + b times i$, where $i$ is a shorthand for the pair $(0,1)$. We will identify all pairs $(a,0)$ with $a in mathbb{R}$ with $mathbb{R}$ itself, and extend the multiplication of pairs to extend the one on the reals.
$endgroup$
– Henno Brandsma
Jul 7 '18 at 5:46
$begingroup$
Just two different ways of writing the same idea. You are free to do this in Mathematics, so long as you clearly define what you mean, and you would only write such things if there's good justification to do so.
$endgroup$
– Antinous
Jul 7 '18 at 6:13