Is $sqrt{-(x^2)}$ discontinous on the reals?
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I know this function does not exist on the reals, but does that mean it is discontinuous on the reals?
real-analysis continuity
asked Nov 18 at 2:24
Jack West
134
add a comment |
up vote
0
down vote
favorite
I know this function does not exist on the reals, but does that mean it is discontinuous on the reals?
real-analysis continuity
asked Nov 18 at 2:24
Jack West
134
Do you mean $sqrt{-(x^2)}$?
– Toby Mak
Nov 18 at 2:26
Yes, I'll edit it now. Apologies.
– Jack West
Nov 18 at 2:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know this function does not exist on the reals, but does that mean it is discontinuous on the reals?
real-analysis continuity
asked Nov 18 at 2:24
Jack West
134
I know this function does not exist on the reals, but does that mean it is discontinuous on the reals?
real-analysis continuity
real-analysis continuity
asked Nov 18 at 2:24
Jack West
134
asked Nov 18 at 2:24
Jack West
134
edited Nov 18 at 2:27
asked Nov 18 at 2:24
Jack West
134
asked Nov 18 at 2:24
Jack West
134
asked Nov 18 at 2:24
Jack West
134
134
Do you mean $sqrt{-(x^2)}$?
– Toby Mak
Nov 18 at 2:26
Yes, I'll edit it now. Apologies.
– Jack West
Nov 18 at 2:27
add a comment |
Do you mean $sqrt{-(x^2)}$?
– Toby Mak
Nov 18 at 2:26
Yes, I'll edit it now. Apologies.
– Jack West
Nov 18 at 2:27
Do you mean $sqrt{-(x^2)}$?
– Toby Mak
Nov 18 at 2:26
Do you mean $sqrt{-(x^2)}$?
– Toby Mak
Nov 18 at 2:26
Yes, I'll edit it now. Apologies.
– Jack West
Nov 18 at 2:27
Yes, I'll edit it now. Apologies.
– Jack West
Nov 18 at 2:27
add a comment |
2 Answers
2
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oldest
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up vote
5
down vote
The domain is ${0 }$, and the function is indeed continuous there. For any fixed $epsilon>0$, choose $delta = epsilon$. Then it is true that $forall x in (-delta, delta): |f(x) - 0| < epsilon$, since the only $x$ there is to check is $x=0$, and at that point we have $|f(x)-0|=|0-0|=0<epsilon$.
answered Nov 18 at 2:33
Ovi
12.1k938108
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up vote
1
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The question cannot be answered. This is would be like saying "If an object doesn't have a color, is it's color a shade of blue?".
The definition of "discontinuous on the reals" assumes that we have a function on the reals. If we do not have such a function, the definition doesn't make sense.
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
I guess what Jack is asking is whether the function is discontinuous {it across} the reals: when tracking the function value as $x$ moves along the imaginary axis in the complex plane, the function switches from i$|x|$ to $-$i$|x|$. It is $0$ at $x=0$, so it is continuous (and exists in one real number)..
– Lucozade
Nov 18 at 2:38
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The domain is ${0 }$, and the function is indeed continuous there. For any fixed $epsilon>0$, choose $delta = epsilon$. Then it is true that $forall x in (-delta, delta): |f(x) - 0| < epsilon$, since the only $x$ there is to check is $x=0$, and at that point we have $|f(x)-0|=|0-0|=0<epsilon$.
answered Nov 18 at 2:33
Ovi
12.1k938108
add a comment |
up vote
5
down vote
The domain is ${0 }$, and the function is indeed continuous there. For any fixed $epsilon>0$, choose $delta = epsilon$. Then it is true that $forall x in (-delta, delta): |f(x) - 0| < epsilon$, since the only $x$ there is to check is $x=0$, and at that point we have $|f(x)-0|=|0-0|=0<epsilon$.
answered Nov 18 at 2:33
Ovi
12.1k938108
add a comment |
up vote
5
down vote
up vote
5
down vote
The domain is ${0 }$, and the function is indeed continuous there. For any fixed $epsilon>0$, choose $delta = epsilon$. Then it is true that $forall x in (-delta, delta): |f(x) - 0| < epsilon$, since the only $x$ there is to check is $x=0$, and at that point we have $|f(x)-0|=|0-0|=0<epsilon$.
