8 replies to this topic

[roytheshort]

Hyper Operator Wikipedia Page.

hyper₀(a, = b + 1
hyper₁(a, = a + b (Addition)
hyper₂(a, = a * b (Multiplication)
hyper₃(a, = a^b (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 hyper_n(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.

[kikjezrous]

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

How about...

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...

[roytheshort]

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

That's what I'm asking.

How about...

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.

[fenyxofshadows]

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

I had one idea, but I misread the information about hyperoperations, so this proof is rubbish

Spoiler

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

hyper_i(2,2) = 2+2 = 4

Step 2) Assuming hyper_i(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-1](2-1)) * 2
hyper_(i+1)(2,2) = (2[i-1]1) * 2
hyper_(i+1)(2,2) = 2 * 2

hyper_(i+1)(2,2) = 4

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

And, lo and behold, it is not true!

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

hyper_i(2,2) = 2+2 = 4

Step 2) Assuming hyper_i(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) = hyper_i(2,2[i-2]2)
hyper_(i+1)(2,2) = hyper_i(2,hyper_i(2,2))
hyper_(i+1)(2,2) = hyper_i(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.

[roytheshort]

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.

[xshortguy]

Define and note that for positive integer b, we have


Observe that for b = 2, this reduces to


When a = 2, this then shows


Since n is arbitrary, for all integers n > 0 it is clear that

Here's a more advanced problem:

Define the following recursive sequence for x > 0: . Show that this sequence converges if and only if

[kikjezrous]

Ugg. Ser:es. Why d:d I not th:nk of that one?
*headdesk*

[roytheshort]

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

hyper_n(x,x) = hyper_n+1(x,2)
hyper_n(x,2) = hyper_n-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*hyper_n(x,y)

[Yourself]

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

This follows directly from the definition of the hyperoperation.

And to simplify this.

x*hyper_n(x,y)

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

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