When do we have to use this technique when writing a proof?
$begingroup$
Say I need to prove that $forall z in Bbb R$, $z^2 - 2z + 7 > 1$.
Now in my solutions, to start the proof it says "let $a in Bbb R$. We need to show that $a^2 - 2a + 7 > 1$ and then just went on to prove it.. but using $a$ instead of $z$. I think it has something to do with showing it works for any arbitrary number or something like that but I'm not sure why. Thanks!
algebra-precalculus
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
asked Dec 8 '18 at 0:13
mingming
3365
$endgroup$
add a comment |
$begingroup$
Say I need to prove that $forall z in Bbb R$, $z^2 - 2z + 7 > 1$.
Now in my solutions, to start the proof it says "let $a in Bbb R$. We need to show that $a^2 - 2a + 7 > 1$ and then just went on to prove it.. but using $a$ instead of $z$. I think it has something to do with showing it works for any arbitrary number or something like that but I'm not sure why. Thanks!
algebra-precalculus
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
asked Dec 8 '18 at 0:13
mingming
3365
$endgroup$
1
$begingroup$
They are just trying to stress that they are working a fixed value for $z$, it's not important. You could just write $z^2-2z+7=z^2-2z+1+6=(z-1)^2+6≥6>1$.
$endgroup$
– lulu
Dec 8 '18 at 0:17
$begingroup$
For future reference, use "in" for $in$ symbol! :)
$endgroup$
– BSplitter
Dec 8 '18 at 0:28
$begingroup$
oh shoot, sorry bout that!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
$begingroup$
Say I need to prove that $forall z in Bbb R$, $z^2 - 2z + 7 > 1$.
Now in my solutions, to start the proof it says "let $a in Bbb R$. We need to show that $a^2 - 2a + 7 > 1$ and then just went on to prove it.. but using $a$ instead of $z$. I think it has something to do with showing it works for any arbitrary number or something like that but I'm not sure why. Thanks!
algebra-precalculus
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
asked Dec 8 '18 at 0:13
mingming
3365
$endgroup$
Say I need to prove that $forall z in Bbb R$, $z^2 - 2z + 7 > 1$.
Now in my solutions, to start the proof it says "let $a in Bbb R$. We need to show that $a^2 - 2a + 7 > 1$ and then just went on to prove it.. but using $a$ instead of $z$. I think it has something to do with showing it works for any arbitrary number or something like that but I'm not sure why. Thanks!
algebra-precalculus
algebra-precalculus
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
asked Dec 8 '18 at 0:13
mingming
3365
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
asked Dec 8 '18 at 0:13
mingming
3365
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
edited Dec 8 '18 at 4:04
Arturo Magidin
262k34586910
262k34586910
asked Dec 8 '18 at 0:13
mingming
3365
asked Dec 8 '18 at 0:13
mingming
3365
asked Dec 8 '18 at 0:13
mingming
3365
3365
1
$begingroup$
They are just trying to stress that they are working a fixed value for $z$, it's not important. You could just write $z^2-2z+7=z^2-2z+1+6=(z-1)^2+6≥6>1$.
$endgroup$
– lulu
Dec 8 '18 at 0:17
$begingroup$
For future reference, use "in" for $in$ symbol! :)
$endgroup$
– BSplitter
Dec 8 '18 at 0:28
$begingroup$
oh shoot, sorry bout that!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
1
$begingroup$
They are just trying to stress that they are working a fixed value for $z$, it's not important. You could just write $z^2-2z+7=z^2-2z+1+6=(z-1)^2+6≥6>1$.
$endgroup$
– lulu
Dec 8 '18 at 0:17
$begingroup$
For future reference, use "in" for $in$ symbol! :)
$endgroup$
– BSplitter
Dec 8 '18 at 0:28
$begingroup$
oh shoot, sorry bout that!
$endgroup$
– ming
Dec 8 '18 at 0:54
1
1
$begingroup$
They are just trying to stress that they are working a fixed value for $z$, it's not important. You could just write $z^2-2z+7=z^2-2z+1+6=(z-1)^2+6≥6>1$.
$endgroup$
– lulu
Dec 8 '18 at 0:17
$begingroup$
They are just trying to stress that they are working a fixed value for $z$, it's not important. You could just write $z^2-2z+7=z^2-2z+1+6=(z-1)^2+6≥6>1$.