answered Nov 18 at 2:33
Ovi
12.1k938108
The domain is ${0 }$, and the function is indeed continuous there. For any fixed $epsilon>0$, choose $delta = epsilon$. Then it is true that $forall x in (-delta, delta): |f(x) - 0| < epsilon$, since the only $x$ there is to check is $x=0$, and at that point we have $|f(x)-0|=|0-0|=0<epsilon$.
answered Nov 18 at 2:33
Ovi
12.1k938108
answered Nov 18 at 2:33
Ovi
12.1k938108
answered Nov 18 at 2:33
Ovi
12.1k938108
answered Nov 18 at 2:33
Ovi
12.1k938108
12.1k938108
add a comment |
add a comment |
up vote
1
down vote
The question cannot be answered. This is would be like saying "If an object doesn't have a color, is it's color a shade of blue?".
The definition of "discontinuous on the reals" assumes that we have a function on the reals. If we do not have such a function, the definition doesn't make sense.
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
I guess what Jack is asking is whether the function is discontinuous {it across} the reals: when tracking the function value as $x$ moves along the imaginary axis in the complex plane, the function switches from i$|x|$ to $-$i$|x|$. It is $0$ at $x=0$, so it is continuous (and exists in one real number)..
– Lucozade
Nov 18 at 2:38
add a comment |
up vote
1
down vote
The question cannot be answered. This is would be like saying "If an object doesn't have a color, is it's color a shade of blue?".
The definition of "discontinuous on the reals" assumes that we have a function on the reals. If we do not have such a function, the definition doesn't make sense.
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
I guess what Jack is asking is whether the function is discontinuous {it across} the reals: when tracking the function value as $x$ moves along the imaginary axis in the complex plane, the function switches from i$|x|$ to $-$i$|x|$. It is $0$ at $x=0$, so it is continuous (and exists in one real number)..
– Lucozade
Nov 18 at 2:38
add a comment |
up vote
1
down vote
up vote
1
down vote
The question cannot be answered. This is would be like saying "If an object doesn't have a color, is it's color a shade of blue?".
The definition of "discontinuous on the reals" assumes that we have a function on the reals. If we do not have such a function, the definition doesn't make sense.
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
The question cannot be answered. This is would be like saying "If an object doesn't have a color, is it's color a shade of blue?".
The definition of "discontinuous on the reals" assumes that we have a function on the reals. If we do not have such a function, the definition doesn't make sense.
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
answered Nov 18 at 2:28
Joey Kilpatrick
1,183422
1,183422
I guess what Jack is asking is whether the function is discontinuous {it across} the reals: when tracking the function value as $x$ moves along the imaginary axis in the complex plane, the function switches from i$|x|$ to $-$i$|x|$. It is $0$ at $x=0$, so it is continuous (and exists in one real number)..
– Lucozade
Nov 18 at 2:38
add a comment |
I guess what Jack is asking is whether the function is discontinuous {it across} the reals: when tracking the function value as $x$ moves along the imaginary axis in the complex plane, the function switches from i$|x|$ to $-$i$|x|$. It is $0$ at $x=0$, so it is continuous (and exists in one real number)..
– Lucozade
Nov 18 at 2:38
I guess what Jack is asking is whether the function is discontinuous {it across} the reals: when tracking the function value as $x$ moves along the imaginary axis in the complex plane, the function switches from i$|x|$ to $-$i$|x|$. It is $0$ at $x=0$, so it is continuous (and exists in one real number)..
– Lucozade
Nov 18 at 2:38
I guess what Jack is asking is whether the function is discontinuous {it across} the reals: when tracking the function value as $x$ moves along the imaginary axis in the complex plane, the function switches from i$|x|$ to $-$i$|x|$. It is $0$ at $x=0$, so it is continuous (and exists in one real number)..
– Lucozade
Nov 18 at 2:38
add a comment |
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Do you mean $sqrt{-(x^2)}$?
– Toby Mak
Nov 18 at 2:26
Yes, I'll edit it now. Apologies.
– Jack West
Nov 18 at 2:27