Mega A Level Maths Thread MK II

Original post by DJMayes

What problem is this? Based on that line it sounds very familiar to a STEP question I've seen before.

Check joostan's post above :smile:

He gave another hint about thinking of factorising x, y, z into primes or something so that's the approach I'm trying :s-smilie:

Reply 3801

Zaphod77

1

Original post by Robbie242

My attempt:

Spoiler

and then I'm stuck any advice felix?

I disagree with your last part.

Spoiler

Reply 3802

Original post by Zaphod77

I disagree with your last part.

Spoiler

Ah yeah definitely my bad!

Reply 3803

username937760

16

Original post by joostan

...

Have I missed something in the question? It seems fairly straightforward...

Spoiler

Reply 3804

Original post by joostan

There's a solution above if you're tempted :tongue:

Hint:

Spoiler

I have got here but I'm kinda stuck on what to do next, I noticed 4 is not a prime number so I can't exploit the same method I used for for the cube root, 2 is a square number, perhaps this is important>?
My attempt:

Spoiler

(edited 10 years ago)

Reply 3805

Original post by Robbie242

I have got here but I'm kinda stuck on what to do next, I noticed 4 is not a prime number so I can't exploit the same method I used for for the cube root, 2 is a square number, perhaps this is important>?

My attempt:

Spoiler

Some of your working is not true. After the bold I recommend you take a look :smile:

EDIT: Ignore that - missed the 2 undeneath :tongue:

That doesn't seem to help much though :redface:

(edited 10 years ago)

Reply 3806

Original post by joostan

Some of your working is not true. After the bold I recommend you take a look :smile:

is it 2a/b ?

Reply 3807

Original post by DJMayes

Have I missed something in the question? It seems fairly straightforward...

Spoiler

Almost identical to my solution, though apparently proof by example is not permitted :redface:

Reply 3808

Original post by joostan

Some of your working is not true. After the bold I recommend you take a look :smile:

EDIT: Ignore that - missed the 2 undeneath :tongue:

That doesn't seem to help much though :redface:

I noticed if I equated them withotu squaring I would have got a=2b anyway :redface:

anymore guidance?

Reply 3809

Revisionbug

1

Why does my calculator straightaway rationalise the denominator i messed up a question due to not knowing root3/3 is just 1/root3 but i didnt know that upon closer inspection my calculator just rationalised the denominator do all calculators do this id prefer not to as im bad at realising when its been rationalised or not forgive me for being such a noob i just dnt kno anything bout calculators im self teaching so have no teachers or any help

Reply 3810

Original post by Robbie242

is it 2a/b ?

Check my edit - Sorry :smile:

\frac{2a}{2b} = \frac{a}{b}

- this doesn't really help :redface:

Reply 3811

Shady778

11

Original post by Robbie242

Depends what type of parents honestly :P My mum went insane when I got a few B's and C's at GCSE Lol, along with my A*'s in media but those don't really matter.

If they are likely to get disappointed, I'd just tell them now tbh, to either add shock value to if the results come out good, or to ensure they don't lose their **** on the day.

Hey bro, Do you think there is a chance of me getting an A grade prediction for A2 maths considering those marks?
Thanks

Reply 3812

Original post by Robbie242

I noticed if I equated them withotu squaring I would have got a=2b anyway :redface:

anymore guidance?

Spoiler

Reply 3813

Original post by Revisionbug

Why does my calculator straightaway rationalise the denominator i messed up a question due to not knowing root3/3 is just 1/root3 but i didnt know that upon closer inspection my calculator just rationalised the denominator do all calculators do this id prefer not to as im bad at realising when its been rationalised or not forgive me for being such a noob i just dnt kno anything bout calculators im self teaching so have no teachers or any help

There's nothing wrong with a rationalised denominator.
Personally I prefer it, but really it makes no difference unless it's a show that it's

\dfrac{1}{\sqrt{3}}

In which case remembering that the two forms are equivalent is necessary.

Reply 3814

tigerz

Original post by joostan

I can't make any such promises

How about . . .
Prove that:

\sqrt[3]{5}

is irrational? :smile:

\sqrt[3]{5}=\frac{a}{b}

5=\frac{a^3}{b^3} \rightarrow 5b^3=a^3

b=5w

as it has to be a multiple of 5 if rational

5=\frac{(5w)^3}{b^3} \therefore 5=\frac{125w^3}{b^3}[br]\rightarrow 5b^3=125w^3\rightarrow b^3=25w^3

this means b^3 is a multiple of 5 so it should be able to simplify
but if a and b was a multiple of 5 then it should be able to simplify further earlier on...

uhm I dunno Clue please? i've confused myself :cry2:

Reply 3815

username937760

16

Original post by joostan

Almost identical to my solution, though apparently proof by example is not permitted :redface:

I don't see why not - it's a valid (and the most sensible!) way of attaining a solution and he can't expect a comprehensive list of solutions as there appear to be infinitely many.

Reply 3816

Original post by tigerz

\sqrt[3]{5}=\frac{a}{b}

5=\frac{a^3}{b^3} \rightarrow 5b^3=a^3

b=5w

as it has to be a multiple of 5 if rational

5=\frac{(5w)^3}{b^3} \therefore 5=\frac{125w^3}{b^3}[br]\rightarrow 5b^3=125w^3\rightarrow b^3=25w^3

this means b^3 is a multiple of 5 so it should be able to simplify
but if a and b was a multiple of 5 then it should be able to simplify further earlier on...

uhm I dunno Clue please? i've confused myself :cry2:

In essence this is a solution.

\dfrac{a}{b}

was said to be in it's simplest form, which it isn't if both are a multiple of 5 :smile:

This is a contradiction, so therefore the cubed root of 5 is irrational.

Reply 3817