Square roots — positive and negative
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It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.
I know that for an equation $x^2 = 9$, the solution is $x = pm 3$. But if simply given $sqrt9$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?
roots radicals
edited Oct 27 '15 at 15:37
Antonio Vargas
20.3k244110
asked Mar 11 '11 at 9:43
wrongusername
4112514
add a comment |Â
up vote
17
down vote
favorite
It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.
I know that for an equation $x^2 = 9$, the solution is $x = pm 3$. But if simply given $sqrt9$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?
roots radicals
edited Oct 27 '15 at 15:37
Antonio Vargas
20.3k244110
asked Mar 11 '11 at 9:43
wrongusername
4112514
I would say this can answer your question. See also this.
– user8101
Mar 11 '11 at 9:49
1
These are often confused because students believe that $sqrtx^2=x$, but actually $sqrtx^2=|x|$. So: $sqrtx^2=sqrt9$ implies $|x|=3$, and so there are two possibilites: $x=3$ or $x=-3$.
– rschwieb
May 21 '13 at 21:06
add a comment |Â
up vote
17
down vote
favorite
up vote
17
down vote
favorite
It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.
I know that for an equation $x^2 = 9$, the solution is $x = pm 3$. But if simply given $sqrt9$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?
roots radicals
edited Oct 27 '15 at 15:37
Antonio Vargas
20.3k244110
asked Mar 11 '11 at 9:43
wrongusername
4112514
It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.
I know that for an equation $x^2 = 9$, the solution is $x = pm 3$. But if simply given $sqrt9$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?
roots radicals
edited Oct 27 '15 at 15:37
Antonio Vargas
20.3k244110
asked Mar 11 '11 at 9:43
wrongusername
4112514
edited Oct 27 '15 at 15:37
Antonio Vargas
20.3k244110
edited Oct 27 '15 at 15:37
Antonio Vargas
20.3k244110
edited Oct 27 '15 at 15:37
Antonio Vargas
20.3k244110
20.3k244110
asked Mar 11 '11 at 9:43
wrongusername
4112514
asked Mar 11 '11 at 9:43
wrongusername
4112514
asked Mar 11 '11 at 9:43
wrongusername
4112514
4112514
I would say this can answer your question. See also this.
– user8101
Mar 11 '11 at 9:49
1
These are often confused because students believe that $sqrtx^2=x$, but actually $sqrtx^2=|x|$. So: $sqrtx^2=sqrt9$ implies $|x|=3$, and so there are two possibilites: $x=3$ or $x=-3$.
– rschwieb
May 21 '13 at 21:06
add a comment |Â
I would say this can answer your question. See also this.
– user8101
Mar 11 '11 at 9:49
1
These are often confused because students believe that $sqrtx^2=x$, but actually $sqrtx^2=|x|$. So: $sqrtx^2=sqrt9$ implies $|x|=3$, and so there are two possibilites: $x=3$ or $x=-3$.
– rschwieb
May 21 '13 at 21:06
I would say this can answer your question. See also this.
– user8101
Mar 11 '11 at 9:49
I would say this can answer your question. See also this.
– user8101
Mar 11 '11 at 9:49
1
1
These are often confused because students believe that $sqrtx^2=x$, but actually $sqrtx^2=|x|$. So: $sqrtx^2=sqrt9$ implies $|x|=3$, and so there are two possibilites: $x=3$ or $x=-3$.
– rschwieb
May 21 '13 at 21:06
These are often confused because students believe that $sqrtx^2=x$, but actually $sqrtx^2=|x|$. So: $sqrtx^2=sqrt9$ implies $|x|=3$, and so there are two possibilites: $x=3$ or $x=-3$.
– rschwieb
May 21 '13 at 21:06
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
17
down vote
accepted
If you want your square-root function $sqrt x$ to be a function, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain. If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.
So to be a function, square-root needs to be single valued; the multi-valued version is really a relation, at which point you might get into issues of principal values.
For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern. Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers.
answered Mar 11 '11 at 10:46
Henry
93.3k470148
add a comment |Â
up vote
5
down vote
For positive real $x$, $sqrt x$ denotes the positive square root of $x$, by definition. Wikipedia agrees with me on this.
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
add a comment |Â
up vote
1
down vote
Consider the similar equation $x^4 = 9$. This one has four solutions. The most obvious solution is $sqrt 3$, which is roughly 1.732. Another solution is $-sqrt 3 approx - 1.732$. The same as the previous solution, just multiplied by $-1$.
There are two other solutions: $x = sqrt-3$ and $x = -sqrt-3$, which are roughly 1.732, the former multiplied by $i$ and the latter multiplied by $-i$. These are all the same distance away from 0.
If you just need one solution, you might as well take the solution that is a positive real number. If you need the other solutions, just multiply the solution you have by units other than 1.
So to solve $x^2 = 9$, the solution $x = 3$ might be enough, but if you need the other solution you just multiply the previous solution by the unit $-1$.
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
If you want your square-root function $sqrt x$ to be a function, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain. If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.
So to be a function, square-root needs to be single valued; the multi-valued version is really a relation, at which point you might get into issues of principal values.
For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern. Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers.
answered Mar 11 '11 at 10:46
Henry
93.3k470148
add a comment |Â
up vote
17
down vote
accepted
If you want your square-root function $sqrt x$ to be a function, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain. If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.
