Mega A Level Maths Thread MK II

Original post by MathsNerd1

What do you mean?

Posted from TSR Mobile

He means the question doesn't require induction, but in my defence, you asked for a proof by induction question :redface:

Reply 3962

Original post by Felix Felicis

He means the question doesn't require induction, but in my defence, you asked for a proof by induction question :redface:

Oh okay and I did indeed, just not quite sure how to even attempt this one :-/

Posted from TSR Mobile

Reply 3963

justinawe

Original post by Felix Felicis

Well yeah but how much effort do you need to put in to write a random number and then copy and paste it xD

Ctrl+V doesn't really take more effort than typing '1' :colone:

Reply 3964

Original post by justinawe

Ctrl+V doesn't really take more effort than typing '1' :colone:

Shush

Fine fine I concede defeat :lol:

Reply 3965

justinawe

Original post by MathsNerd1

Oh okay and I did indeed, just not quite sure how to even attempt this one :-/

Posted from TSR Mobile

Spoiler

Reply 3966

Lord of the Flies

Original post by MathsNerd1

What do you mean?

Ah I should have read the preceding posts, if you're practising induction then carry on. I just meant that induction seems a waste of time on this one, because you can prove the inequality by simply re-writing it.

Reply 3967

Original post by justinawe

Spoiler

Erm (K +1)K! ? I've never really understood how to manipulate factorials :-/

Posted from TSR Mobile

Reply 3968

Original post by Lord of the Flies

...

I was wondering...would you mind having a look at this problem? :s-smilie:

Proof by example's not permitted apparently :s-smilie:

Reply 3969

Original post by MathsNerd1

Erm (K +1)K! ? I've never really understood how to manipulate factorials :-/

Posted from TSR Mobile

Ok, what've you got so far? :smile:

Reply 3970

tigerz

Original post by Felix Felicis

question

rational+irrational=irrational

\rightarrow a+b=c

where

a=x/y

and

c=v/w

as they are rational

this means

a+b=c\rightarrow \frac{x}{y}+b=\frac{v}{w}

a+b=c\rightarrow \frac{x}{y}+b=\frac{v}{w} \rightarrow b=\frac{v}{w}-\frac{x}{y}

b=\frac{vy}{wy}-\frac{xw}{wy}\therefore b=\frac{vy-xw}{wy}[br][br]Since the RHS is rational (as its a fraction with +ive intergers) B must be rational)

However at the start of the eq we stated that b= irrational

\therefore \frac{vy-xw}{wy}

is irrational! This is contradicting so the sum must be irrational

I think I did it right

(edited 10 years ago)

Reply 3971

MAyman12

11

Original post by Felix Felicis

I was wondering...would you mind having a look at this problem? :s-smilie:

Proof by example's not permitted apparently :s-smilie:

I did not say that proof by example is not allowed. As my friend who gave me the question said that there is an another way of solving it. He did give a hint related to complex numbers so this must be the way :redface:

Proof by example is all right, you just want to find an another solution for it.

(edited 10 years ago)

Reply 3972

Original post by Felix Felicis

Ok, what've you got so far? :smile:

I've got pretty much no where as I'm unsure how to prove on side is greater than another :-/ Could you work through a solution to show me how it's done and I'll see if I'm able to follow it?

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Reply 3973

Original post by MAyman12

I did not say that proof by example is not allowed. As my friend who gave me the question said that there is an another way of solving it. He did give a hint related to complex numbers so this must be the way :redface:

Proof by example is all right, you just want to find an another solution for it.

That's what I've tried but I wasted most of today trying to and I just want to put my mind to rest. :redface:

Original post by MathsNerd1

I've got pretty much no where as I'm unsure how to prove on side is greater than another :-/ Could you work through a solution to show me how it's done and I'll see if I'm able to follow it?

Posted from TSR Mobile

I take it you've gotten to the inductive hypothesis that

k! > 2^{k}

?

Well, to complete the inductive step, you want to show that

(k+1)! > 2^{k+1}

Can you think of how to get to

(k+1)!

from

k!

?

Reply 3974

justinawe

Original post by MathsNerd1

Erm (K +1)K! ? I've never really understood how to manipulate factorials :-/

Posted from TSR Mobile

Yes.

It's because,

(k+1)!

= 1 \times 2 \times 3 \times \dots \times k \times (k+1)

= (1 \times 2 \times 3 \times \dots \times k) (k+1) = k!(k+1)

Reply 3975

Kvothe the Arcane

20

Original post by Felix Felicis

That's what I've tried but I wasted most of today trying to and I just want to put my mind to rest. :redface:

I take it you've gotten to the inductive hypothesis that

k! > 2^{k}

?

Well, to complete the inductive step, you want to show that

(k+1)! > 2^{k+1}

Can you think of how to get to

(k+1)!

from

k!

?

Isn't it

Spoiler

?
Edit: Already mentioned :frown:

(edited 10 years ago)

Reply 3976

Original post by Felix Felicis

I take it you've gotten to the inductive hypothesis that

k! > 2^{k}

?

Well, to complete the inductive step, you want to show that

(k+1)! > 2^{k+1}

Can you think of how to get to

(k+1)!

from

k!

?

I keep forgetting that what I've assumed in n=K has to be used in the inductive step too, my mistake :-/

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Reply 3977

Original post by justinawe

Yes.

It's because,

(k+1)!

= 1 \times 2 \times 3 \times \dots \times k \times (k+1)

= (1 \times 2 \times 3 \times \dots \times k) (k+1) = k!(k+1)

Okay, thanks I think I can see how to prove it all now :smile:

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Reply 3978

Original post by Felix Felicis

I take it you've gotten to the inductive hypothesis that

k! > 2^{k}

?

Well, to complete the inductive step, you want to show that

(k+1)! > 2^{k+1}

Can you think of how to get to

(k+1)!

from

k!

?

Okay I thought I got it but I couldn't make any sense of it. To express (K+1)! = K!(K+1) but do I now use the fact that K! > 2^k in my induction step, if so then how exactly as this inequality seems to be making me quite cautious :redface:

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Reply 3979