If the law says “Provide a valid solution for 0x = 50, or go to jail”, can we avoid being jailed?...
$begingroup$
Law Stack Exchange has a question about a hypothetical law that everyone is meant to be guilty of breaking.
What if a law is literally impossible to follow?
While that's an interesting idea, they used an example law which touched on mathematics that I'm not sure would keep our better mathematicians behind bars.
Every citizen is required to provide the government with a valid solution to the equation 0x = 50, and will otherwise be jailed.
I remember $0 cdot infty$ as an indeterminate form, meaning you can't say it is or isn't something without more information. But can't I insist that, whatever $x$ is, it's something that makes this true?
$$0x = 50 iff lim_{|y|toinfty} left(frac{50}{y} cdot y right) = 50$$
I mean, they didn't even say $x$ had to be a real number. Am I crazy, or are mathematicians about to rule the world?
To make this question a little more answerable: What's the simplest way to show that the law isn't fulfilling it's objective?
infinity
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
$endgroup$
closed as off-topic by Lord Shark the Unknown, Eevee Trainer, nbarto, Did, BigbearZzz Dec 16 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – Lord Shark the Unknown, nbarto, Did, BigbearZzz
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 13 more comments
$begingroup$
Law Stack Exchange has a question about a hypothetical law that everyone is meant to be guilty of breaking.
What if a law is literally impossible to follow?
While that's an interesting idea, they used an example law which touched on mathematics that I'm not sure would keep our better mathematicians behind bars.
Every citizen is required to provide the government with a valid solution to the equation 0x = 50, and will otherwise be jailed.
I remember $0 cdot infty$ as an indeterminate form, meaning you can't say it is or isn't something without more information. But can't I insist that, whatever $x$ is, it's something that makes this true?
$$0x = 50 iff lim_{|y|toinfty} left(frac{50}{y} cdot y right) = 50$$
I mean, they didn't even say $x$ had to be a real number. Am I crazy, or are mathematicians about to rule the world?
To make this question a little more answerable: What's the simplest way to show that the law isn't fulfilling it's objective?
infinity
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
$endgroup$
closed as off-topic by Lord Shark the Unknown, Eevee Trainer, nbarto, Did, BigbearZzz Dec 16 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – Lord Shark the Unknown, nbarto, Did, BigbearZzz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
How did you get that limit expression exactly? I fail to see how that limit is analogous to $0 cdot infty$.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:07
1
$begingroup$
Also, a personal note: I think you're overthinking the example. There are absolutely topologies of the real numbers in which "nonsense" expressions like $0 cdot x = 50$ have solutions - indeed, we can even define systems in which division by $0$ is defined (though results in things we do not desire, if just in application). But actually trying to consider the logic of the equation is probably well beyond what the post intended.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:08
$begingroup$
$0x=50implies x=50 because050=50$
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:11
1
$begingroup$
There is no real number $x$ such that $0 cdot x = 50$. If a law required you to produce such a real number, the law could not be obeyed.
$endgroup$
– littleO
Dec 16 '18 at 9:22
1
$begingroup$
One could always artificially introduce a new number (let's call it $alpha$) and extend the definition of multiplication so that $0 times alpha = 50$, but this would be kind of silly (and the resulting number system would not be interesting or useful). Surely the intention was that the law requires citizens to provide a real number $x$ such that $0 cdot x = 50$.
$endgroup$
– littleO
Dec 16 '18 at 9:35
|
show 13 more comments
$begingroup$
Law Stack Exchange has a question about a hypothetical law that everyone is meant to be guilty of breaking.
What if a law is literally impossible to follow?
While that's an interesting idea, they used an example law which touched on mathematics that I'm not sure would keep our better mathematicians behind bars.
Every citizen is required to provide the government with a valid solution to the equation 0x = 50, and will otherwise be jailed.
I remember $0 cdot infty$ as an indeterminate form, meaning you can't say it is or isn't something without more information. But can't I insist that, whatever $x$ is, it's something that makes this true?
$$0x = 50 iff lim_{|y|toinfty} left(frac{50}{y} cdot y right) = 50$$
I mean, they didn't even say $x$ had to be a real number. Am I crazy, or are mathematicians about to rule the world?
To make this question a little more answerable: What's the simplest way to show that the law isn't fulfilling it's objective?
infinity
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
$endgroup$
Law Stack Exchange has a question about a hypothetical law that everyone is meant to be guilty of breaking.
What if a law is literally impossible to follow?