$endgroup$
– lulu
Dec 8 '18 at 0:17
$begingroup$
For future reference, use "in" for $in$ symbol! :)
$endgroup$
– BSplitter
Dec 8 '18 at 0:28
$begingroup$
For future reference, use "in" for $in$ symbol! :)
$endgroup$
– BSplitter
Dec 8 '18 at 0:28
$begingroup$
oh shoot, sorry bout that!
$endgroup$
– ming
Dec 8 '18 at 0:54
$begingroup$
oh shoot, sorry bout that!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's more or less really pointless. Let me give an example: what is the difference between these two expressions?
$$2x + 6 = 12 ;;; text{and} ;;; 2b + 6 = 12$$
The only difference is what we call the variable.
Now, I want to emphasize -- there are some occasions where you don't want to change the name of the variable, which is usually a matter of context and conventions. For example, we normally let $i^2 = -1$ in the complex numbers, but we don't have to say it's $i$. We could name the unit whatever we want: $j, k, theta, epsilon,Gamma$, whatever. It's just a name. But the thing is that $i$ is more or less standard (there are fields that use $j$ instead of $i$ because of $i$ being relegated to a different meaning, for example). Why is it the standard? It's probably just how people initially referred to it, and it has that nice synonymity with "imaginary," since "i" is the first letter of the word. And eventually it just spread and became the standard.
So in that way we say that there are normal standards and conventions. $(x,y)$ denotes a point on the Cartesian plane, $z$ typically a complex number, $pi = 3.14159...$, etc. etc. etc.
I'm emphasizing this exception to "names are arbitrary" strongly because obeying a known convention aids in communication. Let $chi$ represent $pi$ if you want, just don't expect people to understand you, you know? So when I say "names are arbitrary," I mean so with the tenet and warning that conventions are important as well and should be observed where helpful.
But that aside, the name of a variable is arbitrary. You can call a variable whatever you want, it doesn't change the underlying meaning in any inherent way.
The argument can also be made that using $a$ instead of $z$ was done to try and distance oneself from hinging on $z$ or whatever. It's sort of like saying "we take this one real number and call it $a$ instead of $z$, mostly to distinguish that there is no connection or underlying reason we use either naming convention (if there is one at all." And then they prove it for arbitrary real $a$, which implies the theorem for all $z in mathbb{R}$.
Not that the name of the thing actually matters. You could read $a$ or $z$ as "element" (of the reals) and the result would be the same. The proof also could have been done using $z$ as the naming of the variable, or the theorem with $a$ - again, totally arbitrary. I think the author may have just been trying to distance themselves from any potential underlying meanings of $z$ or $a$ by using different names, but it's not really necessary.
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
$endgroup$
$begingroup$
Thanks for the write-up!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
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$begingroup$
It's more or less really pointless. Let me give an example: what is the difference between these two expressions?
$$2x + 6 = 12 ;;; text{and} ;;; 2b + 6 = 12$$
The only difference is what we call the variable.
Now, I want to emphasize -- there are some occasions where you don't want to change the name of the variable, which is usually a matter of context and conventions. For example, we normally let $i^2 = -1$ in the complex numbers, but we don't have to say it's $i$. We could name the unit whatever we want: $j, k, theta, epsilon,Gamma$, whatever. It's just a name. But the thing is that $i$ is more or less standard (there are fields that use $j$ instead of $i$ because of $i$ being relegated to a different meaning, for example). Why is it the standard? It's probably just how people initially referred to it, and it has that nice synonymity with "imaginary," since "i" is the first letter of the word. And eventually it just spread and became the standard.
So in that way we say that there are normal standards and conventions. $(x,y)$ denotes a point on the Cartesian plane, $z$ typically a complex number, $pi = 3.14159...$, etc. etc. etc.
I'm emphasizing this exception to "names are arbitrary" strongly because obeying a known convention aids in communication. Let $chi$ represent $pi$ if you want, just don't expect people to understand you, you know? So when I say "names are arbitrary," I mean so with the tenet and warning that conventions are important as well and should be observed where helpful.
