The root sign and it's relation with 1/2.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$sqrt4=2$
But is it same as writing...
$4^1/2=2$?
Basically I do not understand why $sqrt$-sign equals $1/2$?
roots
edited Aug 11 at 4:53
Cornman
2,55221128
asked Aug 11 at 4:48
Saksham Sharma
10410
add a comment |Â
up vote
1
down vote
favorite
$sqrt4=2$
But is it same as writing...
$4^1/2=2$?
Basically I do not understand why $sqrt$-sign equals $1/2$?
roots
edited Aug 11 at 4:53
Cornman
2,55221128
asked Aug 11 at 4:48
Saksham Sharma
10410
1
For a speculative argument, square√4=2and4^1/2=2and see what you get. If you mean a formal proof, post the definitions you use for $,sqrt,cdot,,$ and $,(,cdot,)^1/2,$.
– dxiv
Aug 11 at 5:00
Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:02
Sorry, but that doesn't make math sense. Replace $4$ in your question with $2$ (or $pi$ for that matter), and it should become more obvious. What's under the $sqrtcdot$ is not necessarily an even integer power of something simple, so how would you halvethe number of times the number is multiplied by itself?
– dxiv
Aug 11 at 5:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$sqrt4=2$
But is it same as writing...
$4^1/2=2$?
Basically I do not understand why $sqrt$-sign equals $1/2$?
roots
edited Aug 11 at 4:53
Cornman
2,55221128
asked Aug 11 at 4:48
Saksham Sharma
10410
$sqrt4=2$
But is it same as writing...
$4^1/2=2$?
Basically I do not understand why $sqrt$-sign equals $1/2$?
roots
edited Aug 11 at 4:53
Cornman
2,55221128
asked Aug 11 at 4:48
Saksham Sharma
10410
edited Aug 11 at 4:53
Cornman
2,55221128
edited Aug 11 at 4:53
Cornman
2,55221128
edited Aug 11 at 4:53
Cornman
2,55221128
2,55221128
asked Aug 11 at 4:48
Saksham Sharma
10410
asked Aug 11 at 4:48
Saksham Sharma
10410
asked Aug 11 at 4:48
Saksham Sharma
10410
10410
1
For a speculative argument, square√4=2and4^1/2=2and see what you get. If you mean a formal proof, post the definitions you use for $,sqrt,cdot,,$ and $,(,cdot,)^1/2,$.
– dxiv
Aug 11 at 5:00
Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:02
Sorry, but that doesn't make math sense. Replace $4$ in your question with $2$ (or $pi$ for that matter), and it should become more obvious. What's under the $sqrtcdot$ is not necessarily an even integer power of something simple, so how would you halvethe number of times the number is multiplied by itself?
– dxiv
Aug 11 at 5:07
add a comment |Â
1
For a speculative argument, square√4=2and4^1/2=2and see what you get. If you mean a formal proof, post the definitions you use for $,sqrt,cdot,,$ and $,(,cdot,)^1/2,$.
– dxiv
Aug 11 at 5:00
Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:02
Sorry, but that doesn't make math sense. Replace $4$ in your question with $2$ (or $pi$ for that matter), and it should become more obvious. What's under the $sqrtcdot$ is not necessarily an even integer power of something simple, so how would you halvethe number of times the number is multiplied by itself?
– dxiv
Aug 11 at 5:07
1
1
For a speculative argument, square
√4=2 and 4^1/2=2 and see what you get. If you mean a formal proof, post the definitions you use for $,sqrt,cdot,,$ and $,(,cdot,)^1/2,$.– dxiv
Aug 11 at 5:00
For a speculative argument, square
√4=2 and 4^1/2=2 and see what you get. If you mean a formal proof, post the definitions you use for $,sqrt,cdot,,$ and $,(,cdot,)^1/2,$.– dxiv
Aug 11 at 5:00
Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:02
Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:02
Sorry, but that doesn't make math sense. Replace $4$ in your question with $2$ (or $pi$ for that matter), and it should become more obvious. What's under the $sqrtcdot$ is not necessarily an even integer power of something simple, so how would you halve
the number of times the number is multiplied by itself?– dxiv
Aug 11 at 5:07
Sorry, but that doesn't make math sense. Replace $4$ in your question with $2$ (or $pi$ for that matter), and it should become more obvious. What's under the $sqrtcdot$ is not necessarily an even integer power of something simple, so how would you halve
the number of times the number is multiplied by itself?– dxiv
Aug 11 at 5:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Basically it is just a definition of the $sqrt$-sign. Note, that the $sqrt$-sign is one of the few examples (maybe the only one?) in mathematics, where you write "nothing" (since $sqrtx=sqrt[2]x$) and actually mean $2$.
