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Mega A Level Maths Thread MK II Watch

1. (Original post by joostan)
Also I should've told you that for any irrational x:

Though it's trivially simple to deduce
Where did x come from, can I just use what you gave me, I feel awfully confused now
2. (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 or e etc.
3. (Original post by joostan)
...
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Lemma
if x is irrational, then is irrational. This is trivial to show by contradiction - if but x is irrational, hence contradiction

We will prove is irrational

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

Assume is irrational.

Consider by the inductive hypothesis, is irrational and hence by lemma is irrational, thereby completing the inductive step, thus the result is true for all n by the principle of mathematical induction.
4. (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!
5. (Original post by Felix Felicis)
Spoiler:
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Lemma
if x is irrational, then is irrational. This is trivial to show by contradiction - if but x is irrational, hence contradiction

We will prove is irrational [tez]\forall n \in \mathbb{N}[/tex]

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

Assume is irrational.

Consider by the inductive hypothesis, is irrational and hence by lemma is irrational, thereby completing the inductive step, thus the result is true for all n by the principle of mathematical induction.
Spoiler:
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Not convinved of the necessity of the modular arithmetic or the lemma, but an outstanding proof nonetheless
You also seem to have abused the
6. (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
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Were looking to prove the is irrational for all natural numbers n

it was seen true for n=1, assume true for n=k such that:

if true for n=k, assume true for n=k+1 such that

so if we assume
by cross multiplying

let 3^k=x

a(3x)=5b

No idea how to proceed

7. (Original post by joostan)
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Not convinved of the necessity of the modular arithmetic or the lemma, but an outstanding proof nonetheless
You also seem to have abused the
I was on a wild goosechase with Mayman's question
8. (Original post by Robbie242)
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Were looking to prove the is irrational for all natural numbers n

it was seen true for n=1, assume true for n=k such that:

if true for n=k, assume true for n=k+1 such that

so if we assume
by cross multiplying

No idea how to proceed

It's

I'm not much good at this problem setting
9. (Original post by joostan)
It's

I'm not much good at this problem setting
fml ok
10. (Original post by Felix Felicis)
I was on a wild goosechase with Mayman's question
Lol. How much time did you say you wasted?
11. (Original post by joostan)
Lol. How much time did you say you wasted?
Like 4 hours
12. (Original post by Felix Felicis)
Like 4 hours
o.O
I'd never do that
13. (Original post by Felix Felicis)
I was on a wild goosechase with Mayman's question
DJ answered it btw
14. (Original post by Felix Felicis)
Like 4 hours
That's commitment!

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15. (Original post by joostan)
It's

I'm not much good at this problem setting
Spoiler:
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Were looking to prove the is irrational for all natural numbers n

it was seen true for n=1, assume true for n=k such that:

if true for n=k, assume true for n=k+1 such that

so if we assume

by multiplying both sides by b

And then I'm stuck, not sure exactly what contradiction I'm supposed to meet
16. (Original post by joostan)
o.O
I'd never do that
(Original post by MathsNerd1)
That's commitment!

Posted from TSR Mobile
I reckon I've a rubix complex Either that or I've been watching too much House

(Original post by MAyman12)
DJ answered it btw
I saw DJ's solution but it's the exact same as the one joostan/ justinawe proposed, it only uses complex numbers
17. (Original post by Robbie242)
Spoiler:
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Were looking to prove the is irrational for all natural numbers n

it was seen true for n=1, assume true for n=k such that:

if true for n=k, assume true for n=k+1 such that

so if we assume

by multiplying both sides by b

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

As the cubed root of any irrational number is irrational . . .
18. (Original post by joostan)
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As the cubed root of any irrational number is irrational . . .
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so could I say if we assume rationality,

But

And as the cube root of any irrational number is irrational, true for n=k+1

Seen true for n=1, if true for n=k, then also seen true for n=k+1, therefore seen true that is irrational for all natural numbers n by mathematical induction.

19. (Original post by Robbie242)
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so could I say if we assume rationality,

But

And as the cube root of any irrational number is irrational, true for n=k+1

Seen true for n=1, if true for n=k, then also seen true for n=k+1, therefore seen true that is irrational for all natural numbers n by mathematical induction.

Precisely
20. (Original post by joostan)
Precisely
One thing I'm algebraically confused on is how

Could you explain this if you don't mind, I know you multiply the inner power by the outer one but can't seem to reach that exact solution

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