While that's an interesting idea, they used an example law which touched on mathematics that I'm not sure would keep our better mathematicians behind bars.
Every citizen is required to provide the government with a valid solution to the equation 0x = 50, and will otherwise be jailed.
I remember $0 cdot infty$ as an indeterminate form, meaning you can't say it is or isn't something without more information. But can't I insist that, whatever $x$ is, it's something that makes this true?
$$0x = 50 iff lim_{|y|toinfty} left(frac{50}{y} cdot y right) = 50$$
I mean, they didn't even say $x$ had to be a real number. Am I crazy, or are mathematicians about to rule the world?
To make this question a little more answerable: What's the simplest way to show that the law isn't fulfilling it's objective?
infinity
infinity
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
edited Dec 16 '18 at 9:36
candied_orange
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
asked Dec 16 '18 at 9:03
candied_orangecandied_orange
466410
466410
closed as off-topic by Lord Shark the Unknown, Eevee Trainer, nbarto, Did, BigbearZzz Dec 16 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – Lord Shark the Unknown, nbarto, Did, BigbearZzz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Lord Shark the Unknown, Eevee Trainer, nbarto, Did, BigbearZzz Dec 16 '18 at 12:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – Lord Shark the Unknown, nbarto, Did, BigbearZzz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
How did you get that limit expression exactly? I fail to see how that limit is analogous to $0 cdot infty$.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:07
1
$begingroup$
Also, a personal note: I think you're overthinking the example. There are absolutely topologies of the real numbers in which "nonsense" expressions like $0 cdot x = 50$ have solutions - indeed, we can even define systems in which division by $0$ is defined (though results in things we do not desire, if just in application). But actually trying to consider the logic of the equation is probably well beyond what the post intended.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:08
$begingroup$
$0x=50implies x=50 because050=50$
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:11
1
$begingroup$
There is no real number $x$ such that $0 cdot x = 50$. If a law required you to produce such a real number, the law could not be obeyed.
$endgroup$
– littleO
Dec 16 '18 at 9:22
1
$begingroup$
One could always artificially introduce a new number (let's call it $alpha$) and extend the definition of multiplication so that $0 times alpha = 50$, but this would be kind of silly (and the resulting number system would not be interesting or useful). Surely the intention was that the law requires citizens to provide a real number $x$ such that $0 cdot x = 50$.
$endgroup$
– littleO
Dec 16 '18 at 9:35
|
show 13 more comments
1
$begingroup$
How did you get that limit expression exactly? I fail to see how that limit is analogous to $0 cdot infty$.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:07
1
$begingroup$
Also, a personal note: I think you're overthinking the example. There are absolutely topologies of the real numbers in which "nonsense" expressions like $0 cdot x = 50$ have solutions - indeed, we can even define systems in which division by $0$ is defined (though results in things we do not desire, if just in application). But actually trying to consider the logic of the equation is probably well beyond what the post intended.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:08
$begingroup$
$0x=50implies x=50 because050=50$
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:11
1
$begingroup$
There is no real number $x$ such that $0 cdot x = 50$. If a law required you to produce such a real number, the law could not be obeyed.
$endgroup$
– littleO
Dec 16 '18 at 9:22
1
$begingroup$
One could always artificially introduce a new number (let's call it $alpha$) and extend the definition of multiplication so that $0 times alpha = 50$, but this would be kind of silly (and the resulting number system would not be interesting or useful). Surely the intention was that the law requires citizens to provide a real number $x$ such that $0 cdot x = 50$.
$endgroup$
– littleO
Dec 16 '18 at 9:35
1
1
$begingroup$
How did you get that limit expression exactly? I fail to see how that limit is analogous to $0 cdot infty$.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:07
$begingroup$
How did you get that limit expression exactly? I fail to see how that limit is analogous to $0 cdot infty$.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:07
1
1
$begingroup$
Also, a personal note: I think you're overthinking the example. There are absolutely topologies of the real numbers in which "nonsense" expressions like $0 cdot x = 50$ have solutions - indeed, we can even define systems in which division by $0$ is defined (though results in things we do not desire, if just in application). But actually trying to consider the logic of the equation is probably well beyond what the post intended.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:08
$begingroup$
Also, a personal note: I think you're overthinking the example. There are absolutely topologies of the real numbers in which "nonsense" expressions like $0 cdot x = 50$ have solutions - indeed, we can even define systems in which division by $0$ is defined (though results in things we do not desire, if just in application). But actually trying to consider the logic of the equation is probably well beyond what the post intended.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:08
$begingroup$
$0x=50implies x=50 because050=50$
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:11
$begingroup$
$0x=50implies x=50 because050=50$
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:11
1
1
$begingroup$
There is no real number $x$ such that $0 cdot x = 50$. If a law required you to produce such a real number, the law could not be obeyed.
