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I just received my first assignment for a mathematical proofs course I am taking this year. We just began the course, and we have so far only covered examples of proofs (how to prove if-then statements in different ways) and the mathematical principle of induction.

Here is the question I am having difficulty with:

"Define a sequence $a_n, n \ge 0,$ inductively by $a_0 = 2,$ and for all $n \ge 0, a_{n+1} = \sqrt{a_n + 1}.$

a) Prove that for every $n \ge 0, a_n > \frac{1+\sqrt{5}}{2}.$

b) Prove that for every $n \ge 0, a_n > a_{n+1}.$ (You may use the fact that the polynomial $x^2 - x - 1 < 0$ if and only if $\frac{1-\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2}.$

What would be the best proof technique for these questions? Should I prove both using induction, or is there a simpler/better way?

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Try induction, see what happens. That's how you learn. –  Gerry Myerson Sep 16 '12 at 6:18

1 Answer 1

up vote 2 down vote accepted

For (a), the induction proof almost writes itself. The base step is easy.

For the induction step, we want to show that if for a certain $k$ we have $a_k\gt \dfrac{1+\sqrt{5}}{2}$, then $a_{k+1}\gt \dfrac{1+\sqrt{5}}{2}$.

To show this, observe that $$a_{k+1}^2=a_k+1\gt \dfrac{1+\sqrt{5}}{2}+1=\dfrac{3+\sqrt{5}}{2}=\left(\dfrac{1+\sqrt{5}}{2}\right)^2.$$

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