So to be a function, square-root needs to be single valued; the multi-valued version is really a relation, at which point you might get into issues of principal values.
For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern. Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers.
answered Mar 11 '11 at 10:46
Henry
93.3k470148
add a comment |Â
up vote
17
down vote
accepted
up vote
17
down vote
accepted
If you want your square-root function $sqrt x$ to be a function, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain. If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.
So to be a function, square-root needs to be single valued; the multi-valued version is really a relation, at which point you might get into issues of principal values.
For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern. Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers.
answered Mar 11 '11 at 10:46
Henry
93.3k470148
If you want your square-root function $sqrt x$ to be a function, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain. If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.
So to be a function, square-root needs to be single valued; the multi-valued version is really a relation, at which point you might get into issues of principal values.
For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern. Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers.
answered Mar 11 '11 at 10:46
Henry
93.3k470148
edited Mar 11 '11 at 15:48
answered Mar 11 '11 at 10:46
Henry
93.3k470148
answered Mar 11 '11 at 10:46
Henry
93.3k470148
answered Mar 11 '11 at 10:46
Henry
93.3k470148
93.3k470148
add a comment |Â
add a comment |Â
up vote
5
down vote
For positive real $x$, $sqrt x$ denotes the positive square root of $x$, by definition. Wikipedia agrees with me on this.
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
add a comment |Â
up vote
5
down vote
For positive real $x$, $sqrt x$ denotes the positive square root of $x$, by definition. Wikipedia agrees with me on this.
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
add a comment |Â
up vote
5
down vote
up vote
5
down vote
For positive real $x$, $sqrt x$ denotes the positive square root of $x$, by definition. Wikipedia agrees with me on this.
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
For positive real $x$, $sqrt x$ denotes the positive square root of $x$, by definition. Wikipedia agrees with me on this.
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
answered Mar 11 '11 at 16:18
TonyK
38.5k346126
38.5k346126
add a comment |Â
add a comment |Â
up vote
1
down vote
Consider the similar equation $x^4 = 9$. This one has four solutions. The most obvious solution is $sqrt 3$, which is roughly 1.732. Another solution is $-sqrt 3 approx - 1.732$. The same as the previous solution, just multiplied by $-1$.
There are two other solutions: $x = sqrt-3$ and $x = -sqrt-3$, which are roughly 1.732, the former multiplied by $i$ and the latter multiplied by $-i$. These are all the same distance away from 0.
If you just need one solution, you might as well take the solution that is a positive real number. If you need the other solutions, just multiply the solution you have by units other than 1.
So to solve $x^2 = 9$, the solution $x = 3$ might be enough, but if you need the other solution you just multiply the previous solution by the unit $-1$.
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
add a comment |Â
up vote
1
down vote
Consider the similar equation $x^4 = 9$. This one has four solutions. The most obvious solution is $sqrt 3$, which is roughly 1.732. Another solution is $-sqrt 3 approx - 1.732$. The same as the previous solution, just multiplied by $-1$.
There are two other solutions: $x = sqrt-3$ and $x = -sqrt-3$, which are roughly 1.732, the former multiplied by $i$ and the latter multiplied by $-i$. These are all the same distance away from 0.
If you just need one solution, you might as well take the solution that is a positive real number. If you need the other solutions, just multiply the solution you have by units other than 1.
So to solve $x^2 = 9$, the solution $x = 3$ might be enough, but if you need the other solution you just multiply the previous solution by the unit $-1$.
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Consider the similar equation $x^4 = 9$. This one has four solutions. The most obvious solution is $sqrt 3$, which is roughly 1.732. Another solution is $-sqrt 3 approx - 1.732$. The same as the previous solution, just multiplied by $-1$.
There are two other solutions: $x = sqrt-3$ and $x = -sqrt-3$, which are roughly 1.732, the former multiplied by $i$ and the latter multiplied by $-i$. These are all the same distance away from 0.
If you just need one solution, you might as well take the solution that is a positive real number. If you need the other solutions, just multiply the solution you have by units other than 1.
So to solve $x^2 = 9$, the solution $x = 3$ might be enough, but if you need the other solution you just multiply the previous solution by the unit $-1$.
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
Consider the similar equation $x^4 = 9$. This one has four solutions. The most obvious solution is $sqrt 3$, which is roughly 1.732. Another solution is $-sqrt 3 approx - 1.732$. The same as the previous solution, just multiplied by $-1$.
There are two other solutions: $x = sqrt-3$ and $x = -sqrt-3$, which are roughly 1.732, the former multiplied by $i$ and the latter multiplied by $-i$. These are all the same distance away from 0.
If you just need one solution, you might as well take the solution that is a positive real number. If you need the other solutions, just multiply the solution you have by units other than 1.
So to solve $x^2 = 9$, the solution $x = 3$ might be enough, but if you need the other solution you just multiply the previous solution by the unit $-1$.
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
answered Aug 10 at 17:12
Robert Soupe
10.1k21947
10.1k21947
add a comment |Â
add a comment |Â
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I would say this can answer your question. See also this.
– user8101
Mar 11 '11 at 9:49
1
These are often confused because students believe that $sqrtx^2=x$, but actually $sqrtx^2=|x|$. So: $sqrtx^2=sqrt9$ implies $|x|=3$, and so there are two possibilites: $x=3$ or $x=-3$.
– rschwieb
May 21 '13 at 21:06