Here's an interesting one:
Prove that $\displaystyle \sqrt{2}$ is irrational by induction
I don't have the time to do this in detail now, but further thought andOriginally Posted by CaptainBlack
sketches of a construction make me fairly sure it can be done using the
continued fraction expansion of $\displaystyle \sqrt2$.
The key idea is that a rational number has a terminating simple cf expansion,
but we can show by induction that the simple cf expansion of $\displaystyle \sqrt 2$ does not
terminate.
RonL
Method #2:Originally Posted by Treadstone 71
Let $\displaystyle P_n$ be the proposition that:
For all $\displaystyle 1\le m \le n$ there is no $\displaystyle k\epsilon \mathbb{Z}_+$, such that $\displaystyle \sqrt2 = \frac{k}{m}$.
This is clearly true for $\displaystyle n=1$ as $\displaystyle \sqrt2$ is not an integer.
The induction step consists of showing that there is no $\displaystyle k\epsilon \mathbb{Z}_+$,
such that $\displaystyle \sqrt2 = \frac{k}{n+1}$, which together with $\displaystyle P_n$ will prove $\displaystyle P_{n+1}$.
This last step I can do, but will not reproduce here.
RonL
Method 3Originally Posted by Treadstone 71
I'm not going to do much on this other than suggest it might be made to work.
Let $\displaystyle P_n$ be something like:
The decimal expansion of $\displaystyle \sqrt2$ does not have a peiodic
tail of any period $\displaystyle \le n$.
RonL
Method 2 is what I have written down originally. It goes slightly differently than what you wrote.
$\displaystyle \sqrt{2}\neq 1$
Suppose $\displaystyle \sqrt{2}\neq \frac{n}{m}$ for some general $\displaystyle n,m\in \mathbb{Z}$ Details are omitted here, but it can be shown that $\displaystyle \sqrt{2}\neq\frac{n+1}{m}$ and $\displaystyle \sqrt{2}\neq\frac{n}{m+1}$ By induction, it does not equal to any rational number.
This is the very fact that you have to prove, because it is equivalent to the definition of an irrational number. You will actually have to compute the decimals of root 2, which comes down to something similar to what you did with the continued fractions.Originally Posted by CaptainBlack
But aren't they all.Originally Posted by Treadstone 71
I was thinking of doing something slightly different. I had no intention ofYou will actually have to compute the decimals of root 2, which comes down to something similar to what you did with the continued fractions.
computing the decimal expansion of $\displaystyle \sqrt 2$, but supposing
that the tail were periodic of period $\displaystyle n$, then writing the sum of
the tail in terms of the replicand (if that is the right word) which we don't
know, and the sum of the geometric series:
$\displaystyle \sum_1^{\infty} \frac{1}{10^{n+1}}$,
and show in some way not yet determined that this lead to a contradiction.
RonL
I dont understand your notation mister Hacker please could you explain?Originally Posted by ThePerfectHacker
In fact i have been teached a very basic and simple method to proove that any integer that is not a square have an irational square root. I am in doubt that any steps of the tree induction method, not developed and using weird notation for some of them (this is the math puzzle section not the advanced question section), so any steps would not use the same or a more complicated argument that the basic method
People that dont know it are invited to search in <<square root>>thread !!!