Mega A Level Maths Thread MK II

Original post by joostan

Also I should've told you that for any irrational x:

\sqrt[3]{x} \not\in \mathbb{Q}

Though it's trivially simple to deduce :smile:

Where did x come from, can I just use what you gave me, I feel awfully confused now

Reply 3861

Original post by Robbie242

Where did x come from, can I just use what you gave me, I feel awfully confused now

Yeah- just where x is an irrational number e.g

\pi

or e etc.

(edited 10 years ago)

Reply 3862

Original post by joostan

...

Spoiler

(edited 10 years ago)

Reply 3863

pureandmodest

Original post by Hunarench95

Ok whenever you have a integration you must differentiate your 'u'

u=x^2+5
du/dx=2x. Therefore du=2x

Take a factor of a half out of your integral. giving 1/2Int(2x(x^2+5)^0.5)dx

You can then swap the 2x*dx with du

Giving 1/2Int(u^0.5)du

Integrating that gives 1/2{(2u^(1.5))/3 + c}

Such in your 'u'

giving ((x^2+5)^1.5)/3 + c

thankyou so much!

Reply 3864

Original post by Felix Felicis

Spoiler

Unparseable latex formula:

5^{1/3} = p/q \Rightarrow p^{3} = 5 q^{3} \Rightarrow p^{3} \0mod5 \Rightarrow a \0mod5

This implies

Unparseable latex formula:

q \0mod5

hence contradiction, so we've proved it for n = 1.

Assume

5^{1/3^{k}}

is irrational.

Consider

5^{1/3^{k+1}} = \left( 5^{1/3^k} \right)^{1/3}

by the inductive hypothesis,

5^{1/3^{k}}

is irrational and hence by lemma

\left( 5^{1/3^k} \right)^{1/3}

is irrational, thereby completing the inductive step, thus the result is true for all n by the principle of mathematical induction.

Spoiler

(edited 10 years ago)

Reply 3865

Original post by joostan

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:

Spoiler

(edited 10 years ago)

Reply 3866

Original post by joostan

Spoiler

I was on a wild goosechase with Mayman's question :s-smilie:

Reply 3867

Original post by Robbie242

Spoiler

It's

5^{1/3^n}

I'm not much good at this problem setting :s-smilie:

Reply 3868

Original post by joostan

It's

5^{1/3^n}

I'm not much good at this problem setting :s-smilie:

fml ok

Reply 3869

Original post by Felix Felicis

I was on a wild goosechase with Mayman's question :s-smilie:

Lol. How much time did you say you wasted? :lol:

Reply 3870

Original post by joostan

Lol. How much time did you say you wasted? :lol:

Like 4 hours :rofl:

Reply 3871

Original post by Felix Felicis

Like 4 hours :rofl:

o.O
I'd never do that :nah:

Reply 3872

MAyman12

Original post by Felix Felicis

I was on a wild goosechase with Mayman's question :s-smilie:

DJ answered it btw :tongue:

Reply 3873

MathsNerd1

Original post by Felix Felicis

Like 4 hours :rofl:

That's commitment!

Posted from TSR Mobile

Reply 3874

Original post by joostan

It's

5^{1/3^n}

I'm not much good at this problem setting :s-smilie:

Spoiler

And then I'm stuck, not sure exactly what contradiction I'm supposed to meet

Reply 3875

Original post by joostan

o.O
I'd never do that :nah:

Original post by MathsNerd1

That's commitment!

Posted from TSR Mobile

I reckon I've a rubix complex :s-smilie:

Either that or I've been watching too much House :ninja:

Original post by MAyman12

DJ answered it btw :tongue:

I saw DJ's solution but it's the exact same as the one joostan/ justinawe proposed, it only uses complex numbers :s-smilie:

Reply 3876

Original post by Robbie242

Spoiler

And then I'm stuck, not sure exactly what contradiction I'm supposed to meet

Spoiler

As the cubed root of any irrational number is irrational . . . :smile:

Reply 3877

Original post by joostan

Spoiler

As the cubed root of any irrational number is irrational . . . :smile:

Spoiler

Reply 3878

Original post by Robbie242

Spoiler

Precisely

Reply 3879