Proof question - Prove hyper[n](2,2) = 4

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#1 roytheshort

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Posted 16 August 2012 - 04:45 PM

hyper0(a, = b + 1
hyper1(a, = a + b (Addition)
hyper2(a, = a * b (Multiplication)
hyper3(a, = ab (Exponention)
hyper4(a, = ba (Tetration)

In other words, the next hyper operator is a repeated occurence of the previous one.

Now I've given a brief explanation of the hyper operator. I have a question.

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GIVEN n ∈ +ℤ AND n ≠ 0

PROVE THAT hypern(2,2) = 4 FOR ALL POSSIBLE VALUES OF N

Simple right? 2+2 = 4, 2*2 = 4, 22 = 4 and 22 = 4. Prove that this is true for all above cases. Or if NOT. Find a counter example. Given that n is in the set of positive integers and n is not equal to 0.
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#2 kikjezrous

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Posted 16 August 2012 - 05:48 PM

I know ::t's true... but what's the proof?

2+2 = 4

2{2+2+2+2....} = 4 (2*2 = 4)

2{2*2*2*2....} = 4 (2^2 = 4)

...

Proof by pattern?
I know ::t's true, but I can't g:ve you a real proof unt:l I take Number Theory.

I also know that 1^1 = 1; 1^^1 = 1; 1^^^1 = 1... etc...
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#3 roytheshort

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Posted 16 August 2012 - 06:14 PM

I know ::t's true... but what's the proof?

2+2 = 4

2{2+2+2+2....} = 4 (2*2 = 4)

2{2*2*2*2....} = 4 (2^2 = 4)

...

Proof by pattern?

Hmm, that's more of a proof by exhaustion that hasn't really got off the ground. You've only proved it for three cases. Where n = 1,2 and 3.

What I would like is a proof that proves that this is true for all cases. (Where n is a positive integer.)

Edited by roytheshort, 16 August 2012 - 08:20 PM.

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coobie

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Posted 16 August 2012 - 08:54 PM

So we need a proof by induction... Hmm... I'm going to try this. I'll edit this post when I'm done.

Spoiler

So, I'm trying a different approach...

And, lo and behold, it is not true!

Prove hyperi(2,2) = 4 for all i as an element of N+
Step 1) Show true for i = 1

hyperi(2,2) = 2+2 = 4

Step 2) Assuming hyperi(2,2) = 4 for all i as an element of N+, show hyper(i+1)(2,2) = 4

hyper(i+1)(2,2) = 2[i-1]2
hyper(i+1)(2,2) = 2[i-2](2[i-2]2)
hyper(i+1)(2,2) = hyperi(2,2[i-2]2)
hyper(i+1)(2,2) = hyperi(2,hyperi(2,2))
hyper(i+1)(2,2) = hyperi(2,4) (using assumption)
hyper(i+1)(2,2) != 4

In about 5 minutes this will get shot down by 2+2 = 4 and 2*2 = 4, but I give up...

Edited by fenyxofshadows, 16 August 2012 - 10:03 PM.

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#5 roytheshort

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Posted 16 August 2012 - 09:24 PM

So we need a proof by induction... Hmm... I'm going to try this. I'll edit this post when I'm done.

If you can't do it. Edit your post to say you can't. Otherwise I'm going to keep checking this every day for a proof. Thanks.

(I'm having a stab at making a proof myself, really is quite difficult.)

Edited by roytheshort, 16 August 2012 - 09:24 PM.

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#6 xshortguy

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Posted 16 August 2012 - 10:41 PM

Define $\textrm{hyper}_n(a, = a \star_n b$ and note that for positive integer b, we have
$a \star_n b = \underbrace{a \star_{n - 1} a \cdots a \star_{n - 1} a }_{b \text{ times}}$

Observe that for b = 2, this reduces to
$a \star_n 2 = a \star_{n - 1} a$

When a = 2, this then shows
$2 \star_n 2 = 2 \star_{n - 1} 2$

Since n is arbitrary, for all integers n > 0 it is clear that $2 \star_n 2 = 2 \star_1 2 = 2 + 2 = 4$

Define the following recursive sequence for x > 0: $x_n = x^{x_{n-1}}$. Show that this sequence converges if and only if $x \in [(1/e)^{e}, e^{(1/e)}]$
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#7 kikjezrous

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Posted 17 August 2012 - 02:36 PM

Ugg. Ser:es. Why d:d I not th:nk of that one?
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#8 roytheshort

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Posted 18 August 2012 - 09:05 PM

Through use of the hyperoperation, I found some algebraic rules that will hold true for all cases where n ∈ +ℤ AND n ≠ 0.

hypern(x,x) = hypern+1(x,2)
hypern(x,2) = hypern-1(x,x)

My next challenge is to prove them. (Shortguy's other proof would be useful to prove this.)

And to simplify this.

x*hypern(x,y)
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#9 Yourself

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Posted 18 August 2012 - 10:07 PM

My next challenge is to prove them. (Shortguy's other proof would be useful to prove this.)

First off, both of those equations are equivalent. If one is true, the other is trivially true. Also, shortguy already proved this:

Observe that for b = 2, this reduces to
$a \star_n 2 = a \star_{n - 1} a$

This follows directly from the definition of the hyperoperation.

And to simplify this.

x*hypern(x,y)

That's as simple as you're going to get.

Edited by Yourself, 18 August 2012 - 10:09 PM.

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