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What is the domain of a division of functions?
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This question is about real functions of real variables.
I think that, in general, if the domain of some function $f(x)$ is A, and the domain of another function $g(x)$ is B, then the domain of $(f/g)(x)$ is A$cap$B and where $gneq0$.
Now, what happens if I have something like $f(x)=2$, $g(x)=1/x$? In this case, $(f/g)(x)=2x$, which seems to be defined for all real numbers. But my statement above (which I think is correct in general) implies that $x=0$ is not allowed. So I'm conflicted.
Can somebody tell me what the domain of $(f/g)(x)$ is in this case? Is it all real numbers, or all real numbers except $0$?
Thanks.
functions combinations real-numbers
asked Aug 15 at 23:29
David S
533
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up vote
8
down vote
favorite
This question is about real functions of real variables.
I think that, in general, if the domain of some function $f(x)$ is A, and the domain of another function $g(x)$ is B, then the domain of $(f/g)(x)$ is A$cap$B and where $gneq0$.
Now, what happens if I have something like $f(x)=2$, $g(x)=1/x$? In this case, $(f/g)(x)=2x$, which seems to be defined for all real numbers. But my statement above (which I think is correct in general) implies that $x=0$ is not allowed. So I'm conflicted.
Can somebody tell me what the domain of $(f/g)(x)$ is in this case? Is it all real numbers, or all real numbers except $0$?
Thanks.
functions combinations real-numbers
asked Aug 15 at 23:29
David S
533
Sorry, I don't understand your comment. If I have a function $f$ that is only defined for $x=6$ and another one ($g$) only defined for $x=3$, then $(f/g)$ is not defined anywhere. I don't see the problem with that.
– David S
Aug 15 at 23:39
1
I agree with the first answer given, which reiterates your second sentence. This will clear up your 3rd block of text: $2/(1/x)$ and $2x$ are different functions. So if I give you the function $2x$, then that's that. And if I give you the function $2/(1/x)$, that's good too. They aren't the same. But if I give you $2x$ and tell you I want the domain to be all reals minus $0$, now the two functions are the same. Without stating the domain, it's implied $2x$ is over all the reals. I could have told you that $2x$ is over the domain $6 < x leq 7$, or whatever else
– DWade64
Aug 15 at 23:47
2
The domain of $f/g$ excludes $0$ based on the definition of $f/g$ you gave. But do not take this too seriously. Math is about communication of ideas more than definitions, so if it is inconvenient to exclude $0$, you can "fill it in" - as long as you do this in a way that is completely clear to yourself and whoever else is interested in your work.
– Jair Taylor
Aug 16 at 0:58
This is correct if you think about them as functions in the set-theoretic sense. But for sufficiently well-behaved families of functions (rational, analytic) you can actually ignore those singularities and it makes perfect sense, even if pointwise you have silly stuff like division by zero. Or, if you treat them as functions on a measurable space, you can ignore what happens on a set of measure $0$, and that's fine, too. In each case, you can think of dividing equivalence classes of functions, rather than actual functions.
– tomasz
Aug 16 at 8:41
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
This question is about real functions of real variables.
I think that, in general, if the domain of some function $f(x)$ is A, and the domain of another function $g(x)$ is B, then the domain of $(f/g)(x)$ is A$cap$B and where $gneq0$.
Now, what happens if I have something like $f(x)=2$, $g(x)=1/x$? In this case, $(f/g)(x)=2x$, which seems to be defined for all real numbers. But my statement above (which I think is correct in general) implies that $x=0$ is not allowed. So I'm conflicted.
Can somebody tell me what the domain of $(f/g)(x)$ is in this case? Is it all real numbers, or all real numbers except $0$?
Thanks.
functions combinations real-numbers
asked Aug 15 at 23:29
David S
533
This question is about real functions of real variables.
I think that, in general, if the domain of some function $f(x)$ is A, and the domain of another function $g(x)$ is B, then the domain of $(f/g)(x)$ is A$cap$B and where $gneq0$.
Now, what happens if I have something like $f(x)=2$, $g(x)=1/x$? In this case, $(f/g)(x)=2x$, which seems to be defined for all real numbers. But my statement above (which I think is correct in general) implies that $x=0$ is not allowed. So I'm conflicted.
Can somebody tell me what the domain of $(f/g)(x)$ is in this case? Is it all real numbers, or all real numbers except $0$?
