Bina-ring
Investigate powers of numbers of the form (1 + sqrt 2).
Age
16 to 18
Challenge level
Consider $(1 + \sqrt2)^n$. This will be of the form $A + B\sqrt2$ where $A$ and $B$ are integers. Decide which entries in the table below are possibe and which are not.
| A even | A odd | |
| B even | ||
| B odd |
What happens for $(a + \sqrt p)^n$ for other values of $p$?
There is more than one way to prove this result.
On way is to define $(1 + \sqrt2)^n = A_n + B_n\sqrt2$ and then find and use $A_{n+1} + B_{n+1}\sqrt 2$.
Thank you Andrei from Tudor Vianu National College, Bucharest, Romania for this solution.
First I looked at the values of $A$ and $B$ for the first values of $n$ in the expansion of $(1 + \sqrt 2)^n = A + B\sqrt 2$. The results I obtained are written in the following table:
| $n$ | $A$ | $B$ |
| 1 | 1 | 1 |
| 2 | 3 | 2 |
| 3 | 7 | 5 |
| 4 | 17 | 12 |
| 5 | 41 | 29 |
In the cases studied, I observe that $A$ is odd for all $n$, and that $B$ is odd for $n$ odd and even for $n$ even. Now, I shall generalise this result by induction for any $n$:
First I assume that $A(n)$ and $B(n)$ are odd and I prove that $A(n+1)$ is odd and $B(n+1)$ is even. This corresponds to $n$ odd. I shall consider $A(n) = 2p+1$ and $B(n) = 2q+1$.
So, $(1 + \sqrt 2)^n = (2p+1) + (2q+1)\sqrt 2$. $$(1 + \sqrt 2)^{n+1} = (1 + \sqrt 2)^n (1+\sqrt 2) = ((2p+1) + (2q+1)\sqrt 2)(1+ \sqrt 2) = (2p+1 + 2(2q+1)) + \sqrt 2 (2q+1+2p+1).$$ So, $A(n+1) = 2(p+2q+1) + 1$ and $B(n+1) = 2(p+q+1).$ These values that I have obtained for A(n+1) and B(n+1) confirm my prediction.
In a similar manner I assume $A(n)$ odd and $B(n)$ even:
$A(n) = 2p+1$, $B(n) = 2q$. I obtain:
$A(n+1) = 2(p+2q) + 1$ and $B(n+1) = 2(p+q) + 1$.
Now, my proof by induction is complete.
The results are summarized in the following table:
| $A$ even | $A$ odd | |
| $B$ even | - | $n$ even |
| $B$ odd | - | $n$ odd |
Working in a similar manner for $(a + \sqrt p)^n$, I summarize my results in the following table:
| $n$ | $A$ | $B$ | |
| $a$ odd, $p$ odd | all $n$ |
even
exception $n=1, A=1$
|
even
exception $n=1,
B=1$
|
| $a$ even, $p$ even | all $n$ | even |
even
exception $n=1,
B=1$
|
| $a$ odd, $p$ even |
odd
even
|
odd
odd
|
odd
even
|
| $a$ even, $p$ odd |
odd
even
|
even
odd
|
odd
even
|
The set of numbers of the form $a + b\sqrt 2$ where $a$ and $b$ are integers form a mathematical structure called a ring . It is easy to show that $R$ is closed for addition, that 0 belongs to $R$ and is the additive identity and that every number in $R$ has an additive inverse which is in $R$. Also addition of numbers in $R$ is associative so this is an additive group .
What about multiplication?
Again it is easy show that $R$ is closed for multiplication, that 1 belongs to $R$ and is the multiplicative identity and that multiplication of numbers in $R$ is associative. However it is also easy to find a counter example to show that not every number in $R$ has a multiplicative inverse which is in $R$
(Try this for youself, for example look for an inverse for $(2 + 3\sqrt 2)$ and you will find that it would have to be $({-1\over 7} + {3\sqrt 2\over 14})$ but this number is not in $R$. NB. $R$ is the set of numbers $a + b\sqrt 2$ where $a$ and $b$ are integers ).
So we can add, subtract and multiply these numbers. If $u, v$ and $w$ are in the set $R$ it is easy to show that the distributive property holds: $u(v + w) = uv + uw$. So we have some of the same structure as the arithmetic of real numbers but without division. This structure is called a ring .