$endgroup$
– littleO
Dec 16 '18 at 9:22
$begingroup$
There is no real number $x$ such that $0 cdot x = 50$. If a law required you to produce such a real number, the law could not be obeyed.
$endgroup$
– littleO
Dec 16 '18 at 9:22
1
1
$begingroup$
One could always artificially introduce a new number (let's call it $alpha$) and extend the definition of multiplication so that $0 times alpha = 50$, but this would be kind of silly (and the resulting number system would not be interesting or useful). Surely the intention was that the law requires citizens to provide a real number $x$ such that $0 cdot x = 50$.
$endgroup$
– littleO
Dec 16 '18 at 9:35
$begingroup$
One could always artificially introduce a new number (let's call it $alpha$) and extend the definition of multiplication so that $0 times alpha = 50$, but this would be kind of silly (and the resulting number system would not be interesting or useful). Surely the intention was that the law requires citizens to provide a real number $x$ such that $0 cdot x = 50$.
$endgroup$
– littleO
Dec 16 '18 at 9:35
|
show 13 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Don't confuse the two next equations :
$$0 text{ multiplied by }x=50 tag 1$$
$$0 text{ concatenated with }x=50 tag 2$$
Both are loosely written as $0x=50$. That's the trap.
Of course $(1)$ has no finite solution.
Obviously the solution of $(2)$ is $x=50$ because
$$0 text{ concatenated with }50=050=50.$$
The ambiguous question implicitly refers to $(2)$, not to $(1)$ and the expected answer is $x=50$.
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
$endgroup$
1
$begingroup$
I gave the same answer but someone other than the OP edited the question to make it $0cdot x$, invalidating my(our) response.
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:36
$begingroup$
This is not at all what I had in mind and that's wonderful. I've edited the question (twice now) to be faithful with the post on law. The only issue is now my question needs better tags.
$endgroup$
– candied_orange
Dec 16 '18 at 9:55
$begingroup$
@ Shubham Johri. I fully agree with you. That's a drawback of StackExchange forum : Somebody can modify the wording of a question or of an answer without always fully understanding what really is the meaning. Fortunately it is not often that happens.
$endgroup$
– JJacquelin
Dec 16 '18 at 10:02
$begingroup$
@ShubhamJohri, et al. I edited the question title and another instance to match what was (at the time) the equation in the yellow "quotation box". I should've examined the question history to see if that notation had previously been changed from OP's original intent. That said ... I prefer the version with the "$cdot$", as it leaves open the possibility of defining the operation (say, as concatenation, or addition, or the "second operand" function $acdot b =b$, or, or, or, ...). :)
$endgroup$
– Blue
Dec 16 '18 at 10:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Don't confuse the two next equations :
$$0 text{ multiplied by }x=50 tag 1$$
$$0 text{ concatenated with }x=50 tag 2$$
Both are loosely written as $0x=50$. That's the trap.
Of course $(1)$ has no finite solution.
Obviously the solution of $(2)$ is $x=50$ because
$$0 text{ concatenated with }50=050=50.$$
The ambiguous question implicitly refers to $(2)$, not to $(1)$ and the expected answer is $x=50$.
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
$endgroup$
1
$begingroup$
I gave the same answer but someone other than the OP edited the question to make it $0cdot x$, invalidating my(our) response.
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:36
$begingroup$
This is not at all what I had in mind and that's wonderful. I've edited the question (twice now) to be faithful with the post on law. The only issue is now my question needs better tags.
$endgroup$
– candied_orange
Dec 16 '18 at 9:55
$begingroup$
@ Shubham Johri. I fully agree with you. That's a drawback of StackExchange forum : Somebody can modify the wording of a question or of an answer without always fully understanding what really is the meaning. Fortunately it is not often that happens.