Thanks.
functions combinations real-numbers
asked Aug 15 at 23:29
David S
533
asked Aug 15 at 23:29
David S
533
asked Aug 15 at 23:29
David S
533
asked Aug 15 at 23:29
David S
533
533
Sorry, I don't understand your comment. If I have a function $f$ that is only defined for $x=6$ and another one ($g$) only defined for $x=3$, then $(f/g)$ is not defined anywhere. I don't see the problem with that.
– David S
Aug 15 at 23:39
1
I agree with the first answer given, which reiterates your second sentence. This will clear up your 3rd block of text: $2/(1/x)$ and $2x$ are different functions. So if I give you the function $2x$, then that's that. And if I give you the function $2/(1/x)$, that's good too. They aren't the same. But if I give you $2x$ and tell you I want the domain to be all reals minus $0$, now the two functions are the same. Without stating the domain, it's implied $2x$ is over all the reals. I could have told you that $2x$ is over the domain $6 < x leq 7$, or whatever else
– DWade64
Aug 15 at 23:47
2
The domain of $f/g$ excludes $0$ based on the definition of $f/g$ you gave. But do not take this too seriously. Math is about communication of ideas more than definitions, so if it is inconvenient to exclude $0$, you can "fill it in" - as long as you do this in a way that is completely clear to yourself and whoever else is interested in your work.
– Jair Taylor
Aug 16 at 0:58
This is correct if you think about them as functions in the set-theoretic sense. But for sufficiently well-behaved families of functions (rational, analytic) you can actually ignore those singularities and it makes perfect sense, even if pointwise you have silly stuff like division by zero. Or, if you treat them as functions on a measurable space, you can ignore what happens on a set of measure $0$, and that's fine, too. In each case, you can think of dividing equivalence classes of functions, rather than actual functions.
– tomasz
Aug 16 at 8:41
add a comment |Â
Sorry, I don't understand your comment. If I have a function $f$ that is only defined for $x=6$ and another one ($g$) only defined for $x=3$, then $(f/g)$ is not defined anywhere. I don't see the problem with that.
– David S
Aug 15 at 23:39
1
I agree with the first answer given, which reiterates your second sentence. This will clear up your 3rd block of text: $2/(1/x)$ and $2x$ are different functions. So if I give you the function $2x$, then that's that. And if I give you the function $2/(1/x)$, that's good too. They aren't the same. But if I give you $2x$ and tell you I want the domain to be all reals minus $0$, now the two functions are the same. Without stating the domain, it's implied $2x$ is over all the reals. I could have told you that $2x$ is over the domain $6 < x leq 7$, or whatever else
– DWade64
Aug 15 at 23:47
2
The domain of $f/g$ excludes $0$ based on the definition of $f/g$ you gave. But do not take this too seriously. Math is about communication of ideas more than definitions, so if it is inconvenient to exclude $0$, you can "fill it in" - as long as you do this in a way that is completely clear to yourself and whoever else is interested in your work.
– Jair Taylor
Aug 16 at 0:58
This is correct if you think about them as functions in the set-theoretic sense. But for sufficiently well-behaved families of functions (rational, analytic) you can actually ignore those singularities and it makes perfect sense, even if pointwise you have silly stuff like division by zero. Or, if you treat them as functions on a measurable space, you can ignore what happens on a set of measure $0$, and that's fine, too. In each case, you can think of dividing equivalence classes of functions, rather than actual functions.
– tomasz
Aug 16 at 8:41
Sorry, I don't understand your comment. If I have a function $f$ that is only defined for $x=6$ and another one ($g$) only defined for $x=3$, then $(f/g)$ is not defined anywhere. I don't see the problem with that.
– David S
Aug 15 at 23:39
Sorry, I don't understand your comment. If I have a function $f$ that is only defined for $x=6$ and another one ($g$) only defined for $x=3$, then $(f/g)$ is not defined anywhere. I don't see the problem with that.
