Mixing
A Question About Square Roots And Exponent Laws
Clash Royale CLAN TAG#URR8PPP
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Why is it in math, $sqrtab$=$sqrta$$sqrtb$? I get why this is the case for any other power instead of $1/2$. For instance, if the power was for, then $(ab)^4$=$(a)^4$$(b)^4$ because on both sides, there will be 4 a's and 4 b's. But somehow, this proof doesn't seem as intuitive when used to prove that $sqrtab$=$sqrta$$sqrtb$. So can someone please prove/explain to me why $sqrtab$=$sqrta$$sqrtb$? Please don't do this too rigorously, just explain it at the level of a high school pre-calc student please.
algebra-precalculus exponentiation radicals
edited Jul 31 at 15:42
Abcd
2,3151624
asked Jul 31 at 7:58
Ethan Chan
598322
add a comment |Â
up vote
2
down vote
favorite
Why is it in math, $sqrtab$=$sqrta$$sqrtb$? I get why this is the case for any other power instead of $1/2$. For instance, if the power was for, then $(ab)^4$=$(a)^4$$(b)^4$ because on both sides, there will be 4 a's and 4 b's. But somehow, this proof doesn't seem as intuitive when used to prove that $sqrtab$=$sqrta$$sqrtb$. So can someone please prove/explain to me why $sqrtab$=$sqrta$$sqrtb$? Please don't do this too rigorously, just explain it at the level of a high school pre-calc student please.
algebra-precalculus exponentiation radicals
edited Jul 31 at 15:42
Abcd
2,3151624
asked Jul 31 at 7:58
Ethan Chan
598322
Perhaps because each has the same square?
– Dave L. Renfro
Jul 31 at 8:04
Fundamentally this depends on one of the properties of index that you studied about $(ab)^2=a^2b^2$
– daruma
Jul 31 at 8:04
Convince yourself that $(sqrtasqrtb)^2 = ab$.
– Good Morning Captain
Jul 31 at 8:04
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Why is it in math, $sqrtab$=$sqrta$$sqrtb$? I get why this is the case for any other power instead of $1/2$. For instance, if the power was for, then $(ab)^4$=$(a)^4$$(b)^4$ because on both sides, there will be 4 a's and 4 b's. But somehow, this proof doesn't seem as intuitive when used to prove that $sqrtab$=$sqrta$$sqrtb$. So can someone please prove/explain to me why $sqrtab$=$sqrta$$sqrtb$? Please don't do this too rigorously, just explain it at the level of a high school pre-calc student please.
algebra-precalculus exponentiation radicals
edited Jul 31 at 15:42
Abcd
2,3151624
asked Jul 31 at 7:58
Ethan Chan
598322
Why is it in math, $sqrtab$=$sqrta$$sqrtb$? I get why this is the case for any other power instead of $1/2$. For instance, if the power was for, then $(ab)^4$=$(a)^4$$(b)^4$ because on both sides, there will be 4 a's and 4 b's. But somehow, this proof doesn't seem as intuitive when used to prove that $sqrtab$=$sqrta$$sqrtb$. So can someone please prove/explain to me why $sqrtab$=$sqrta$$sqrtb$? Please don't do this too rigorously, just explain it at the level of a high school pre-calc student please.
algebra-precalculus exponentiation radicals
edited Jul 31 at 15:42
Abcd
2,3151624
asked Jul 31 at 7:58
Ethan Chan
598322
edited Jul 31 at 15:42
Abcd
2,3151624
edited Jul 31 at 15:42
Abcd
2,3151624
edited Jul 31 at 15:42
Abcd
2,3151624
2,3151624
asked Jul 31 at 7:58
Ethan Chan
598322
asked Jul 31 at 7:58
Ethan Chan
598322
asked Jul 31 at 7:58
Ethan Chan
598322
598322
Perhaps because each has the same square?
– Dave L. Renfro
Jul 31 at 8:04
Fundamentally this depends on one of the properties of index that you studied about $(ab)^2=a^2b^2$
– daruma
Jul 31 at 8:04
Convince yourself that $(sqrtasqrtb)^2 = ab$.
– Good Morning Captain
Jul 31 at 8:04
add a comment |Â
Perhaps because each has the same square?
– Dave L. Renfro
Jul 31 at 8:04
Fundamentally this depends on one of the properties of index that you studied about $(ab)^2=a^2b^2$
– daruma
Jul 31 at 8:04
Convince yourself that $(sqrtasqrtb)^2 = ab$.
– Good Morning Captain
Jul 31 at 8:04
Perhaps because each has the same square?
