Mega A Level Maths Thread MK II

Jheeze Robbie's on this irrationality maths shiz!

time for me to attempt Felix's prove

\log_{3}2

is irrational :/

Reply 3882

username937760

16

Original post by Felix Felicis

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

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

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

The question relies on one specific result for generating sets, although there are probably other ways you can obtain the result to the ones we've been through. However, you cannot really go further with it due to the infinite number of potential solutions. By the way, I'm not entirely convinced your use of modular arithmetic is correct:

Spoiler

(Also, you think you've been watching too much? Started watching it a month or two ago, now halfway through the fifth season. )

Reply 3883

Original post by joostan

Sure:

\dfrac{1}{3^{k+1}}=\dfrac{1}{3} \times \dfrac{1}{3^k}

Rearrange a little bit more and voíla :smile:

Not entirely sure actually, if I multiplied them I'd get the same solution so should I perhaps try dividing by 1 third?

Reply 3884

Original post by Robbie242

Not entirely sure actually, if I multiplied them I'd get the same solution so should I perhaps try dividing by 1 third?

Well in the power you get the above result, then you can split it up as I did.
Then using the cubic root result, you get the desired proof. :smile:

(edited 10 years ago)

Reply 3885

\log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2

I somehow managed to simplify this to:

2b=3a

is this correct?

Reply 3886

Original post by joostan

Well in the power you get the above result, then you can split it up as I did, using the cubic root result. :smile:

eh I've only managed to prove

\frac{1}{3^{k}}=\frac{1}{3^{k}}

my algebra is weak

Reply 3887

L'Evil Fish

OP

18

Gonna disappear from this thread for a bit I think :ninja:

I'll be back before my S1 exam, I don't think there are any maths exams before that, but if there are good luck! (actually there won't be, it's half term :lol:

)

Anyway, adios!

Reply 3888

Original post by tigerz

\log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2

I somehow managed to simplify this to:

2b=3a

is this correct?

Try raising your

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

to the power of b, and then you may be able to draw some conclusions

Reply 3889

Original post by tigerz

\log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2

I somehow managed to simplify this to:

2b=3a

is this correct?

2^b = 3^a

Original post by L'Evil Fish

Gonna disappear from this thread for a bit I think :ninja:

I'll be back before my S1 exam, I don't think there are any maths exams before that, but if there are good luck! (actually there won't be, it's half term :lol:

)

Anyway, adios!

Aufwiedersehen :ninja:

Reply 3890

Original post by Robbie242

eh I've only managed to prove

\frac{1}{3^{k}}=\frac{1}{3^{k}}

my algebra is weak

Spoiler

Reply 3891

Original post by joostan

2^b = 3^a

Aufwiedersehen :ninja:

nvm managed to figure it out, the visual thing you gave me makes sense when multiplying the powers, your just expressing it in a different way.

Reply 3892

Original post by Robbie242

nvm managed to figure it out, the visual thing you gave me makes sense when multiplying the powers, your just expressing it in a different way.

Coolio

Reply 3893

Original post by Robbie242

Try raising your

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

to the power of b, and then you may be able to draw some conclusions

Original post by joostan

2^b = 3^a

Aufwiedersehen :ninja:

Tex error,my bad thats what I meant :P right so I can continue from there :smile:

Reply 3894

MathsNerd1

17

Original post by L'Evil Fish

Gonna disappear from this thread for a bit I think :ninja:

I'll be back before my S1 exam, I don't think there are any maths exams before that, but if there are good luck! (actually there won't be, it's half term :lol:

)

Anyway, adios!

I should do the same to get this revision done that I really should've done today! :-/ See you some time in the future :biggrin:

Posted from TSR Mobile

Reply 3895

MathsNerd1

17

Can anyone run through the basics of induction and then give me some relatively easy questions so I can work at it, then slowly bump up the difficulty so I'm able to try a few STEP questions on the topic, I just feel like I've forgotten too much of it to do anything that would be any good :-/

Posted from TSR Mobile

Reply 3896

ThatRandomGuy

1

Anyone who is doing/has done D2, in game theory, when you are working out the value of the game in a 3x2 matrix and you have 3 intersection points on a probability graph, how do you know which one is optimal?

Reply 3897

ThatRandomGuy

1

And for anyone who did S2 last week, for the question on the number of games in stock, did you end up with 12 or 13? No one seems to know which one is the correct answer :s-smilie:

Reply 3898

\log_{3}2=\frac{a}{b} \therefore 3^\frac{a}{b}=2

this means

2^b=3^a

We know that

2^b

will be even so can be represented by

2w

We also know that

3^a

will be odd so is

2x+1

This means

2^b=3^a\equiv 2w=2x+1\therefore w=x+1

URM i'm lost now, I need to link it back to the original question but confused :confused:

Reply 3899