– David S
Aug 15 at 23:39
1
1
I agree with the first answer given, which reiterates your second sentence. This will clear up your 3rd block of text: $2/(1/x)$ and $2x$ are different functions. So if I give you the function $2x$, then that's that. And if I give you the function $2/(1/x)$, that's good too. They aren't the same. But if I give you $2x$ and tell you I want the domain to be all reals minus $0$, now the two functions are the same. Without stating the domain, it's implied $2x$ is over all the reals. I could have told you that $2x$ is over the domain $6 < x leq 7$, or whatever else
– DWade64
Aug 15 at 23:47
I agree with the first answer given, which reiterates your second sentence. This will clear up your 3rd block of text: $2/(1/x)$ and $2x$ are different functions. So if I give you the function $2x$, then that's that. And if I give you the function $2/(1/x)$, that's good too. They aren't the same. But if I give you $2x$ and tell you I want the domain to be all reals minus $0$, now the two functions are the same. Without stating the domain, it's implied $2x$ is over all the reals. I could have told you that $2x$ is over the domain $6 < x leq 7$, or whatever else
– DWade64
Aug 15 at 23:47
2
2
The domain of $f/g$ excludes $0$ based on the definition of $f/g$ you gave. But do not take this too seriously. Math is about communication of ideas more than definitions, so if it is inconvenient to exclude $0$, you can "fill it in" - as long as you do this in a way that is completely clear to yourself and whoever else is interested in your work.
– Jair Taylor
Aug 16 at 0:58
The domain of $f/g$ excludes $0$ based on the definition of $f/g$ you gave. But do not take this too seriously. Math is about communication of ideas more than definitions, so if it is inconvenient to exclude $0$, you can "fill it in" - as long as you do this in a way that is completely clear to yourself and whoever else is interested in your work.
– Jair Taylor
Aug 16 at 0:58
This is correct if you think about them as functions in the set-theoretic sense. But for sufficiently well-behaved families of functions (rational, analytic) you can actually ignore those singularities and it makes perfect sense, even if pointwise you have silly stuff like division by zero. Or, if you treat them as functions on a measurable space, you can ignore what happens on a set of measure $0$, and that's fine, too. In each case, you can think of dividing equivalence classes of functions, rather than actual functions.
– tomasz
Aug 16 at 8:41
This is correct if you think about them as functions in the set-theoretic sense. But for sufficiently well-behaved families of functions (rational, analytic) you can actually ignore those singularities and it makes perfect sense, even if pointwise you have silly stuff like division by zero. Or, if you treat them as functions on a measurable space, you can ignore what happens on a set of measure $0$, and that's fine, too. In each case, you can think of dividing equivalence classes of functions, rather than actual functions.
– tomasz
Aug 16 at 8:41
add a comment |Â
3 Answers
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Your mistake is in thinking that
$$frac21/xquadhboxandquad 2x$$
are always equal. They're not. To carefully prove that they are equal, we have
$$frac21/x=frac21/x,1=frac21/xfrac xx=frac2x1=2x .$$
But this is not correct when $x=0$, because $frac00$ is not equal to $1$. So we have to consider $x=0$ separately. In that case we have $2x=0$, but
$$frac21/x=frac21/0=frac2hboxnonsense=hboxnonsense .$$
So in your example, the domain of $f/g$ must exclude $0$.
answered Aug 16 at 0:15
David
66.2k663125
5
I loved your answer; super simple and clear. Great name too, btw.
– David S
Aug 16 at 1:25
1
@DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$frac21/x$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not.
– user21820
Aug 16 at 10:50
@user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$frac21/0=cdots=hboxnonsense$"? Just asking.
– David
Aug 16 at 23:55
@David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$frac21/x$" to be well-defined we must have "$1/x$" well-defined, which requires $x ne 0$. Note the use of quotes to distinguish expressions from values.
– user21820
Aug 17 at 5:55
Your second chain of equalities is not mathematical equalities- well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that.
– David
Aug 17 at 6:23
 |Â
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up vote
6
down vote
If the domain of $f$ is $A$ and the domain of $g$ is $B$ then the domain of $f/g$ is $$Acap Bsetminusg(x)ne 0.$$ (Of course, we must assume that $Acap Bsetminusg(x)ne 0ne emptyset$. In other case $f/g$ doesn't make sense.)
In your example, $f(x)=2$ and $g(x)=1/x.$ We have that
$$dfracf(x)g(x)=2x, forall xinmathbbRsetminus0.$$ Why? Note that $g(0)$ doesn't exist. So we can't consider
$$dfracf(0)g(0).$$
answered Aug 15 at 23:33
mfl
25.1k12141
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up vote
0
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It is all real numbers without zero i think.