– Dave L. Renfro
Jul 31 at 8:04
Perhaps because each has the same square?
– Dave L. Renfro
Jul 31 at 8:04
Fundamentally this depends on one of the properties of index that you studied about $(ab)^2=a^2b^2$
– daruma
Jul 31 at 8:04
Fundamentally this depends on one of the properties of index that you studied about $(ab)^2=a^2b^2$
– daruma
Jul 31 at 8:04
Convince yourself that $(sqrtasqrtb)^2 = ab$.
– Good Morning Captain
Jul 31 at 8:04
Convince yourself that $(sqrtasqrtb)^2 = ab$.
– Good Morning Captain
Jul 31 at 8:04
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The statement is only true if $a, b geq 0$.
Let $a, b geq 0$. Let $u = sqrta$; let $v = sqrtb$. Since $sqrtx$ is the principal (nonnegative) square root of $x$, $u, v geq 0$, $u^2 = a$, $v^2 = b$.
beginalign*
sqrtab & = sqrtu^2v^2 && textsubstitution\
& = sqrtuuvv && textby definition\
& = sqrtu(uv)v && textassociativity of multiplication\
& = sqrtu(vu)v && textcommutativity of multiplication\
& = sqrt(uv)(uv) && textassociativity of multiplication\
& = sqrt(uv)^2 && textby definition\
& = |uv| && texttake principal square root\
& = uv && text = uv$\
& = sqrtasqrtb && textsubstitution
endalign*
answered Jul 31 at 8:26
N. F. Taussig
38k93053
1
Thank you, I get it now.
– Ethan Chan
Jul 31 at 8:28
add a comment |Â
up vote
2
down vote
Let $a,bge0$ (the version with complex numbers of this identity does not hold). By definition, you want to prove that $alpha=sqrt asqrt b$ is a non-negative real number such that $alpha^2=ab$. Since $sqrt a$ and $sqrt b$ are non-negative real numbers, $alpha$ is too. Finally, $alpha^2=left(sqrt asqrt bright)^2=left(sqrt aright)^2left(sqrt bright)^2=ab$.
answered Jul 31 at 8:04
Saucy O'Path
2,394217
But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same.
– Ethan Chan
Jul 31 at 8:09
@EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $sqrt[2n]bullet$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $xmapsto x^2$ (or $xmapsto x^2n$ with $nge 1$) is strictly increasing on $[0,infty)$, there is at most one such number in this interval.
– Saucy O'Path
Jul 31 at 8:13
Oh right, never mind. But then, how about for something like a cube root?
– Ethan Chan
Jul 31 at 8:14
1
math.stackexchange.com/questions/1700731/…
– Saucy O'Path
Jul 31 at 10:22
add a comment |Â
up vote
0
down vote
First of all: it is not always $sqrtab=sqrtasqrtb$, there are some conditions you have to check first e.g. you want to have $a,bgeq 0$ or even more trivial $a,binmathbb R$ as the root of complex numbers turns out to be...well, rather complex to handle.
Assuming that all of these conditions are met and you accept the fact that $(xcdot y)^n=x^ncdot y^n$ for $ninmathbb N$ maybe the following helps you:
$$left(sqrt[n]xcdotsqrt[n]yright)^n = sqrt[n]xcdotsqrt[n]y = xcdot y=left(sqrt[n]xcdot yright)^n$$
Thus we have
$$left(sqrt[n]xcdotsqrt[n]yright)^n =left(sqrt[n]xcdot yright)^n$$
which yields
$$sqrt[n]xcdotsqrt[n]y=sqrt[n]xcdot y.$$
answered Jul 31 at 8:07
Hirshy
4,32021336
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The statement is only true if $a, b geq 0$.
Let $a, b geq 0$. Let $u = sqrta$; let $v = sqrtb$. Since $sqrtx$ is the principal (nonnegative) square root of $x$, $u, v geq 0$, $u^2 = a$, $v^2 = b$.
beginalign*
sqrtab & = sqrtu^2v^2 && textsubstitution\
& = sqrtuuvv && textby definition\
& = sqrtu(uv)v && textassociativity of multiplication\
& = sqrtu(vu)v && textcommutativity of multiplication\
& = sqrt(uv)(uv) && textassociativity of multiplication\
& = sqrt(uv)^2 && textby definition\
& = |uv| && texttake principal square root\
& = uv && text = uv$\
& = sqrtasqrtb && textsubstitution
endalign*
answered Jul 31 at 8:26
N. F. Taussig
38k93053
1
Thank you, I get it now.
