This is actually from an Analysis text but i feel its a set theory question.

Proposition for ever rational number $\epsilon > 0$ there exists a non-negative number x s.t $x^2 < 2 < (x+ \epsilon )^2 $

It provides a proof that im having trouble understanding.

Proof: let $ \epsilon >0$ be rational. Suppose for contradiction sake that there is no on-negative rational number x that $x^2 < 2 < (x+ \epsilon )^2 $ holds.

ie when every $ x^2 < 2$ the statement $(x+ \epsilon )^2 <2 $

It states by a previous proposition that $(x+ \epsilon )^2 $ cannot equal 2.

Then it states "Since $0^2 < 2$ we thus have $ \epsilon ^2 < 2$ which then implies that $ (2\epsilon )^2 < 2$ and indeed a simple induction shows that $ (n\epsilon )^2 < 2$ for every natural number n." Which is what i cant understand.

The rest of the proof is strange as well im fine with the statement $ \epsilon ^2 < 2$ as it clearly follows that $ \epsilon ^2 < (x+ \epsilon )^2 $ as x is positive and $ \epsilon ^2$ is on both sides of the expression.

If i was proving it then i would rewrite

$ \epsilon ^2 = n $ $ \epsilon' $ s.t $n \in \mathbb {N} $ and $ \epsilon' \in \mathbb {Q} $

i would then use the Archimedean property to prove this is a contradiction.

If anyone can follow/explain what the bold text means i would greatly appreciate it.

  • 4
    $\begingroup$ "i feel its a set theory question." It isn't. $\endgroup$ – Andrés E. Caicedo Aug 25 '16 at 22:20
  • $\begingroup$ if you like, i proved it for you in a different way so that you can see a more transparent reasoning behind that statement. $\endgroup$ – Francesco Alem. Aug 25 '16 at 23:08

Before the bolded passage, you've concluded that if the statement you're trying to prove fails, then it must be the case that $x^2<2$ implies $(x+\epsilon)^2<2$.

Now, just take $x=0$. $0^2=0<2$, so we must have $(0+\epsilon)^2=\epsilon^2<2$.

So $\epsilon^2<2$. Now take $x=\epsilon$. Since $\epsilon^2<2$, we have $(\epsilon+\epsilon)^2=(2\epsilon)^2<2$.

Now take $x=2\epsilon$ . . .

More generally, having shown that $(n\epsilon)^2<2$, we may now take $x=n\epsilon$ and conclude that $(x+\epsilon)^2=((n+1)\epsilon)^2<2$. So, by induction, $(n\epsilon)^2<2$ for every natural number $n$. Since $\epsilon>0$, this contradicts the Archimedean property.

From your question, I think you're a little confused about what exactly is being proved by induction here. The goal is to prove $$(*)\quad \mbox{For each natural number $n$, $(n\epsilon)^2<2$.}$$ Note that this is not an induction on $\epsilon$ - it's an induction on the coefficient of $\epsilon$. The base case is $n=1$ (that is, showing $\epsilon^2<2$), and the inductive step is as described above.

  • $\begingroup$ Your a genius thanks. $\endgroup$ – Faust Aug 25 '16 at 22:16

The statement is true. here's a constructive proof for ya.

let $\epsilon\in \mathbf{Q}\,|\, \epsilon >0$

choose $k \in \mathbf{N}$ such that $k>2\epsilon$ so that $\frac{\epsilon}{k}<\frac{1}{2}$

choose $x \in [\sqrt 2 -\frac{\epsilon}{k};\,\,\sqrt2 - \frac{\epsilon}{2k}] \cap\mathbf{Q}$ arbitrarily (such an $x$ exists since rationals are dense in the reals, a consequence of the archimedean property).

$x$ is chosen in such a way that certainly makes it positive and lower than $\sqrt 2$ since $x<\sqrt 2 -\frac{\epsilon}{2k}<\sqrt 2$, hence it must hold $x^2<2$

also $x+\epsilon$ is certainly above $\sqrt 2$ since $x+\epsilon>\sqrt 2 -\frac{\epsilon}{k}+\epsilon>\sqrt 2$ hence it must hold $(x+\epsilon)^2>2$

it follows that $x^2<2<(x+\epsilon)^2$


We assumed that if $x^2<2$ then $(x+ϵ)^2<2$ (*).

Lets start with $x=0$: $x^2 = 0^2 = 0 < 2$, so we conclude that $(x+ϵ)^2 = ϵ^2<2$.

Now we have learned that $ϵ^2<2$, so we can use (*) again, with $x=ϵ$. We conclude that $(x+ϵ)^2=(ϵ+ϵ)^2=(2ϵ)^2<2$

Now we have learned that $(2ϵ)^2<2$, so we can use (*) again ... and again .. and again.


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