$endgroup$
– JJacquelin
Dec 16 '18 at 10:02
$begingroup$
@ShubhamJohri, et al. I edited the question title and another instance to match what was (at the time) the equation in the yellow "quotation box". I should've examined the question history to see if that notation had previously been changed from OP's original intent. That said ... I prefer the version with the "$cdot$", as it leaves open the possibility of defining the operation (say, as concatenation, or addition, or the "second operand" function $acdot b =b$, or, or, or, ...). :)
$endgroup$
– Blue
Dec 16 '18 at 10:18
add a comment |
$begingroup$
Don't confuse the two next equations :
$$0 text{ multiplied by }x=50 tag 1$$
$$0 text{ concatenated with }x=50 tag 2$$
Both are loosely written as $0x=50$. That's the trap.
Of course $(1)$ has no finite solution.
Obviously the solution of $(2)$ is $x=50$ because
$$0 text{ concatenated with }50=050=50.$$
The ambiguous question implicitly refers to $(2)$, not to $(1)$ and the expected answer is $x=50$.
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
$endgroup$
1
$begingroup$
I gave the same answer but someone other than the OP edited the question to make it $0cdot x$, invalidating my(our) response.
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:36
$begingroup$
This is not at all what I had in mind and that's wonderful. I've edited the question (twice now) to be faithful with the post on law. The only issue is now my question needs better tags.
$endgroup$
– candied_orange
Dec 16 '18 at 9:55
$begingroup$
@ Shubham Johri. I fully agree with you. That's a drawback of StackExchange forum : Somebody can modify the wording of a question or of an answer without always fully understanding what really is the meaning. Fortunately it is not often that happens.
$endgroup$
– JJacquelin
Dec 16 '18 at 10:02
$begingroup$
@ShubhamJohri, et al. I edited the question title and another instance to match what was (at the time) the equation in the yellow "quotation box". I should've examined the question history to see if that notation had previously been changed from OP's original intent. That said ... I prefer the version with the "$cdot$", as it leaves open the possibility of defining the operation (say, as concatenation, or addition, or the "second operand" function $acdot b =b$, or, or, or, ...). :)
$endgroup$
– Blue
Dec 16 '18 at 10:18
add a comment |
$begingroup$
Don't confuse the two next equations :
$$0 text{ multiplied by }x=50 tag 1$$
$$0 text{ concatenated with }x=50 tag 2$$
Both are loosely written as $0x=50$. That's the trap.
Of course $(1)$ has no finite solution.
Obviously the solution of $(2)$ is $x=50$ because
$$0 text{ concatenated with }50=050=50.$$
The ambiguous question implicitly refers to $(2)$, not to $(1)$ and the expected answer is $x=50$.
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
$endgroup$
Don't confuse the two next equations :
$$0 text{ multiplied by }x=50 tag 1$$
$$0 text{ concatenated with }x=50 tag 2$$
Both are loosely written as $0x=50$. That's the trap.
Of course $(1)$ has no finite solution.
Obviously the solution of $(2)$ is $x=50$ because
$$0 text{ concatenated with }50=050=50.$$
The ambiguous question implicitly refers to $(2)$, not to $(1)$ and the expected answer is $x=50$.
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
edited Dec 16 '18 at 14:52
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
answered Dec 16 '18 at 9:34
JJacquelinJJacquelin
43.8k21853
43.8k21853
1
$begingroup$
I gave the same answer but someone other than the OP edited the question to make it $0cdot x$, invalidating my(our) response.
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:36
$begingroup$
This is not at all what I had in mind and that's wonderful. I've edited the question (twice now) to be faithful with the post on law. The only issue is now my question needs better tags.
$endgroup$
– candied_orange
Dec 16 '18 at 9:55
$begingroup$
@ Shubham Johri. I fully agree with you. That's a drawback of StackExchange forum : Somebody can modify the wording of a question or of an answer without always fully understanding what really is the meaning. Fortunately it is not often that happens.
$endgroup$
– JJacquelin
Dec 16 '18 at 10:02
$begingroup$
@ShubhamJohri, et al. I edited the question title and another instance to match what was (at the time) the equation in the yellow "quotation box". I should've examined the question history to see if that notation had previously been changed from OP's original intent. That said ... I prefer the version with the "$cdot$", as it leaves open the possibility of defining the operation (say, as concatenation, or addition, or the "second operand" function $acdot b =b$, or, or, or, ...). :)
$endgroup$
– Blue
Dec 16 '18 at 10:18
add a comment |
1
$begingroup$
I gave the same answer but someone other than the OP edited the question to make it $0cdot x$, invalidating my(our) response.
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:36
$begingroup$
This is not at all what I had in mind and that's wonderful. I've edited the question (twice now) to be faithful with the post on law. The only issue is now my question needs better tags.