– Ethan Chan
Jul 31 at 8:28
add a comment |Â
up vote
2
down vote
accepted
The statement is only true if $a, b geq 0$.
Let $a, b geq 0$. Let $u = sqrta$; let $v = sqrtb$. Since $sqrtx$ is the principal (nonnegative) square root of $x$, $u, v geq 0$, $u^2 = a$, $v^2 = b$.
beginalign*
sqrtab & = sqrtu^2v^2 && textsubstitution\
& = sqrtuuvv && textby definition\
& = sqrtu(uv)v && textassociativity of multiplication\
& = sqrtu(vu)v && textcommutativity of multiplication\
& = sqrt(uv)(uv) && textassociativity of multiplication\
& = sqrt(uv)^2 && textby definition\
& = |uv| && texttake principal square root\
& = uv && text = uv$\
& = sqrtasqrtb && textsubstitution
endalign*
answered Jul 31 at 8:26
N. F. Taussig
38k93053
1
Thank you, I get it now.
– Ethan Chan
Jul 31 at 8:28
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The statement is only true if $a, b geq 0$.
Let $a, b geq 0$. Let $u = sqrta$; let $v = sqrtb$. Since $sqrtx$ is the principal (nonnegative) square root of $x$, $u, v geq 0$, $u^2 = a$, $v^2 = b$.
beginalign*
sqrtab & = sqrtu^2v^2 && textsubstitution\
& = sqrtuuvv && textby definition\
& = sqrtu(uv)v && textassociativity of multiplication\
& = sqrtu(vu)v && textcommutativity of multiplication\
& = sqrt(uv)(uv) && textassociativity of multiplication\
& = sqrt(uv)^2 && textby definition\
& = |uv| && texttake principal square root\
& = uv && text = uv$\
& = sqrtasqrtb && textsubstitution
endalign*
answered Jul 31 at 8:26
N. F. Taussig
38k93053
The statement is only true if $a, b geq 0$.
Let $a, b geq 0$. Let $u = sqrta$; let $v = sqrtb$. Since $sqrtx$ is the principal (nonnegative) square root of $x$, $u, v geq 0$, $u^2 = a$, $v^2 = b$.
beginalign*
sqrtab & = sqrtu^2v^2 && textsubstitution\
& = sqrtuuvv && textby definition\
& = sqrtu(uv)v && textassociativity of multiplication\
& = sqrtu(vu)v && textcommutativity of multiplication\
& = sqrt(uv)(uv) && textassociativity of multiplication\
& = sqrt(uv)^2 && textby definition\
& = |uv| && texttake principal square root\
& = uv && text = uv$\
& = sqrtasqrtb && textsubstitution
endalign*
answered Jul 31 at 8:26
N. F. Taussig
38k93053
answered Jul 31 at 8:26
N. F. Taussig
38k93053
answered Jul 31 at 8:26
N. F. Taussig
38k93053
answered Jul 31 at 8:26
N. F. Taussig
38k93053
38k93053
1
Thank you, I get it now.
– Ethan Chan
Jul 31 at 8:28
add a comment |Â
1
Thank you, I get it now.
– Ethan Chan
Jul 31 at 8:28
1
1
Thank you, I get it now.
– Ethan Chan
Jul 31 at 8:28
Thank you, I get it now.
– Ethan Chan
Jul 31 at 8:28
add a comment |Â
up vote
2
down vote
Let $a,bge0$ (the version with complex numbers of this identity does not hold). By definition, you want to prove that $alpha=sqrt asqrt b$ is a non-negative real number such that $alpha^2=ab$. Since $sqrt a$ and $sqrt b$ are non-negative real numbers, $alpha$ is too. Finally, $alpha^2=left(sqrt asqrt bright)^2=left(sqrt aright)^2left(sqrt bright)^2=ab$.
answered Jul 31 at 8:04
Saucy O'Path
2,394217
But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same.
– Ethan Chan
Jul 31 at 8:09
@EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $sqrt[2n]bullet$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $xmapsto x^2$ (or $xmapsto x^2n$ with $nge 1$) is strictly increasing on $[0,infty)$, there is at most one such number in this interval.
– Saucy O'Path
Jul 31 at 8:13
Oh right, never mind. But then, how about for something like a cube root?