Check out this similar example:
$$F(x)=dfrac(x-2)(x-1)(x-2)$$
You can simplify this fraction to $F(x)=(x-1)$ ONLY when 2 is excluded from the domain
So we say that $F$ is defined $$forall xinmathbbRsetminus2.$$
Even though $F(x)=(x-1)$
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Your mistake is in thinking that
$$frac21/xquadhboxandquad 2x$$
are always equal. They're not. To carefully prove that they are equal, we have
$$frac21/x=frac21/x,1=frac21/xfrac xx=frac2x1=2x .$$
But this is not correct when $x=0$, because $frac00$ is not equal to $1$. So we have to consider $x=0$ separately. In that case we have $2x=0$, but
$$frac21/x=frac21/0=frac2hboxnonsense=hboxnonsense .$$
So in your example, the domain of $f/g$ must exclude $0$.
answered Aug 16 at 0:15
David
66.2k663125
5
I loved your answer; super simple and clear. Great name too, btw.
– David S
Aug 16 at 1:25
1
@DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$frac21/x$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not.
– user21820
Aug 16 at 10:50
@user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$frac21/0=cdots=hboxnonsense$"? Just asking.
– David
Aug 16 at 23:55
@David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$frac21/x$" to be well-defined we must have "$1/x$" well-defined, which requires $x ne 0$. Note the use of quotes to distinguish expressions from values.
– user21820
Aug 17 at 5:55
Your second chain of equalities is not mathematical equalities- well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that.
– David
Aug 17 at 6:23
 |Â
show 1 more comment
up vote
9
down vote
accepted
Your mistake is in thinking that
$$frac21/xquadhboxandquad 2x$$
are always equal. They're not. To carefully prove that they are equal, we have
$$frac21/x=frac21/x,1=frac21/xfrac xx=frac2x1=2x .$$
But this is not correct when $x=0$, because $frac00$ is not equal to $1$. So we have to consider $x=0$ separately. In that case we have $2x=0$, but
$$frac21/x=frac21/0=frac2hboxnonsense=hboxnonsense .$$
So in your example, the domain of $f/g$ must exclude $0$.
answered Aug 16 at 0:15
David
66.2k663125
5
I loved your answer; super simple and clear. Great name too, btw.
– David S
Aug 16 at 1:25
1
@DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$frac21/x$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not.
– user21820
Aug 16 at 10:50
@user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$frac21/0=cdots=hboxnonsense$"? Just asking.
– David
Aug 16 at 23:55
@David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$frac21/x$" to be well-defined we must have "$1/x$" well-defined, which requires $x ne 0$. Note the use of quotes to distinguish expressions from values.
– user21820
Aug 17 at 5:55
Your second chain of equalities is not mathematical equalities- well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that.
– David
Aug 17 at 6:23
 |Â
show 1 more comment
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Your mistake is in thinking that
$$frac21/xquadhboxandquad 2x$$
are always equal. They're not. To carefully prove that they are equal, we have
$$frac21/x=frac21/x,1=frac21/xfrac xx=frac2x1=2x .$$
But this is not correct when $x=0$, because $frac00$ is not equal to $1$. So we have to consider $x=0$ separately. In that case we have $2x=0$, but
$$frac21/x=frac21/0=frac2hboxnonsense=hboxnonsense .$$
So in your example, the domain of $f/g$ must exclude $0$.
answered Aug 16 at 0:15
David
66.2k663125
Your mistake is in thinking that
$$frac21/xquadhboxandquad 2x$$
are always equal. They're not. To carefully prove that they are equal, we have
$$frac21/x=frac21/x,1=frac21/xfrac xx=frac2x1=2x .$$
But this is not correct when $x=0$, because $frac00$ is not equal to $1$. So we have to consider $x=0$ separately. In that case we have $2x=0$, but
$$frac21/x=frac21/0=frac2hboxnonsense=hboxnonsense .$$
So in your example, the domain of $f/g$ must exclude $0$.
answered Aug 16 at 0:15
David
66.2k663125
answered Aug 16 at 0:15
David
66.2k663125
answered Aug 16 at 0:15
David
66.2k663125
answered Aug 16 at 0:15
David
66.2k663125
66.2k663125
5
I loved your answer; super simple and clear. Great name too, btw.
– David S
Aug 16 at 1:25
1
@DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$frac21/x$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not.
– user21820
Aug 16 at 10:50
@user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$frac21/0=cdots=hboxnonsense$"? Just asking.
– David
Aug 16 at 23:55
@David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$frac21/x$" to be well-defined we must have "$1/x$" well-defined, which requires $x ne 0$. Note the use of quotes to distinguish expressions from values.
