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I'm having difficulty with a mathematical problem.

I've got the following;

The basis is:

 -1 ∈ V

And the induction is;

x ∈ V → x/(1-x) ∈ V

Now, I have made 4 statements and I want to get to know if they're either true, or false.

The statements are;

1. All elements are negative. (under zero)
2. All elements are between -2 and 0
3. -1/7 ∈ V 
4. -2/3 ∈ V

There's also an exception; There are no elements of V that cannot be acquired by applying statement 1 and 2.

Now, I'm not really sure how to start here. How can I effectively sort out elements in a induction definition?

MichaelP
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    The 1st statement says $-1$ is in $V$. Using that, the 2nd statement tells you another element of $V$ --- what is it? Then, using that new element, the 2nd statement gives you yet another element. And so on. When you have computed a few, step back, look at them, and look at the 4 statements again. – Gerry Myerson Sep 30 '13 at 10:00

2 Answers2

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HINT: It never hurts to gather some numerical data. Since $-1\in V$, you know that

$$\frac{-1}{1-(-1)}=-\frac12\in V\;.$$

Now use the fact that $-\frac12\in V$ to see that

$$\frac{-1/2}{1-\left(-\frac12\right)}=-\frac{1/2}{3/2}=-\frac13\in V\;.$$

Again:

$$\frac{-1/3}{1-\left(-\frac13\right)}=-\frac{1/3}{4/3}=-\frac14\in V\;.$$

From the pattern of these examples you can probably make a good conjecture about exactly what numbers are in $V$ and then prove it by induction. As a little extra hint, three of the four statements are true.

Brian M. Scott
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You can proceed as follows :

  1. Start by computing the first few terms : $x_0 = -1$, then $$ x_1 = -1/(1+1) = -1/2 $$ $$ x_2 = -1/2/(1+1/2) = -1/3 $$ $$ x_3 = -1/3/(1+1/3) = -1/4 $$ This gives you an idea as to what $x_n$ should be. ie. $x_n = -1/(n+1)$

  2. Prove your guess by induction :

a) $x_0 = -1 = -1/(0+1)$. Hence your guess is true when $n=0$

b) Assume $x_{n-1} = -1/n$, then $$ x_n = -1/n/(1+1/n) = -1/(n+1) $$ Hence, your guess is true for all $n\in \mathbb{N}$.

By your exception, it follows that $$ V = \{-1, -1/2, -1/3, \ldots \} $$

Prahlad Vaidyanathan
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