$endgroup$
– candied_orange
Dec 16 '18 at 9:55
$begingroup$
@ Shubham Johri. I fully agree with you. That's a drawback of StackExchange forum : Somebody can modify the wording of a question or of an answer without always fully understanding what really is the meaning. Fortunately it is not often that happens.
$endgroup$
– JJacquelin
Dec 16 '18 at 10:02
$begingroup$
@ShubhamJohri, et al. I edited the question title and another instance to match what was (at the time) the equation in the yellow "quotation box". I should've examined the question history to see if that notation had previously been changed from OP's original intent. That said ... I prefer the version with the "$cdot$", as it leaves open the possibility of defining the operation (say, as concatenation, or addition, or the "second operand" function $acdot b =b$, or, or, or, ...). :)
$endgroup$
– Blue
Dec 16 '18 at 10:18
1
1
$begingroup$
I gave the same answer but someone other than the OP edited the question to make it $0cdot x$, invalidating my(our) response.
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:36
$begingroup$
I gave the same answer but someone other than the OP edited the question to make it $0cdot x$, invalidating my(our) response.
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:36
$begingroup$
This is not at all what I had in mind and that's wonderful. I've edited the question (twice now) to be faithful with the post on law. The only issue is now my question needs better tags.
$endgroup$
– candied_orange
Dec 16 '18 at 9:55
$begingroup$
This is not at all what I had in mind and that's wonderful. I've edited the question (twice now) to be faithful with the post on law. The only issue is now my question needs better tags.
$endgroup$
– candied_orange
Dec 16 '18 at 9:55
$begingroup$
@ Shubham Johri. I fully agree with you. That's a drawback of StackExchange forum : Somebody can modify the wording of a question or of an answer without always fully understanding what really is the meaning. Fortunately it is not often that happens.
$endgroup$
– JJacquelin
Dec 16 '18 at 10:02
$begingroup$
@ Shubham Johri. I fully agree with you. That's a drawback of StackExchange forum : Somebody can modify the wording of a question or of an answer without always fully understanding what really is the meaning. Fortunately it is not often that happens.
$endgroup$
– JJacquelin
Dec 16 '18 at 10:02
$begingroup$
@ShubhamJohri, et al. I edited the question title and another instance to match what was (at the time) the equation in the yellow "quotation box". I should've examined the question history to see if that notation had previously been changed from OP's original intent. That said ... I prefer the version with the "$cdot$", as it leaves open the possibility of defining the operation (say, as concatenation, or addition, or the "second operand" function $acdot b =b$, or, or, or, ...). :)
$endgroup$
– Blue
Dec 16 '18 at 10:18
$begingroup$
@ShubhamJohri, et al. I edited the question title and another instance to match what was (at the time) the equation in the yellow "quotation box". I should've examined the question history to see if that notation had previously been changed from OP's original intent. That said ... I prefer the version with the "$cdot$", as it leaves open the possibility of defining the operation (say, as concatenation, or addition, or the "second operand" function $acdot b =b$, or, or, or, ...). :)
$endgroup$
– Blue
Dec 16 '18 at 10:18
add a comment |
1
$begingroup$
How did you get that limit expression exactly? I fail to see how that limit is analogous to $0 cdot infty$.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:07
1
$begingroup$
Also, a personal note: I think you're overthinking the example. There are absolutely topologies of the real numbers in which "nonsense" expressions like $0 cdot x = 50$ have solutions - indeed, we can even define systems in which division by $0$ is defined (though results in things we do not desire, if just in application). But actually trying to consider the logic of the equation is probably well beyond what the post intended.
$endgroup$
– Eevee Trainer
Dec 16 '18 at 9:08
$begingroup$
$0x=50implies x=50 because050=50$
$endgroup$
– Shubham Johri
Dec 16 '18 at 9:11
1
$begingroup$
There is no real number $x$ such that $0 cdot x = 50$. If a law required you to produce such a real number, the law could not be obeyed.
$endgroup$
– littleO
Dec 16 '18 at 9:22
1
$begingroup$
One could always artificially introduce a new number (let's call it $alpha$) and extend the definition of multiplication so that $0 times alpha = 50$, but this would be kind of silly (and the resulting number system would not be interesting or useful). Surely the intention was that the law requires citizens to provide a real number $x$ such that $0 cdot x = 50$.
$endgroup$
– littleO
Dec 16 '18 at 9:35