How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?
$begingroup$
Let $x, r in mathbb{Q}$.
I need to find the conditions on $ x, r$ so that the value of $ large (1+x)^r$ is a rational number.
Which $x, r$ makes $(1+x)^r$ a rational number?
Answer:
If I take $x=frac{16}{9}$ and $r=frac{1}{2}$, then $ (1+x)^r=(1+frac{16}{9})^{frac{1}{2}}=sqrt{frac{25}{9}}=pm frac{5}{3} in mathbb{Q}$,
If I take $x=frac{19}{8}$ and $r=frac{1}{3}$, then $ (1+x)^r=(1+frac{19}{8})^{frac{1}{3}}=large sqrt[3]{frac{27}{8}}= frac{3}{2} in mathbb{Q}$,
and so on $ cdots $
How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?
Can you give me the general form of $x$ and $r$ so that $ (1+x)^r$ becomes a rational number?
number-theory elementary-number-theory binomial-theorem
asked Jan 2 at 11:28
arifamatharifamath
1176
$endgroup$
add a comment |
$begingroup$
Let $x, r in mathbb{Q}$.
I need to find the conditions on $ x, r$ so that the value of $ large (1+x)^r$ is a rational number.
Which $x, r$ makes $(1+x)^r$ a rational number?
Answer:
If I take $x=frac{16}{9}$ and $r=frac{1}{2}$, then $ (1+x)^r=(1+frac{16}{9})^{frac{1}{2}}=sqrt{frac{25}{9}}=pm frac{5}{3} in mathbb{Q}$,
If I take $x=frac{19}{8}$ and $r=frac{1}{3}$, then $ (1+x)^r=(1+frac{19}{8})^{frac{1}{3}}=large sqrt[3]{frac{27}{8}}= frac{3}{2} in mathbb{Q}$,
and so on $ cdots $
How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?
Can you give me the general form of $x$ and $r$ so that $ (1+x)^r$ becomes a rational number?
number-theory elementary-number-theory binomial-theorem
asked Jan 2 at 11:28
arifamatharifamath
1176
$endgroup$
4
$begingroup$
$x=(a/b)^n-1,r=m/n$ with $a,b,m,n$ integers
$endgroup$
– Wojowu
Jan 2 at 11:32
$begingroup$
@Wojowu, Does it cover all possible choices? How do you match this formula for $x=frac{16}{9}, r=frac{1}{2}$ ?
$endgroup$
– arifamath
Jan 2 at 13:02
1
$begingroup$
$x=(5/3)^2-1,r=1/2$.
$endgroup$
– Wojowu
Jan 2 at 13:12
add a comment |
$begingroup$
Let $x, r in mathbb{Q}$.
I need to find the conditions on $ x, r$ so that the value of $ large (1+x)^r$ is a rational number.
Which $x, r$ makes $(1+x)^r$ a rational number?
Answer:
If I take $x=frac{16}{9}$ and $r=frac{1}{2}$, then $ (1+x)^r=(1+frac{16}{9})^{frac{1}{2}}=sqrt{frac{25}{9}}=pm frac{5}{3} in mathbb{Q}$,
If I take $x=frac{19}{8}$ and $r=frac{1}{3}$, then $ (1+x)^r=(1+frac{19}{8})^{frac{1}{3}}=large sqrt[3]{frac{27}{8}}= frac{3}{2} in mathbb{Q}$,
and so on $ cdots $
How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?
Can you give me the general form of $x$ and $r$ so that $ (1+x)^r$ becomes a rational number?
number-theory elementary-number-theory binomial-theorem
asked Jan 2 at 11:28
arifamatharifamath
1176
$endgroup$
Let $x, r in mathbb{Q}$.
I need to find the conditions on $ x, r$ so that the value of $ large (1+x)^r$ is a rational number.
Which $x, r$ makes $(1+x)^r$ a rational number?
Answer:
If I take $x=frac{16}{9}$ and $r=frac{1}{2}$, then $ (1+x)^r=(1+frac{16}{9})^{frac{1}{2}}=sqrt{frac{25}{9}}=pm frac{5}{3} in mathbb{Q}$,
If I take $x=frac{19}{8}$ and $r=frac{1}{3}$, then $ (1+x)^r=(1+frac{19}{8})^{frac{1}{3}}=large sqrt[3]{frac{27}{8}}= frac{3}{2} in mathbb{Q}$,
and so on $ cdots $
How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?
Can you give me the general form of $x$ and $r$ so that $ (1+x)^r$ becomes a rational number?
number-theory elementary-number-theory binomial-theorem
number-theory elementary-number-theory binomial-theorem
asked Jan 2 at 11:28
arifamatharifamath
1176
asked Jan 2 at 11:28
arifamatharifamath
1176
asked Jan 2 at 11:28
arifamatharifamath
1176
asked Jan 2 at 11:28
arifamatharifamath
1176
asked Jan 2 at 11:28
arifamatharifamath
1176
1176
4
$begingroup$
$x=(a/b)^n-1,r=m/n$ with $a,b,m,n$ integers
$endgroup$
– Wojowu
Jan 2 at 11:32
$begingroup$
@Wojowu, Does it cover all possible choices? How do you match this formula for $x=frac{16}{9}, r=frac{1}{2}$ ?
$endgroup$
– arifamath
Jan 2 at 13:02
1
$begingroup$
$x=(5/3)^2-1,r=1/2$.
$endgroup$
– Wojowu
Jan 2 at 13:12
add a comment |
4
$begingroup$
$x=(a/b)^n-1,r=m/n$ with $a,b,m,n$ integers
$endgroup$
– Wojowu
Jan 2 at 11:32
$begingroup$
@Wojowu, Does it cover all possible choices? How do you match this formula for $x=frac{16}{9}, r=frac{1}{2}$ ?
$endgroup$
– arifamath
Jan 2 at 13:02
1
$begingroup$
$x=(5/3)^2-1,r=1/2$.
$endgroup$
– Wojowu
Jan 2 at 13:12
4
4
$begingroup$
$x=(a/b)^n-1,r=m/n$ with $a,b,m,n$ integers
$endgroup$
– Wojowu
Jan 2 at 11:32
$begingroup$
$x=(a/b)^n-1,r=m/n$ with $a,b,m,n$ integers
$endgroup$
– Wojowu
Jan 2 at 11:32
$begingroup$
@Wojowu, Does it cover all possible choices? How do you match this formula for $x=frac{16}{9}, r=frac{1}{2}$ ?
$endgroup$
– arifamath
Jan 2 at 13:02
$begingroup$
@Wojowu, Does it cover all possible choices? How do you match this formula for $x=frac{16}{9}, r=frac{1}{2}$ ?
$endgroup$
– arifamath
Jan 2 at 13:02
1
1
$begingroup$
$x=(5/3)^2-1,r=1/2$.
$endgroup$
– Wojowu
Jan 2 at 13:12
$begingroup$
$x=(5/3)^2-1,r=1/2$.
$endgroup$
– Wojowu
Jan 2 at 13:12
add a comment |
0
active
oldest
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4
$begingroup$
$x=(a/b)^n-1,r=m/n$ with $a,b,m,n$ integers
$endgroup$
– Wojowu
Jan 2 at 11:32
$begingroup$
@Wojowu, Does it cover all possible choices? How do you match this formula for $x=frac{16}{9}, r=frac{1}{2}$ ?
$endgroup$
– arifamath
Jan 2 at 13:02
1
$begingroup$
$x=(5/3)^2-1,r=1/2$.
$endgroup$
– Wojowu
Jan 2 at 13:12