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Please note that I am exceptionally talented in over-complicating even the simplest of topics, however this may be worth a read. Also, the following operations are only performed with positive integers.

The simplest operation that increases value is addition. $x+y$ can be represented as $x+(1+1+1+1...)$, "$1$" repeating $y$ times. The next operation that increases is multiplication. Similarly, $x*y$ can be represented as $x+x+x+x...$ repeating $y$ times. This pattern follows with exponentiation: $x^y=x*x*x*x...$ Using this pattern, one can find tetration. $x^{x^{x^x...}}$ $y$ times. Now let's label every item in the pattern with some degree:

{Addition:$0$, Multiplication:$1$, Exponentiation:$2$, Tetration:$3$}. Now to find some universal method to represent any degree in the continued list given two operands.

$\lambda$ seems like a good idea as it has 3 lines protruding from i. We can represent any of these operations like so:

Imgur $n$ being the degree of operation.

$x^y$ can be represented as Imgur because exponentiation is the second degree.

Now consider the following:

Imgur

TLDR (not a sufficient one)

If the degree of iteration is infinite, and the operands are integers greater than one, will the output always be infinity?

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    $\begingroup$ $~^n_x\lambda_y$ will output $~^n_x\lambda_y$ $\endgroup$ – JMoravitz Oct 9 '17 at 17:42
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    $\begingroup$ As to your question, you should be able to easily prove by induction that $~^{n+1}_x\lambda_y>~^n_x\lambda_y$ and for natural numbers, $n,x,y$ will again result in a natural number. This will directly imply that the limit as $n$ approaches infinity of $~^n_x\lambda_y$ will approach infinity. You have not properly defined your lambda notation to accept any of $n,x,y$ as anything but finite natural numbers and the output will again always be finite, so it is incorrect to talk about "output being infinity" without using limits to do so. $\endgroup$ – JMoravitz Oct 9 '17 at 17:46
  • $\begingroup$ "infinite" is not a natural number. You should write $\lim \limits_{n \to \inf}$ and replace the inf symbol on top of the lambda for an "n" $\endgroup$ – Francisco José Letterio Oct 9 '17 at 17:53
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    $\begingroup$ $2+2=4;\;\;\;2*2=4;\;\;\;2^2=2\uparrow 2=4;\;\;\;2\uparrow\uparrow 2=4;$ That's tetration, using Knuth's uparrow convention. The pattern continues for iterated tetration (pentation). $2 \uparrow \uparrow \uparrow 2 = 4$ etc. So if you use the same definition as Knuth, then it remains bounded for $2 \uparrow^n 2=4$ $\endgroup$ – Sheldon L Oct 9 '17 at 21:24
  • $\begingroup$ I think it is much better to write the iterations always from a starting value, (perhaps from some neutral element), so for $ADD_b^h(x)=x + h \cdot b$ where if the base for the operation $b=1$ we have simply $ADD_1^h (x)=x+h$ and using the neutral element $ADD_b^h(0) = b \cdot h$. Similarly for multiplication $MUL_b^h(x)=x*b*b*...*b=x*b^h$ having $1$ as neutral element gives $MUL_b^h(1)=b^h$ and for exponentiation $EXP_b^h(x)=b^{b^{...b^x}} $ giving with the neutral element $EXP_b^h(1)=b^{b^{...b^1}}=^h b $ Keeping the formal $x$ in the notation clarifies the functional aspect. $\endgroup$ – Gottfried Helms Oct 10 '17 at 6:20

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