So yes. It is $sqrt4=4^1/2=2$
To make it more clear, we have $sqrt[n]x=x^1/n$, where $sqrt[n]x$ notes the $n-th$ root of $x$. For example $sqrt[3]8=2$, because $2cdot 2cdot 2=2^3=8$.
answered Aug 11 at 4:57
Cornman
2,55221128
I am actually confused. Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:00
2
You missphrase it. $sqrtx$ "asks" for the number which you have to multipy with itself to get $x$. So for $sqrt16$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$.
– Cornman
Aug 11 at 5:02
1
@SakshamSharma Yes, that's correct.
– Toby Mak
Aug 11 at 5:06
1
@Saksham Sharma No. You missphrase it again. $x^1/2$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $sqrt$, $sqrt[2]$ or ^(1/2)
– Cornman
Aug 11 at 5:08
1
@SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^1/2 is $x^1/2$. If you want to write the sqrt-sign with an power, you have to go like this: sqrt[n]x to get $sqrt[n]x$
– Cornman
Aug 11 at 5:10
 |Â
show 5 more comments
up vote
2
down vote
Unfortunately, I think, roots were studied before powers with arbitrary exponents were dared explored, so that the (ugly, in my opinion) symbol $sqrt(cdot)$ came much earlier than the other notation $(cdot)^1/2,$ which is a more natural notation as it falls out of the extension of exponentiation to rational numbers.
Specifically, a motivation for why $(cdot)^1/2$ is considered to be the same as the operator $sqrt(cdot)$ is this. The (positive) square root $x$ of a positive number $r$ is the number (why this $x$ is unique is proved in calculus) satisfying the relation $x^2=r.$ Once we grasp this characteristic property of square roots, it becomes easy to see that if we want to extend exponentiation to rational exponents so that the usual laws (especially in this case that $(a^m)^n=(a^n)^m=a^mn$ and $a^1=a$) are preserved, then we must have $$(x^2)^1/2=r^1/2,$$ whose left hand side becomes $x$ under our assumptions, so that we have the square root $x$ of $r$ given by $r^1/2.$
answered Aug 11 at 10:13
Allawonder
1,657414
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Basically it is just a definition of the $sqrt$-sign. Note, that the $sqrt$-sign is one of the few examples (maybe the only one?) in mathematics, where you write "nothing" (since $sqrtx=sqrt[2]x$) and actually mean $2$.
So yes. It is $sqrt4=4^1/2=2$
To make it more clear, we have $sqrt[n]x=x^1/n$, where $sqrt[n]x$ notes the $n-th$ root of $x$. For example $sqrt[3]8=2$, because $2cdot 2cdot 2=2^3=8$.
answered Aug 11 at 4:57
Cornman
2,55221128
I am actually confused. Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:00
2
You missphrase it. $sqrtx$ "asks" for the number which you have to multipy with itself to get $x$. So for $sqrt16$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$.
– Cornman
Aug 11 at 5:02
1
@SakshamSharma Yes, that's correct.
– Toby Mak
Aug 11 at 5:06
1
@Saksham Sharma No. You missphrase it again. $x^1/2$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $sqrt$, $sqrt[2]$ or ^(1/2)
– Cornman
Aug 11 at 5:08
1
@SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^1/2 is $x^1/2$. If you want to write the sqrt-sign with an power, you have to go like this: sqrt[n]x to get $sqrt[n]x$
– Cornman
Aug 11 at 5:10
 |Â
show 5 more comments
up vote
4
down vote
accepted
Basically it is just a definition of the $sqrt$-sign. Note, that the $sqrt$-sign is one of the few examples (maybe the only one?) in mathematics, where you write "nothing" (since $sqrtx=sqrt[2]x$) and actually mean $2$.
So yes. It is $sqrt4=4^1/2=2$
To make it more clear, we have $sqrt[n]x=x^1/n$, where $sqrt[n]x$ notes the $n-th$ root of $x$. For example $sqrt[3]8=2$, because $2cdot 2cdot 2=2^3=8$.
answered Aug 11 at 4:57
Cornman
2,55221128
I am actually confused. Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:00
2
You missphrase it. $sqrtx$ "asks" for the number which you have to multipy with itself to get $x$. So for $sqrt16$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$.
– Cornman
Aug 11 at 5:02
1
@SakshamSharma Yes, that's correct.
– Toby Mak
Aug 11 at 5:06
1
@Saksham Sharma No. You missphrase it again. $x^1/2$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $sqrt$, $sqrt[2]$ or ^(1/2)
– Cornman
Aug 11 at 5:08
1
@SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^1/2 is $x^1/2$. If you want to write the sqrt-sign with an power, you have to go like this: sqrt[n]x to get $sqrt[n]x$
– Cornman
Aug 11 at 5:10
 |Â
show 5 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Basically it is just a definition of the $sqrt$-sign. Note, that the $sqrt$-sign is one of the few examples (maybe the only one?) in mathematics, where you write "nothing" (since $sqrtx=sqrt[2]x$) and actually mean $2$.