But that aside, the name of a variable is arbitrary. You can call a variable whatever you want, it doesn't change the underlying meaning in any inherent way.
The argument can also be made that using $a$ instead of $z$ was done to try and distance oneself from hinging on $z$ or whatever. It's sort of like saying "we take this one real number and call it $a$ instead of $z$, mostly to distinguish that there is no connection or underlying reason we use either naming convention (if there is one at all." And then they prove it for arbitrary real $a$, which implies the theorem for all $z in mathbb{R}$.
Not that the name of the thing actually matters. You could read $a$ or $z$ as "element" (of the reals) and the result would be the same. The proof also could have been done using $z$ as the naming of the variable, or the theorem with $a$ - again, totally arbitrary. I think the author may have just been trying to distance themselves from any potential underlying meanings of $z$ or $a$ by using different names, but it's not really necessary.
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
$endgroup$
$begingroup$
Thanks for the write-up!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
$begingroup$
It's more or less really pointless. Let me give an example: what is the difference between these two expressions?
$$2x + 6 = 12 ;;; text{and} ;;; 2b + 6 = 12$$
The only difference is what we call the variable.
Now, I want to emphasize -- there are some occasions where you don't want to change the name of the variable, which is usually a matter of context and conventions. For example, we normally let $i^2 = -1$ in the complex numbers, but we don't have to say it's $i$. We could name the unit whatever we want: $j, k, theta, epsilon,Gamma$, whatever. It's just a name. But the thing is that $i$ is more or less standard (there are fields that use $j$ instead of $i$ because of $i$ being relegated to a different meaning, for example). Why is it the standard? It's probably just how people initially referred to it, and it has that nice synonymity with "imaginary," since "i" is the first letter of the word. And eventually it just spread and became the standard.
So in that way we say that there are normal standards and conventions. $(x,y)$ denotes a point on the Cartesian plane, $z$ typically a complex number, $pi = 3.14159...$, etc. etc. etc.
I'm emphasizing this exception to "names are arbitrary" strongly because obeying a known convention aids in communication. Let $chi$ represent $pi$ if you want, just don't expect people to understand you, you know? So when I say "names are arbitrary," I mean so with the tenet and warning that conventions are important as well and should be observed where helpful.
But that aside, the name of a variable is arbitrary. You can call a variable whatever you want, it doesn't change the underlying meaning in any inherent way.
The argument can also be made that using $a$ instead of $z$ was done to try and distance oneself from hinging on $z$ or whatever. It's sort of like saying "we take this one real number and call it $a$ instead of $z$, mostly to distinguish that there is no connection or underlying reason we use either naming convention (if there is one at all." And then they prove it for arbitrary real $a$, which implies the theorem for all $z in mathbb{R}$.
Not that the name of the thing actually matters. You could read $a$ or $z$ as "element" (of the reals) and the result would be the same. The proof also could have been done using $z$ as the naming of the variable, or the theorem with $a$ - again, totally arbitrary. I think the author may have just been trying to distance themselves from any potential underlying meanings of $z$ or $a$ by using different names, but it's not really necessary.
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
$endgroup$
$begingroup$
Thanks for the write-up!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
$begingroup$
It's more or less really pointless. Let me give an example: what is the difference between these two expressions?
$$2x + 6 = 12 ;;; text{and} ;;; 2b + 6 = 12$$
The only difference is what we call the variable.
Now, I want to emphasize -- there are some occasions where you don't want to change the name of the variable, which is usually a matter of context and conventions. For example, we normally let $i^2 = -1$ in the complex numbers, but we don't have to say it's $i$. We could name the unit whatever we want: $j, k, theta, epsilon,Gamma$, whatever. It's just a name. But the thing is that $i$ is more or less standard (there are fields that use $j$ instead of $i$ because of $i$ being relegated to a different meaning, for example). Why is it the standard? It's probably just how people initially referred to it, and it has that nice synonymity with "imaginary," since "i" is the first letter of the word. And eventually it just spread and became the standard.
So in that way we say that there are normal standards and conventions. $(x,y)$ denotes a point on the Cartesian plane, $z$ typically a complex number, $pi = 3.14159...$, etc. etc. etc.