– user21820
Aug 17 at 5:55
Your second chain of equalities is not mathematical equalities- well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that.
– David
Aug 17 at 6:23
 |Â
show 1 more comment
5
I loved your answer; super simple and clear. Great name too, btw.
– David S
Aug 16 at 1:25
1
@DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$frac21/x$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not.
– user21820
Aug 16 at 10:50
@user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$frac21/0=cdots=hboxnonsense$"? Just asking.
– David
Aug 16 at 23:55
@David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$frac21/x$" to be well-defined we must have "$1/x$" well-defined, which requires $x ne 0$. Note the use of quotes to distinguish expressions from values.
– user21820
Aug 17 at 5:55
Your second chain of equalities is not mathematical equalities- well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that.
– David
Aug 17 at 6:23
5
5
I loved your answer; super simple and clear. Great name too, btw.
– David S
Aug 16 at 1:25
I loved your answer; super simple and clear. Great name too, btw.
– David S
Aug 16 at 1:25
1
1
@DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$frac21/x$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not.
– user21820
Aug 16 at 10:50
@DavidS: Unfortunately, this answer is wrong. mfl's answer is correct. The error is that "$frac21/x$" is not even defined when $x = 0$, simply because "$1/x$" is not defined when $x = 0$. One cannot manipulate expressions as if they are defined, when they are actually not.
– user21820
Aug 16 at 10:50
@user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$frac21/0=cdots=hboxnonsense$"? Just asking.
– David
Aug 16 at 23:55
@user21820 Basically, that's exactly what I said in my answer. What did you think I meant by "$frac21/0=cdots=hboxnonsense$"? Just asking.
– David
Aug 16 at 23:55
@David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$frac21/x$" to be well-defined we must have "$1/x$" well-defined, which requires $x ne 0$. Note the use of quotes to distinguish expressions from values.
– user21820
Aug 17 at 5:55
@David: I object to your explanation, especially the way it implies that the problem is that "$0/0$ is not equal to $1$". But that is not the real problem. "$1/x$" and anything containing it is already meaningless (or syntactically malformed, or a type error) if $x = 0$. For the same reason, your second chain of equalities is not mathematical equalities, but rather reductions. In particular, the correct reasoning is that for "$frac21/x$" to be well-defined we must have "$1/x$" well-defined, which requires $x ne 0$. Note the use of quotes to distinguish expressions from values.
– user21820
Aug 17 at 5:55
Your second chain of equalities is not mathematical equalities - well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that.– David
Aug 17 at 6:23
Your second chain of equalities is not mathematical equalities - well of course not, and that is exactly my point. The problem that many beginners have is that they think $0/0$ is equal to $1$. Any question on this site needs to be answered on a level appropriate to the question, and judging by the OP's comment, I have successfully done that.– David
Aug 17 at 6:23
 |Â
show 1 more comment
up vote
6
down vote
If the domain of $f$ is $A$ and the domain of $g$ is $B$ then the domain of $f/g$ is $$Acap Bsetminusg(x)ne 0.$$ (Of course, we must assume that $Acap Bsetminusg(x)ne 0ne emptyset$. In other case $f/g$ doesn't make sense.)
In your example, $f(x)=2$ and $g(x)=1/x.$ We have that
$$dfracf(x)g(x)=2x, forall xinmathbbRsetminus0.$$ Why? Note that $g(0)$ doesn't exist. So we can't consider
$$dfracf(0)g(0).$$
answered Aug 15 at 23:33
mfl
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If the domain of $f$ is $A$ and the domain of $g$ is $B$ then the domain of $f/g$ is $$Acap Bsetminusg(x)ne 0.$$ (Of course, we must assume that $Acap Bsetminusg(x)ne 0ne emptyset$. In other case $f/g$ doesn't make sense.)
In your example, $f(x)=2$ and $g(x)=1/x.$ We have that
$$dfracf(x)g(x)=2x, forall xinmathbbRsetminus0.$$ Why? Note that $g(0)$ doesn't exist. So we can't consider
$$dfracf(0)g(0).$$
answered Aug 15 at 23:33
mfl
25.1k12141
add a comment |Â
up vote
6
down vote
up vote
6
down vote
If the domain of $f$ is $A$ and the domain of $g$ is $B$ then the domain of $f/g$ is $$Acap Bsetminusg(x)ne 0.$$ (Of course, we must assume that $Acap Bsetminusg(x)ne 0ne emptyset$. In other case $f/g$ doesn't make sense.)
