Mega A Level Maths Thread MK II

Original post by DJMayes

Well, alternatively:

Spoiler

Also:

I believe that Robbie's idea of prime numbers is actually a better way of tackling the question, because it can be extended to cases where both numbers are odd or both even. for example:

Spoiler

Well, you're actually right. :biggrin:

Reply 3841

Original post by Robbie242

What about the upsidedown A?

For all

Reply 3842

Original post by Robbie242

What about the upsidedown A?

For all

Reply 3843

tigerz

Original post by joostan

Yep

Can I have a similar one later on as a step up and to check if I get it properly lols

Reply 3844

Original post by joostan

For all

Prove that

Unparseable latex formula:

5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}

so its basically saying Prove that

5^{\frac{1}{3^n}}

is in the set of rationals is not rational for all natural numbers n

totally confused, do I have to prove it is not rational for all natural numbers n

Reply 3845

Original post by tigerz

Can I have a similar one later on as a step up and to check if I get it properly lols

Sure:
Why not have a go at Felix's problem:
Prove that:

\log_32

is irrational :smile:

Reply 3846

Original post by Robbie242

Prove that

Unparseable latex formula:

5^{\frac{1}{3^n}} \not\in \matbb{Q}\ \forall n \in \mathbb{N}

so its basically saying Prove that

5^{\frac{1}{3^n}}

is in the set of rationals is not rational for all natural numbers n

totally confused, do I have to prove it is not rational for all natural numbers n

Yes to the bold :smile:

Reply 3847

Proof that π+e is an irrational.

I really don't know if that is easy or hard, just thought about it.

Reply 3848

Original post by joostan

Yes to the bold :smile:

Spoiler

Reply 3849

tigerz

Original post by joostan

Sure:
Why not have a go at Felix's problem:
Prove that:

\log_32

is irrational :smile:

Okays

Ima eat dinner and feed my niece then i'll attempt it, woohoo logs :tongue:

Reply 3850

username937760

Original post by MAyman12

Proof that π+e is an irrational.

I really don't know if that is easy or hard, just thought about it.

I don't think there is even a proof for this - there is a set of about 5 combinations of pi and e, which have not been proven rational or irrational, only that only so many can be rational or irrational.

Reply 3851

Felix Felicis

Original post by MAyman12

Proof that π+e is an irrational.

I really don't know if that is easy or hard, just thought about it.

Are you trolling :curious:

It's not known where

\pi + e

is irrational...

Reply 3852

Original post by DJMayes

Original post by Felix Felicis

Are you trolling :curious:

It's not known where

\pi + e

is irrational...

I just thought about it, didn't even search it. :colondollar:

Reply 3853

Original post by MAyman12

Proof that π+e is an irrational.

I really don't know if that is easy or hard, just thought about it.

Assuming that we know both

\pi

and

e

are rational, and that

\pi

and

e

are transcendental you can easily prove that either

\pi+e

e\pi

or both are irrational :s-smilie:

But a proper solution can't be found :cool:

(edited 10 years ago)

Reply 3854

Original post by Robbie242

Spoiler

Sorry - the tex was probably hard to read.
It's 1/3^n :smile:

Reply 3855

Original post by joostan

Sorry - the tex was probably hard to read.
It's 1/3^n :smile:

aha, any general say 3 points I should aim to do when tackling this question, and maybe I can figure the rest out?

Am I aiming to find a contradiction for n=k+1?

Reply 3856

Original post by Robbie242

aha, any general say 3 points I should aim to do when tackling this question, and maybe I can figure the rest out?

Am I aiming to find a contradiction for n=k+1?

No - it's induction you want. Your basis case is n=1, which you've already proved :smile:

Reply 3857

Original post by joostan

No - it's induction you want. Your basis case is n=1, which you've already proved :smile:

Right but was I on the right path last time? I set it equal to a/b when we are trying to prove it to be irrational? Or is the contradiction itself the inductive proof?

Reply 3858

Original post by Robbie242

Right but was I on the right path last time? I set it equal to a/b when we are trying to prove it to be irrational? Or is the contradiction itself the inductive proof?

You may be able to do so, but a standard proof by induction is all that's required.
Basis case. (already done)
Assume true for n=k
Show that it's consistent with n=k+1 :smile:

Reply 3859