Calculate the inverse of an element in $mathbb{Q}(sqrt 3)$












0












$begingroup$


I want to find the inverse of an element in $mathbb{Q}(sqrt 3)$.



For example $a=2-sqrt3$.



I was considering at first to find a $b= {1 over a}$ and from the calculation emerge that $b= 2+sqrt3$.
But is this process right or wrong?



Thanks






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    What process? You did not specify what you actually did.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:24
















0












$begingroup$


I want to find the inverse of an element in $mathbb{Q}(sqrt 3)$.



For example $a=2-sqrt3$.



I was considering at first to find a $b= {1 over a}$ and from the calculation emerge that $b= 2+sqrt3$.
But is this process right or wrong?



Thanks






abstract-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    What process? You did not specify what you actually did.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:24














0












0








0





$begingroup$


I want to find the inverse of an element in $mathbb{Q}(sqrt 3)$.



For example $a=2-sqrt3$.



I was considering at first to find a $b= {1 over a}$ and from the calculation emerge that $b= 2+sqrt3$.
But is this process right or wrong?



Thanks






abstract-algebra







share|cite|improve this question









$endgroup$




I want to find the inverse of an element in $mathbb{Q}(sqrt 3)$.



For example $a=2-sqrt3$.



I was considering at first to find a $b= {1 over a}$ and from the calculation emerge that $b= 2+sqrt3$.
But is this process right or wrong?



Thanks






abstract-algebra




abstract-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 8:20









AlessarAlessar

313115




313115












  • $begingroup$
    What process? You did not specify what you actually did.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:24


















  • $begingroup$
    What process? You did not specify what you actually did.
    $endgroup$
    – Tobias Kildetoft
    Dec 13 '18 at 8:24
















$begingroup$
What process? You did not specify what you actually did.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:24




$begingroup$
What process? You did not specify what you actually did.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 8:24










1 Answer
1






active

oldest

votes


















1












$begingroup$

Assume $a, bin Bbb Q$ and $a + bsqrt3neq 0$. Expanding fractions gives us
$$
frac{1}{a + bsqrt3} = frac{a - bsqrt3}{a^2-3b^2}\
= frac{a}{a^2-3b^2} - frac{b}{a^2-3b^2}sqrt3
$$
Note that $sqrt3$ being irrational is precisely what we need to see that the denominators aren't $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So in my example $a=2$ and $b=-1$, so having a $sqrt3$ factor in the fraction expansion assure us that the inverse is indeed in $mathbb{Q}(sqrt3)$, am I right?
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    @Alessar The thing that ensures that the inverse is indeed in $Bbb Q(sqrt3)$ is that $frac{a}{a^2-3b^2}$ and $frac{b}{a^2-3b^2}$ are rational. I'm expanding by $a-bsqrt3$ for the simple reason that that makes the denominator rational and much easier to work with.
    $endgroup$
    – Arthur
    Dec 13 '18 at 8:29












  • $begingroup$
    Ok got it, thanks for the clarification Arthur
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:36











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1 Answer
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1 Answer
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active

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1












$begingroup$

Assume $a, bin Bbb Q$ and $a + bsqrt3neq 0$. Expanding fractions gives us
$$
frac{1}{a + bsqrt3} = frac{a - bsqrt3}{a^2-3b^2}\
= frac{a}{a^2-3b^2} - frac{b}{a^2-3b^2}sqrt3
$$
Note that $sqrt3$ being irrational is precisely what we need to see that the denominators aren't $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So in my example $a=2$ and $b=-1$, so having a $sqrt3$ factor in the fraction expansion assure us that the inverse is indeed in $mathbb{Q}(sqrt3)$, am I right?
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    @Alessar The thing that ensures that the inverse is indeed in $Bbb Q(sqrt3)$ is that $frac{a}{a^2-3b^2}$ and $frac{b}{a^2-3b^2}$ are rational. I'm expanding by $a-bsqrt3$ for the simple reason that that makes the denominator rational and much easier to work with.
    $endgroup$
    – Arthur
    Dec 13 '18 at 8:29












  • $begingroup$
    Ok got it, thanks for the clarification Arthur
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:36
















1












$begingroup$

Assume $a, bin Bbb Q$ and $a + bsqrt3neq 0$. Expanding fractions gives us
$$
frac{1}{a + bsqrt3} = frac{a - bsqrt3}{a^2-3b^2}\
= frac{a}{a^2-3b^2} - frac{b}{a^2-3b^2}sqrt3
$$
Note that $sqrt3$ being irrational is precisely what we need to see that the denominators aren't $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So in my example $a=2$ and $b=-1$, so having a $sqrt3$ factor in the fraction expansion assure us that the inverse is indeed in $mathbb{Q}(sqrt3)$, am I right?
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    @Alessar The thing that ensures that the inverse is indeed in $Bbb Q(sqrt3)$ is that $frac{a}{a^2-3b^2}$ and $frac{b}{a^2-3b^2}$ are rational. I'm expanding by $a-bsqrt3$ for the simple reason that that makes the denominator rational and much easier to work with.
    $endgroup$
    – Arthur
    Dec 13 '18 at 8:29












