1

Prove, by induction, that all positive integers $n$, $(1-{\sqrt 5})^n$ can be written in the form $a-b{\sqrt 5}$ where $a$ and $b$ are positive integers.

I understand these idea of proof by induction and this was a different type of question that I'm used too and wasn't sure on how to approach it as I'm not entirely confident with proving things with induction yet.

Any help would be appreciated.

Lorenzo B.
  • 2,252
H.Linkhorn
  • 1,283

4 Answers4

2

1-Base Step: $$For\ n=1,\ {{\left( 1-\sqrt{5} \right)}^{1}}=a-b\sqrt{5},\ with\ a=1\,and\ b=1$$ 2-Inductive step: Assume that ${{\left( 1-\sqrt{5} \right)}^{n}}=a-b\sqrt{5},\ Consider\ {{\left( 1-\sqrt{5} \right)}^{n+1}}$

$\begin{align} & {{\left( 1-\sqrt{5} \right)}^{n+1}}=\left( 1-\sqrt{5} \right){{\left( 1-\sqrt{5} \right)}^{n}} \\ & \quad \quad \quad \quad \ \ =\left( 1-\sqrt{5} \right)\left( a-\sqrt{5}b \right) \\ & \quad \quad \quad \quad \ \ =a-a\sqrt{5}+5b-b\sqrt{5} \\ & \quad \quad \quad \quad \ \ =\left( a+5b \right)-\left( a+b \right)\sqrt{5} \\ \end{align}$

So the inductive case holds. Now by induction we see that the assumption is true.

0

For a direct proof let use binomial theorem for $(1-{\sqrt 5})^n$ that is

$$(x+y)^n = {n \choose 0}x^n y^0 + {n \choose 1}x^{n-1}y^1 + {n \choose 2}x^{n-2}y^2 + \cdots + {n \choose n-1}x^1 y^{n-1} + {n \choose n}x^0 y^n$$

By induction let consider

  • base case: for $n=1\implies (1-{\sqrt 5})^1=1-\sqrt 5$
  • induction step: assume that $(1-{\sqrt 5})^n=a-b{\sqrt 5}$ is true then

$$(1-{\sqrt 5})^{n+1}=(1-{\sqrt 5})^n(1-{\sqrt 5})\stackrel{I.H.}=(a-b{\sqrt 5})(1-{\sqrt 5})=\\=a-a\sqrt 5-b\sqrt 5+5b=a+5b-(a+b)\sqrt 5$$

that is

$$(1-{\sqrt 5})^{n+1}=a+5b-(a+b)\sqrt 5$$

user
  • 154,566
  • i understand the binomial theorem, but sorry but my knowledge of proof by induction is that good yet so could you give the next step which i might be able to go from. – H.Linkhorn Apr 05 '18 at 18:29
  • You don’t need induction, simply note that x=1, binomial coefficients are integers and $(\sqrt 5)^k$ is an integer for k even and is in the form $b_k(\sqrt 5)^k$ for k odd thus we can collect all terms in the given form. – user Apr 05 '18 at 18:33
  • the question says i have to prove it through induction – H.Linkhorn Apr 05 '18 at 18:37
  • If you want to prove by induction simply consider the base case for n=1 which is trivial then for induction step assume that $(1-{\sqrt 5})^n=a-b{\sqrt 5}$ and then consider $(1-{\sqrt 5})^{n+1}=(1-{\sqrt 5})^n(1-{\sqrt 5})=(a-b{\sqrt 5})(1-{\sqrt 5})$ – user Apr 05 '18 at 18:38
0

Hint: Note that $x^{n+1} = x^n\cdot x$.

Hagen von Eitzen
  • 374,180
0

Hint: $\;(1-\sqrt{5})(a-b\sqrt{5})=a+5b-(a+b)\sqrt{5}\,$.

dxiv
  • 76,497