Using these conditions to imply that $(1/π )arctan(1/3)$ is irrational
Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?
I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.
complex-analysis
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
asked Dec 4 '18 at 1:21
Rey
12
add a comment |
Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?
I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.
complex-analysis
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
asked Dec 4 '18 at 1:21
Rey
12
Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
– David
Dec 4 '18 at 1:30
add a comment |
Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?
I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.
complex-analysis
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
asked Dec 4 '18 at 1:21
Rey
12
Suppose $x_n equiv 3 pmod 5$ and $y_n equiv 1 pmod 5$ and that $(3 + i)^n = x_n + iy_n$, with n as an integer greater than or equal to $1$. How does this imply that $frac{1}{pi}arctan(frac{1}{3})$ is irrational?
I have found so far that $x_n+1 = 3x_n - y_n$
and $y_n+1 = x_n + 3y_n$,
where $x$ is the real part and $y$ is the imaginary part.
complex-analysis
complex-analysis
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
asked Dec 4 '18 at 1:21
Rey
12
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
asked Dec 4 '18 at 1:21
Rey
12
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
edited Dec 4 '18 at 10:06
Tianlalu
3,09621038
3,09621038
asked Dec 4 '18 at 1:21
Rey
12
asked Dec 4 '18 at 1:21
Rey
12
asked Dec 4 '18 at 1:21
Rey
12
12
Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
– David
Dec 4 '18 at 1:30
add a comment |
Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
– David
Dec 4 '18 at 1:30
Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
– David
Dec 4 '18 at 1:30
Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
– David
Dec 4 '18 at 1:30
add a comment |
1 Answer
1
active
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You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.
Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
add a comment |
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1 Answer
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You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.
Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
add a comment |
You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.
Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
add a comment |
You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.
Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
You have $x_1equiv 3, y_1equiv 1bmod 5$. Put in your recursive relations, $bmod 5$, and see what residues you get for $x_2, y_2$, then $x_3, y_3$, and so on. Prove by mathematical induction that $x_nequiv 3, y_nequiv 1bmod 5$ for all $n$.
Then $(3+i)^n$ can never be a real number and its argument, $narctan(1/3)$, can never be a multiple of $pi$.
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
answered Dec 4 '18 at 1:43
Oscar Lanzi
12.1k12036
12.1k12036
add a comment |
add a comment |
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Welcome to MSE. Without a bit of background on this problem I doubt that you will get many useful answers. Please see here.
– David
Dec 4 '18 at 1:30