I'm emphasizing this exception to "names are arbitrary" strongly because obeying a known convention aids in communication. Let $chi$ represent $pi$ if you want, just don't expect people to understand you, you know? So when I say "names are arbitrary," I mean so with the tenet and warning that conventions are important as well and should be observed where helpful.
But that aside, the name of a variable is arbitrary. You can call a variable whatever you want, it doesn't change the underlying meaning in any inherent way.
The argument can also be made that using $a$ instead of $z$ was done to try and distance oneself from hinging on $z$ or whatever. It's sort of like saying "we take this one real number and call it $a$ instead of $z$, mostly to distinguish that there is no connection or underlying reason we use either naming convention (if there is one at all." And then they prove it for arbitrary real $a$, which implies the theorem for all $z in mathbb{R}$.
Not that the name of the thing actually matters. You could read $a$ or $z$ as "element" (of the reals) and the result would be the same. The proof also could have been done using $z$ as the naming of the variable, or the theorem with $a$ - again, totally arbitrary. I think the author may have just been trying to distance themselves from any potential underlying meanings of $z$ or $a$ by using different names, but it's not really necessary.
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
$endgroup$
It's more or less really pointless. Let me give an example: what is the difference between these two expressions?
$$2x + 6 = 12 ;;; text{and} ;;; 2b + 6 = 12$$
The only difference is what we call the variable.
Now, I want to emphasize -- there are some occasions where you don't want to change the name of the variable, which is usually a matter of context and conventions. For example, we normally let $i^2 = -1$ in the complex numbers, but we don't have to say it's $i$. We could name the unit whatever we want: $j, k, theta, epsilon,Gamma$, whatever. It's just a name. But the thing is that $i$ is more or less standard (there are fields that use $j$ instead of $i$ because of $i$ being relegated to a different meaning, for example). Why is it the standard? It's probably just how people initially referred to it, and it has that nice synonymity with "imaginary," since "i" is the first letter of the word. And eventually it just spread and became the standard.
So in that way we say that there are normal standards and conventions. $(x,y)$ denotes a point on the Cartesian plane, $z$ typically a complex number, $pi = 3.14159...$, etc. etc. etc.
I'm emphasizing this exception to "names are arbitrary" strongly because obeying a known convention aids in communication. Let $chi$ represent $pi$ if you want, just don't expect people to understand you, you know? So when I say "names are arbitrary," I mean so with the tenet and warning that conventions are important as well and should be observed where helpful.
But that aside, the name of a variable is arbitrary. You can call a variable whatever you want, it doesn't change the underlying meaning in any inherent way.
The argument can also be made that using $a$ instead of $z$ was done to try and distance oneself from hinging on $z$ or whatever. It's sort of like saying "we take this one real number and call it $a$ instead of $z$, mostly to distinguish that there is no connection or underlying reason we use either naming convention (if there is one at all." And then they prove it for arbitrary real $a$, which implies the theorem for all $z in mathbb{R}$.
Not that the name of the thing actually matters. You could read $a$ or $z$ as "element" (of the reals) and the result would be the same. The proof also could have been done using $z$ as the naming of the variable, or the theorem with $a$ - again, totally arbitrary. I think the author may have just been trying to distance themselves from any potential underlying meanings of $z$ or $a$ by using different names, but it's not really necessary.
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
answered Dec 8 '18 at 0:50
Eevee TrainerEevee Trainer
5,7981936
5,7981936
$begingroup$
Thanks for the write-up!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
$begingroup$
Thanks for the write-up!
$endgroup$
– ming
Dec 8 '18 at 0:54
$begingroup$
Thanks for the write-up!
$endgroup$
– ming
Dec 8 '18 at 0:54
$begingroup$
Thanks for the write-up!
$endgroup$
– ming
Dec 8 '18 at 0:54
add a comment |
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$begingroup$
They are just trying to stress that they are working a fixed value for $z$, it's not important. You could just write $z^2-2z+7=z^2-2z+1+6=(z-1)^2+6≥6>1$.
$endgroup$
– lulu
Dec 8 '18 at 0:17
$begingroup$
For future reference, use "in" for $in$ symbol! :)
$endgroup$
– BSplitter
Dec 8 '18 at 0:28
$begingroup$
oh shoot, sorry bout that!
$endgroup$
– ming
Dec 8 '18 at 0:54