– Ethan Chan
Jul 31 at 8:14
1
math.stackexchange.com/questions/1700731/…
– Saucy O'Path
Jul 31 at 10:22
add a comment |Â
up vote
2
down vote
Let $a,bge0$ (the version with complex numbers of this identity does not hold). By definition, you want to prove that $alpha=sqrt asqrt b$ is a non-negative real number such that $alpha^2=ab$. Since $sqrt a$ and $sqrt b$ are non-negative real numbers, $alpha$ is too. Finally, $alpha^2=left(sqrt asqrt bright)^2=left(sqrt aright)^2left(sqrt bright)^2=ab$.
answered Jul 31 at 8:04
Saucy O'Path
2,394217
But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same.
– Ethan Chan
Jul 31 at 8:09
@EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $sqrt[2n]bullet$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $xmapsto x^2$ (or $xmapsto x^2n$ with $nge 1$) is strictly increasing on $[0,infty)$, there is at most one such number in this interval.
– Saucy O'Path
Jul 31 at 8:13
Oh right, never mind. But then, how about for something like a cube root?
– Ethan Chan
Jul 31 at 8:14
1
math.stackexchange.com/questions/1700731/…
– Saucy O'Path
Jul 31 at 10:22
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $a,bge0$ (the version with complex numbers of this identity does not hold). By definition, you want to prove that $alpha=sqrt asqrt b$ is a non-negative real number such that $alpha^2=ab$. Since $sqrt a$ and $sqrt b$ are non-negative real numbers, $alpha$ is too. Finally, $alpha^2=left(sqrt asqrt bright)^2=left(sqrt aright)^2left(sqrt bright)^2=ab$.
answered Jul 31 at 8:04
Saucy O'Path
2,394217
Let $a,bge0$ (the version with complex numbers of this identity does not hold). By definition, you want to prove that $alpha=sqrt asqrt b$ is a non-negative real number such that $alpha^2=ab$. Since $sqrt a$ and $sqrt b$ are non-negative real numbers, $alpha$ is too. Finally, $alpha^2=left(sqrt asqrt bright)^2=left(sqrt aright)^2left(sqrt bright)^2=ab$.
answered Jul 31 at 8:04
Saucy O'Path
2,394217
answered Jul 31 at 8:04
Saucy O'Path
2,394217
answered Jul 31 at 8:04
Saucy O'Path
2,394217
answered Jul 31 at 8:04
Saucy O'Path
2,394217
2,394217
But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same.
– Ethan Chan
Jul 31 at 8:09
@EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $sqrt[2n]bullet$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $xmapsto x^2$ (or $xmapsto x^2n$ with $nge 1$) is strictly increasing on $[0,infty)$, there is at most one such number in this interval.
– Saucy O'Path
Jul 31 at 8:13
Oh right, never mind. But then, how about for something like a cube root?
– Ethan Chan
Jul 31 at 8:14
1
math.stackexchange.com/questions/1700731/…
– Saucy O'Path
Jul 31 at 10:22
add a comment |Â
But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same.
– Ethan Chan
Jul 31 at 8:09
@EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $sqrt[2n]bullet$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $xmapsto x^2$ (or $xmapsto x^2n$ with $nge 1$) is strictly increasing on $[0,infty)$, there is at most one such number in this interval.
– Saucy O'Path
Jul 31 at 8:13
Oh right, never mind. But then, how about for something like a cube root?
– Ethan Chan
Jul 31 at 8:14
1
math.stackexchange.com/questions/1700731/…
– Saucy O'Path
Jul 31 at 10:22
But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same.
– Ethan Chan
Jul 31 at 8:09
But how does the fact that when they are squared, they yield the same results prove that they are the same? Because when -2 and 2 are squared, they are both equal to 2? But that doesn't mean they are the same.
– Ethan Chan
Jul 31 at 8:09
@EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $sqrt[2n]bullet$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $xmapsto x^2$ (or $xmapsto x^2n$ with $nge 1$) is strictly increasing on $[0,infty)$, there is at most one such number in this interval.
– Saucy O'Path
Jul 31 at 8:13
@EthanChan That's why I kept writing "non-negative" all those times, and why the definition of square root - or $sqrt[2n]bullet$, for that matter - explicitly requires the root to be the non-negative solution to the equation $t^2=a$. Since the map $xmapsto x^2$ (or $xmapsto x^2n$ with $nge 1$) is strictly increasing on $[0,infty)$, there is at most one such number in this interval.
– Saucy O'Path
Jul 31 at 8:13
Oh right, never mind. But then, how about for something like a cube root?
– Ethan Chan
Jul 31 at 8:14
Oh right, never mind. But then, how about for something like a cube root?