So yes. It is $sqrt4=4^1/2=2$
To make it more clear, we have $sqrt[n]x=x^1/n$, where $sqrt[n]x$ notes the $n-th$ root of $x$. For example $sqrt[3]8=2$, because $2cdot 2cdot 2=2^3=8$.
answered Aug 11 at 4:57
Cornman
2,55221128
Basically it is just a definition of the $sqrt$-sign. Note, that the $sqrt$-sign is one of the few examples (maybe the only one?) in mathematics, where you write "nothing" (since $sqrtx=sqrt[2]x$) and actually mean $2$.
So yes. It is $sqrt4=4^1/2=2$
To make it more clear, we have $sqrt[n]x=x^1/n$, where $sqrt[n]x$ notes the $n-th$ root of $x$. For example $sqrt[3]8=2$, because $2cdot 2cdot 2=2^3=8$.
answered Aug 11 at 4:57
Cornman
2,55221128
edited Aug 11 at 5:00
answered Aug 11 at 4:57
Cornman
2,55221128
answered Aug 11 at 4:57
Cornman
2,55221128
answered Aug 11 at 4:57
Cornman
2,55221128
2,55221128
I am actually confused. Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:00
2
You missphrase it. $sqrtx$ "asks" for the number which you have to multipy with itself to get $x$. So for $sqrt16$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$.
– Cornman
Aug 11 at 5:02
1
@SakshamSharma Yes, that's correct.
– Toby Mak
Aug 11 at 5:06
1
@Saksham Sharma No. You missphrase it again. $x^1/2$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $sqrt$, $sqrt[2]$ or ^(1/2)
– Cornman
Aug 11 at 5:08
1
@SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^1/2 is $x^1/2$. If you want to write the sqrt-sign with an power, you have to go like this: sqrt[n]x to get $sqrt[n]x$
– Cornman
Aug 11 at 5:10
 |Â
show 5 more comments
I am actually confused. Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:00
2
You missphrase it. $sqrtx$ "asks" for the number which you have to multipy with itself to get $x$. So for $sqrt16$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$.
– Cornman
Aug 11 at 5:02
1
@SakshamSharma Yes, that's correct.
– Toby Mak
Aug 11 at 5:06
1
@Saksham Sharma No. You missphrase it again. $x^1/2$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $sqrt$, $sqrt[2]$ or ^(1/2)
– Cornman
Aug 11 at 5:08
1
@SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^1/2 is $x^1/2$. If you want to write the sqrt-sign with an power, you have to go like this: sqrt[n]x to get $sqrt[n]x$
– Cornman
Aug 11 at 5:10
I am actually confused. Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:00
I am actually confused. Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:00
2
2
You missphrase it. $sqrtx$ "asks" for the number which you have to multipy with itself to get $x$. So for $sqrt16$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$.
– Cornman
Aug 11 at 5:02
You missphrase it. $sqrtx$ "asks" for the number which you have to multipy with itself to get $x$. So for $sqrt16$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$.
– Cornman
Aug 11 at 5:02
1
1
@SakshamSharma Yes, that's correct.
– Toby Mak
Aug 11 at 5:06
@SakshamSharma Yes, that's correct.
– Toby Mak
Aug 11 at 5:06
1
1
@Saksham Sharma No. You missphrase it again. $x^1/2$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $sqrt$, $sqrt[2]$ or ^(1/2)
– Cornman
Aug 11 at 5:08
@Saksham Sharma No. You missphrase it again. $x^1/2$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $sqrt$, $sqrt[2]$ or ^(1/2)
– Cornman
Aug 11 at 5:08
1
1
@SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^1/2 is $x^1/2$. If you want to write the sqrt-sign with an power, you have to go like this: sqrt[n]x to get $sqrt[n]x$
– Cornman
Aug 11 at 5:10
@SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^1/2 is $x^1/2$. If you want to write the sqrt-sign with an power, you have to go like this: sqrt[n]x to get $sqrt[n]x$
– Cornman
Aug 11 at 5:10
 |Â
show 5 more comments
up vote
2
down vote
Unfortunately, I think, roots were studied before powers with arbitrary exponents were dared explored, so that the (ugly, in my opinion) symbol $sqrt(cdot)$ came much earlier than the other notation $(cdot)^1/2,$ which is a more natural notation as it falls out of the extension of exponentiation to rational numbers.
