Complicated square root problem.












3












$begingroup$


I was wondering the general method to solve




What is the value of $sqrt{a-bsqrt{c}}?$




The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)






number-theory radicals nested-radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    "The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
    $endgroup$
    – fleablood
    1 hour ago












  • $begingroup$
    By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
    $endgroup$
    – fleablood
    1 hour ago










  • $begingroup$
    Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    @fleablood I know, complex numbers. How should I write it then?
    $endgroup$
    – Max0815
    1 hour ago
















3












$begingroup$


I was wondering the general method to solve




What is the value of $sqrt{a-bsqrt{c}}?$




The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)






number-theory radicals nested-radicals







share|cite|improve this question











$endgroup$












  • $begingroup$
    "The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
    $endgroup$
    – fleablood
    1 hour ago












  • $begingroup$
    By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
    $endgroup$
    – fleablood
    1 hour ago










  • $begingroup$
    Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    @fleablood I know, complex numbers. How should I write it then?
    $endgroup$
    – Max0815
    1 hour ago














3












3








3


2



$begingroup$


I was wondering the general method to solve




What is the value of $sqrt{a-bsqrt{c}}?$




The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)






number-theory radicals nested-radicals







share|cite|improve this question











$endgroup$




I was wondering the general method to solve




What is the value of $sqrt{a-bsqrt{c}}?$




The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)






number-theory radicals nested-radicals




number-theory radicals nested-radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 30 mins ago









Michael Rozenberg

104k1891196




104k1891196










asked 2 hours ago









Max0815Max0815

67118




67118












  • $begingroup$
    "The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
    $endgroup$
    – fleablood
    1 hour ago












  • $begingroup$
    By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
    $endgroup$
    – fleablood
    1 hour ago










  • $begingroup$
    Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    @fleablood I know, complex numbers. How should I write it then?
    $endgroup$
    – Max0815
    1 hour ago


















  • $begingroup$
    "The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
    $endgroup$
    – fleablood
    1 hour ago












  • $begingroup$
    By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
    $endgroup$
    – fleablood
    1 hour ago










  • $begingroup$
    Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
    $endgroup$
    – JMoravitz
    1 hour ago










  • $begingroup$
    @fleablood I know, complex numbers. How should I write it then?
    $endgroup$
    – Max0815
    1 hour ago
















$begingroup$
"The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
$endgroup$
– fleablood
1 hour ago






$begingroup$
"The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
$endgroup$
– fleablood
1 hour ago














$begingroup$
By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
$endgroup$
– fleablood
1 hour ago




$begingroup$
By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
$endgroup$
– fleablood
1 hour ago












$begingroup$
Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
$endgroup$
– JMoravitz
1 hour ago




$begingroup$
Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
$endgroup$
– JMoravitz
1 hour ago












$begingroup$
@fleablood I know, complex numbers. How should I write it then?
$endgroup$
– Max0815
1 hour ago




$begingroup$
@fleablood I know, complex numbers. How should I write it then?
$endgroup$
– Max0815
1 hour ago










2 Answers
2






active

oldest

votes


















1












$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)



To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.



Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes. thanx!!!!!
    $endgroup$
    – Max0815
    1 hour ago










  • $begingroup$
    If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
    $endgroup$
    – Stan Tendijck
    1 hour ago












  • $begingroup$
    yes I believe so too.
    $endgroup$
    – Max0815
    48 mins ago



















5












$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.



For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is interesting. I had never seen the identities you begin with.
    $endgroup$
    – Lubin
    49 mins ago












  • $begingroup$
    @Lubin same with me too.
    $endgroup$
    – Max0815
    48 mins ago










  • $begingroup$
    We can prove it. It's not hard.
    $endgroup$
    – Michael Rozenberg
    33 mins ago










  • $begingroup$
    How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy.
    $endgroup$
    – Max0815
    6 secs ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)



To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.



Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes. thanx!!!!!
    $endgroup$
    – Max0815
    1 hour ago










  • $begingroup$
    If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
    $endgroup$
    – Stan Tendijck
    1 hour ago












  • $begingroup$
    yes I believe so too.
    $endgroup$
    – Max0815
    48 mins ago
















1












$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)



To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.



Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes. thanx!!!!!
    $endgroup$
    – Max0815
    1 hour ago










  • $begingroup$
    If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
    $endgroup$
    – Stan Tendijck
    1 hour ago












  • $begingroup$
    yes I believe so too.
    $endgroup$
    – Max0815
    48 mins ago














1












1








1





$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)



To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.



Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?






share|cite|improve this answer











$endgroup$



One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)



To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.



Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









Stan TendijckStan Tendijck

1,826311




1,826311












  • $begingroup$
    Yes. thanx!!!!!
    $endgroup$
    – Max0815
    1 hour ago










  • $begingroup$
    If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
    $endgroup$
    – Stan Tendijck
    1 hour ago












  • $begingroup$
    yes I believe so too.
    $endgroup$
    – Max0815
    48 mins ago


















  • $begingroup$
    Yes. thanx!!!!!
    $endgroup$
    – Max0815
    1 hour ago










  • $begingroup$
    If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
    $endgroup$
    – Stan Tendijck
    1 hour ago












  • $begingroup$
    yes I believe so too.
    $endgroup$
    – Max0815
    48 mins ago
















$begingroup$
Yes. thanx!!!!!
$endgroup$
– Max0815
1 hour ago




$begingroup$
Yes. thanx!!!!!
$endgroup$
– Max0815
1 hour ago












$begingroup$
If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
$endgroup$
– Stan Tendijck
1 hour ago






$begingroup$
If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
$endgroup$
– Stan Tendijck
1 hour ago














$begingroup$
yes I believe so too.
$endgroup$
– Max0815
48 mins ago




$begingroup$
yes I believe so too.
$endgroup$
– Max0815
48 mins ago











5












$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.



For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is interesting. I had never seen the identities you begin with.
    $endgroup$
    – Lubin
    49 mins ago












  • $begingroup$
    @Lubin same with me too.
    $endgroup$
    – Max0815
    48 mins ago










  • $begingroup$
    We can prove it. It's not hard.
    $endgroup$
    – Michael Rozenberg
    33 mins ago










  • $begingroup$
    How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy.
    $endgroup$
    – Max0815
    6 secs ago
















5












$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.



For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is interesting. I had never seen the identities you begin with.
    $endgroup$
    – Lubin
    49 mins ago












  • $begingroup$
    @Lubin same with me too.
    $endgroup$
    – Max0815
    48 mins ago










  • $begingroup$
    We can prove it. It's not hard.
    $endgroup$
    – Michael Rozenberg
    33 mins ago










  • $begingroup$
    How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy.
    $endgroup$
    – Max0815
    6 secs ago














5












5








5





$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.



For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.






share|cite|improve this answer











$endgroup$



There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.



For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.







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edited 1 hour ago

























answered 1 hour ago









Michael RozenbergMichael Rozenberg

104k1891196




104k1891196












  • $begingroup$
    This is interesting. I had never seen the identities you begin with.
    $endgroup$
    – Lubin
    49 mins ago












  • $begingroup$
    @Lubin same with me too.
    $endgroup$
    – Max0815
    48 mins ago










  • $begingroup$
    We can prove it. It's not hard.
    $endgroup$
    – Michael Rozenberg
    33 mins ago










  • $begingroup$
    How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy.
    $endgroup$
    – Max0815
    6 secs ago


















  • $begingroup$
    This is interesting. I had never seen the identities you begin with.
    $endgroup$
    – Lubin
    49 mins ago












  • $begingroup$
    @Lubin same with me too.
    $endgroup$
    – Max0815
    48 mins ago










  • $begingroup$
    We can prove it. It's not hard.
    $endgroup$
    – Michael Rozenberg
    33 mins ago










  • $begingroup$
    How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy.
    $endgroup$
    – Max0815
    6 secs ago
















$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
49 mins ago






$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
49 mins ago














$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
48 mins ago




$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
48 mins ago












$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
33 mins ago




$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
33 mins ago












$begingroup$
How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy.
$endgroup$
– Max0815
6 secs ago




$begingroup$
How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy.
$endgroup$
– Max0815
6 secs ago


















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