– Ethan Chan
Jul 31 at 8:14
1
1
math.stackexchange.com/questions/1700731/…
– Saucy O'Path
Jul 31 at 10:22
math.stackexchange.com/questions/1700731/…
– Saucy O'Path
Jul 31 at 10:22
add a comment |Â
up vote
0
down vote
First of all: it is not always $sqrtab=sqrtasqrtb$, there are some conditions you have to check first e.g. you want to have $a,bgeq 0$ or even more trivial $a,binmathbb R$ as the root of complex numbers turns out to be...well, rather complex to handle.
Assuming that all of these conditions are met and you accept the fact that $(xcdot y)^n=x^ncdot y^n$ for $ninmathbb N$ maybe the following helps you:
$$left(sqrt[n]xcdotsqrt[n]yright)^n = sqrt[n]xcdotsqrt[n]y = xcdot y=left(sqrt[n]xcdot yright)^n$$
Thus we have
$$left(sqrt[n]xcdotsqrt[n]yright)^n =left(sqrt[n]xcdot yright)^n$$
which yields
$$sqrt[n]xcdotsqrt[n]y=sqrt[n]xcdot y.$$
answered Jul 31 at 8:07
Hirshy
4,32021336
add a comment |Â
up vote
0
down vote
First of all: it is not always $sqrtab=sqrtasqrtb$, there are some conditions you have to check first e.g. you want to have $a,bgeq 0$ or even more trivial $a,binmathbb R$ as the root of complex numbers turns out to be...well, rather complex to handle.
Assuming that all of these conditions are met and you accept the fact that $(xcdot y)^n=x^ncdot y^n$ for $ninmathbb N$ maybe the following helps you:
$$left(sqrt[n]xcdotsqrt[n]yright)^n = sqrt[n]xcdotsqrt[n]y = xcdot y=left(sqrt[n]xcdot yright)^n$$
Thus we have
$$left(sqrt[n]xcdotsqrt[n]yright)^n =left(sqrt[n]xcdot yright)^n$$
which yields
$$sqrt[n]xcdotsqrt[n]y=sqrt[n]xcdot y.$$
answered Jul 31 at 8:07
Hirshy
4,32021336
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First of all: it is not always $sqrtab=sqrtasqrtb$, there are some conditions you have to check first e.g. you want to have $a,bgeq 0$ or even more trivial $a,binmathbb R$ as the root of complex numbers turns out to be...well, rather complex to handle.
Assuming that all of these conditions are met and you accept the fact that $(xcdot y)^n=x^ncdot y^n$ for $ninmathbb N$ maybe the following helps you:
$$left(sqrt[n]xcdotsqrt[n]yright)^n = sqrt[n]xcdotsqrt[n]y = xcdot y=left(sqrt[n]xcdot yright)^n$$
Thus we have
$$left(sqrt[n]xcdotsqrt[n]yright)^n =left(sqrt[n]xcdot yright)^n$$
which yields
$$sqrt[n]xcdotsqrt[n]y=sqrt[n]xcdot y.$$
answered Jul 31 at 8:07
Hirshy
4,32021336
First of all: it is not always $sqrtab=sqrtasqrtb$, there are some conditions you have to check first e.g. you want to have $a,bgeq 0$ or even more trivial $a,binmathbb R$ as the root of complex numbers turns out to be...well, rather complex to handle.
Assuming that all of these conditions are met and you accept the fact that $(xcdot y)^n=x^ncdot y^n$ for $ninmathbb N$ maybe the following helps you:
$$left(sqrt[n]xcdotsqrt[n]yright)^n = sqrt[n]xcdotsqrt[n]y = xcdot y=left(sqrt[n]xcdot yright)^n$$
Thus we have
$$left(sqrt[n]xcdotsqrt[n]yright)^n =left(sqrt[n]xcdot yright)^n$$
which yields
$$sqrt[n]xcdotsqrt[n]y=sqrt[n]xcdot y.$$
answered Jul 31 at 8:07
Hirshy
4,32021336
answered Jul 31 at 8:07
Hirshy
4,32021336
answered Jul 31 at 8:07
Hirshy
4,32021336
answered Jul 31 at 8:07
Hirshy
4,32021336
4,32021336
add a comment |Â
add a comment |Â
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Perhaps because each has the same square?
– Dave L. Renfro
Jul 31 at 8:04
Fundamentally this depends on one of the properties of index that you studied about $(ab)^2=a^2b^2$
– daruma
Jul 31 at 8:04
Convince yourself that $(sqrtasqrtb)^2 = ab$.
– Good Morning Captain
Jul 31 at 8:04