In your example, $f(x)=2$ and $g(x)=1/x.$ We have that
$$dfracf(x)g(x)=2x, forall xinmathbbRsetminus0.$$ Why? Note that $g(0)$ doesn't exist. So we can't consider
$$dfracf(0)g(0).$$
answered Aug 15 at 23:33
mfl
25.1k12141
If the domain of $f$ is $A$ and the domain of $g$ is $B$ then the domain of $f/g$ is $$Acap Bsetminusg(x)ne 0.$$ (Of course, we must assume that $Acap Bsetminusg(x)ne 0ne emptyset$. In other case $f/g$ doesn't make sense.)
In your example, $f(x)=2$ and $g(x)=1/x.$ We have that
$$dfracf(x)g(x)=2x, forall xinmathbbRsetminus0.$$ Why? Note that $g(0)$ doesn't exist. So we can't consider
$$dfracf(0)g(0).$$
answered Aug 15 at 23:33
mfl
25.1k12141
answered Aug 15 at 23:33
mfl
25.1k12141
answered Aug 15 at 23:33
mfl
25.1k12141
answered Aug 15 at 23:33
mfl
25.1k12141
25.1k12141
add a comment |Â
add a comment |Â
up vote
0
down vote
It is all real numbers without zero i think.
Check out this similar example:
$$F(x)=dfrac(x-2)(x-1)(x-2)$$
You can simplify this fraction to $F(x)=(x-1)$ ONLY when 2 is excluded from the domain
So we say that $F$ is defined $$forall xinmathbbRsetminus2.$$
Even though $F(x)=(x-1)$
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up vote
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It is all real numbers without zero i think.
Check out this similar example:
$$F(x)=dfrac(x-2)(x-1)(x-2)$$
You can simplify this fraction to $F(x)=(x-1)$ ONLY when 2 is excluded from the domain
So we say that $F$ is defined $$forall xinmathbbRsetminus2.$$
Even though $F(x)=(x-1)$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It is all real numbers without zero i think.
Check out this similar example:
$$F(x)=dfrac(x-2)(x-1)(x-2)$$
You can simplify this fraction to $F(x)=(x-1)$ ONLY when 2 is excluded from the domain
So we say that $F$ is defined $$forall xinmathbbRsetminus2.$$
Even though $F(x)=(x-1)$
It is all real numbers without zero i think.
Check out this similar example:
$$F(x)=dfrac(x-2)(x-1)(x-2)$$
You can simplify this fraction to $F(x)=(x-1)$ ONLY when 2 is excluded from the domain
So we say that $F$ is defined $$forall xinmathbbRsetminus2.$$
Even though $F(x)=(x-1)$
edited Aug 15 at 23:56
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Fareed AF
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Sorry, I don't understand your comment. If I have a function $f$ that is only defined for $x=6$ and another one ($g$) only defined for $x=3$, then $(f/g)$ is not defined anywhere. I don't see the problem with that.
– David S
Aug 15 at 23:39
1
I agree with the first answer given, which reiterates your second sentence. This will clear up your 3rd block of text: $2/(1/x)$ and $2x$ are different functions. So if I give you the function $2x$, then that's that. And if I give you the function $2/(1/x)$, that's good too. They aren't the same. But if I give you $2x$ and tell you I want the domain to be all reals minus $0$, now the two functions are the same. Without stating the domain, it's implied $2x$ is over all the reals. I could have told you that $2x$ is over the domain $6 < x leq 7$, or whatever else
– DWade64
Aug 15 at 23:47
2
The domain of $f/g$ excludes $0$ based on the definition of $f/g$ you gave. But do not take this too seriously. Math is about communication of ideas more than definitions, so if it is inconvenient to exclude $0$, you can "fill it in" - as long as you do this in a way that is completely clear to yourself and whoever else is interested in your work.
– Jair Taylor
Aug 16 at 0:58
This is correct if you think about them as functions in the set-theoretic sense. But for sufficiently well-behaved families of functions (rational, analytic) you can actually ignore those singularities and it makes perfect sense, even if pointwise you have silly stuff like division by zero. Or, if you treat them as functions on a measurable space, you can ignore what happens on a set of measure $0$, and that's fine, too. In each case, you can think of dividing equivalence classes of functions, rather than actual functions.
– tomasz
Aug 16 at 8:41