Specifically, a motivation for why $(cdot)^1/2$ is considered to be the same as the operator $sqrt(cdot)$ is this. The (positive) square root $x$ of a positive number $r$ is the number (why this $x$ is unique is proved in calculus) satisfying the relation $x^2=r.$ Once we grasp this characteristic property of square roots, it becomes easy to see that if we want to extend exponentiation to rational exponents so that the usual laws (especially in this case that $(a^m)^n=(a^n)^m=a^mn$ and $a^1=a$) are preserved, then we must have $$(x^2)^1/2=r^1/2,$$ whose left hand side becomes $x$ under our assumptions, so that we have the square root $x$ of $r$ given by $r^1/2.$
answered Aug 11 at 10:13
Allawonder
1,657414
add a comment |Â
up vote
2
down vote
Unfortunately, I think, roots were studied before powers with arbitrary exponents were dared explored, so that the (ugly, in my opinion) symbol $sqrt(cdot)$ came much earlier than the other notation $(cdot)^1/2,$ which is a more natural notation as it falls out of the extension of exponentiation to rational numbers.
Specifically, a motivation for why $(cdot)^1/2$ is considered to be the same as the operator $sqrt(cdot)$ is this. The (positive) square root $x$ of a positive number $r$ is the number (why this $x$ is unique is proved in calculus) satisfying the relation $x^2=r.$ Once we grasp this characteristic property of square roots, it becomes easy to see that if we want to extend exponentiation to rational exponents so that the usual laws (especially in this case that $(a^m)^n=(a^n)^m=a^mn$ and $a^1=a$) are preserved, then we must have $$(x^2)^1/2=r^1/2,$$ whose left hand side becomes $x$ under our assumptions, so that we have the square root $x$ of $r$ given by $r^1/2.$
answered Aug 11 at 10:13
Allawonder
1,657414
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Unfortunately, I think, roots were studied before powers with arbitrary exponents were dared explored, so that the (ugly, in my opinion) symbol $sqrt(cdot)$ came much earlier than the other notation $(cdot)^1/2,$ which is a more natural notation as it falls out of the extension of exponentiation to rational numbers.
Specifically, a motivation for why $(cdot)^1/2$ is considered to be the same as the operator $sqrt(cdot)$ is this. The (positive) square root $x$ of a positive number $r$ is the number (why this $x$ is unique is proved in calculus) satisfying the relation $x^2=r.$ Once we grasp this characteristic property of square roots, it becomes easy to see that if we want to extend exponentiation to rational exponents so that the usual laws (especially in this case that $(a^m)^n=(a^n)^m=a^mn$ and $a^1=a$) are preserved, then we must have $$(x^2)^1/2=r^1/2,$$ whose left hand side becomes $x$ under our assumptions, so that we have the square root $x$ of $r$ given by $r^1/2.$
answered Aug 11 at 10:13
Allawonder
1,657414
Unfortunately, I think, roots were studied before powers with arbitrary exponents were dared explored, so that the (ugly, in my opinion) symbol $sqrt(cdot)$ came much earlier than the other notation $(cdot)^1/2,$ which is a more natural notation as it falls out of the extension of exponentiation to rational numbers.
Specifically, a motivation for why $(cdot)^1/2$ is considered to be the same as the operator $sqrt(cdot)$ is this. The (positive) square root $x$ of a positive number $r$ is the number (why this $x$ is unique is proved in calculus) satisfying the relation $x^2=r.$ Once we grasp this characteristic property of square roots, it becomes easy to see that if we want to extend exponentiation to rational exponents so that the usual laws (especially in this case that $(a^m)^n=(a^n)^m=a^mn$ and $a^1=a$) are preserved, then we must have $$(x^2)^1/2=r^1/2,$$ whose left hand side becomes $x$ under our assumptions, so that we have the square root $x$ of $r$ given by $r^1/2.$
answered Aug 11 at 10:13
Allawonder
1,657414
edited Aug 11 at 10:23
answered Aug 11 at 10:13
Allawonder
1,657414
answered Aug 11 at 10:13
Allawonder
1,657414
answered Aug 11 at 10:13
Allawonder
1,657414
1,657414
add a comment |Â
add a comment |Â
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1
For a speculative argument, square
√4=2and4^1/2=2and see what you get. If you mean a formal proof, post the definitions you use for $,sqrt,cdot,,$ and $,(,cdot,)^1/2,$.– dxiv
Aug 11 at 5:00
Does √ halves the number of times the number is multiplied by itself?
– Saksham Sharma
Aug 11 at 5:02
Sorry, but that doesn't make math sense. Replace $4$ in your question with $2$ (or $pi$ for that matter), and it should become more obvious. What's under the $sqrtcdot$ is not necessarily an even integer power of something simple, so how would you halve
the number of times the number is multiplied by itself?– dxiv
Aug 11 at 5:07