<p>Then lets prove that if P is irrational then P/2 is also irrational.
Suppose P/2 is rational, i.e. P/2 = (a/b), then P = 2a/b – a contradiction.</p>
<p>Let’s prove the existence of Jupiter now. Gentlemen, shall we?</p>
<p>@kenhungkk: Oh, so we count 0 as an irrational number. I don’t remember much about the definition; it’s been a long time 
@MrPanda: You’ve just written it down.</p>
<p>@12npm12: 0 is not an irrational number.</p>
<p>
</p>
<p>@MrPropaganda: Yes. Not when multiplied by 1/2 though. If it did, then you could multiply the new ‘rational’ by 2 to get back an irrational…</p>
<p>@MSGM16: Sorry, I meant rational number :D</p>
<p>
Good idea
This is trivial, I was actually interested the infinite set of irrational numbers a,b,c,…t
where “a” is not the generator of the others. I mean the entire set is not a multiple of one number (generator).
Not like 2sqrt(2), 4sqrt(2)…
But
Forexample sqrt(2), sqrt(10) sqrt(14) …</p>
<p>Assume that sqrt 2n is rational. where n=odd
sqrt2n can be expressed by p/q as a simplest form where p and q belong to Z.
sqrt2n = p / q
2n = p2 / q2
p^2 = 2nq^2
p^2 is an even number.
p is an even number.
p can be expressed by 2s where s belong to Z.
nq^2 = p^2 / 2
nq^2 = 4s^2 / 2
nq^2 = 2s^2 nq^2 is even but remember n=odd<br>
Therefore q^2 is an even number.
q is an even number.
both p and q are even numbers therefore not in simplest form!
This contradicts what we have assumed at the beginning of this proof and so the assumption sqrt2 is a rational number is false.
sqrt2 is an irrational number.</p>
<p>if n=1 we have sqrt(2), if n=3 we have sqrt(6) …
I am not sure if this is true though, you guys can check.</p>
<p>
</p>
<p>Inspired by the above quote
Similar stuff
proof that there exist two irrational numbers a and b such that b^a= rational
Those numbers does exist just find a counter example
hint: again you can use sqrt(2) and some “b” as a counterexample!</p>
<p>You can take:
b = sqrt(2)
a = log base 2 (9)</p>
<p>Well, all roots of primes are irrational, using ‘if prime p divides n^2 then p divides n’ argument.</p>
<p>^ what are you trying to prove in here?</p>
<p>@ MSGM16
are you saying (sqrt(2))^(ln(9)/ln(2))= rational
I think you mean e^ln(x)=x ?</p>
<p>First of all “a” is a root of a polynomial so is “b”.</p>
<p>@Inconclusive: Yes, I’m saying (sqrt(2))^(ln(9)/ln(2))= 3.</p>
<p>
IF
“a” and “b” are roots of a polynomial.</p>
<p>let b=a=sqrt(2)</p>
<p>(sqrt(2))^(sqrt(2))= rational then we have a counterexample if (sqrt(2))^(sqrt(2)) is not a rational. Then it is irrational</p>
<p>Therefore {(sqrt(2))^(sqrt(2))}^sqrt(2)=sqrt(2))^{(sqrt(2)*sqrt(2)}=sqrt(2)^2=2 then again we have a counter example</p>