  • $begingroup$
    Ok got it, thanks for the clarification Arthur
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:36














1












1








1





$begingroup$

Assume $a, bin Bbb Q$ and $a + bsqrt3neq 0$. Expanding fractions gives us
$$
frac{1}{a + bsqrt3} = frac{a - bsqrt3}{a^2-3b^2}\
= frac{a}{a^2-3b^2} - frac{b}{a^2-3b^2}sqrt3
$$
Note that $sqrt3$ being irrational is precisely what we need to see that the denominators aren't $0$.






share|cite|improve this answer









$endgroup$



Assume $a, bin Bbb Q$ and $a + bsqrt3neq 0$. Expanding fractions gives us
$$
frac{1}{a + bsqrt3} = frac{a - bsqrt3}{a^2-3b^2}\
= frac{a}{a^2-3b^2} - frac{b}{a^2-3b^2}sqrt3
$$
Note that $sqrt3$ being irrational is precisely what we need to see that the denominators aren't $0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 8:24









ArthurArthur

116k7116199




116k7116199












  • $begingroup$
    So in my example $a=2$ and $b=-1$, so having a $sqrt3$ factor in the fraction expansion assure us that the inverse is indeed in $mathbb{Q}(sqrt3)$, am I right?
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    @Alessar The thing that ensures that the inverse is indeed in $Bbb Q(sqrt3)$ is that $frac{a}{a^2-3b^2}$ and $frac{b}{a^2-3b^2}$ are rational. I'm expanding by $a-bsqrt3$ for the simple reason that that makes the denominator rational and much easier to work with.
    $endgroup$
    – Arthur
    Dec 13 '18 at 8:29












  • $begingroup$
    Ok got it, thanks for the clarification Arthur
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:36


















  • $begingroup$
    So in my example $a=2$ and $b=-1$, so having a $sqrt3$ factor in the fraction expansion assure us that the inverse is indeed in $mathbb{Q}(sqrt3)$, am I right?
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:27






  • 1




    $begingroup$
    @Alessar The thing that ensures that the inverse is indeed in $Bbb Q(sqrt3)$ is that $frac{a}{a^2-3b^2}$ and $frac{b}{a^2-3b^2}$ are rational. I'm expanding by $a-bsqrt3$ for the simple reason that that makes the denominator rational and much easier to work with.
    $endgroup$
    – Arthur
    Dec 13 '18 at 8:29












  • $begingroup$
    Ok got it, thanks for the clarification Arthur
    $endgroup$
    – Alessar
    Dec 13 '18 at 8:36
















$begingroup$
So in my example $a=2$ and $b=-1$, so having a $sqrt3$ factor in the fraction expansion assure us that the inverse is indeed in $mathbb{Q}(sqrt3)$, am I right?
$endgroup$
– Alessar
Dec 13 '18 at 8:27




$begingroup$
So in my example $a=2$ and $b=-1$, so having a $sqrt3$ factor in the fraction expansion assure us that the inverse is indeed in $mathbb{Q}(sqrt3)$, am I right?
$endgroup$
– Alessar
Dec 13 '18 at 8:27




1




1




$begingroup$
@Alessar The thing that ensures that the inverse is indeed in $Bbb Q(sqrt3)$ is that $frac{a}{a^2-3b^2}$ and $frac{b}{a^2-3b^2}$ are rational. I'm expanding by $a-bsqrt3$ for the simple reason that that makes the denominator rational and much easier to work with.
$endgroup$
– Arthur
Dec 13 '18 at 8:29






$begingroup$
@Alessar The thing that ensures that the inverse is indeed in $Bbb Q(sqrt3)$ is that $frac{a}{a^2-3b^2}$ and $frac{b}{a^2-3b^2}$ are rational. I'm expanding by $a-bsqrt3$ for the simple reason that that makes the denominator rational and much easier to work with.
$endgroup$
– Arthur
Dec 13 '18 at 8:29














$begingroup$
Ok got it, thanks for the clarification Arthur
$endgroup$
– Alessar
Dec 13 '18 at 8:36




$begingroup$
Ok got it, thanks for the clarification Arthur
$endgroup$
– Alessar
Dec 13 '18 at 8:36


















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