a
↑
n
b
{\displaystyle a\uparrow ^{n}b}
is defined for a, b and n are integers
n
≥
1
{\displaystyle n\geq 1}
and
b
≥
0
{\displaystyle b\geq 0}
.
a
↑
1
b
=
a
b
{\displaystyle a\uparrow ^{1}b=a^{b}}
a
↑
n
0
=
1
{\displaystyle a\uparrow ^{n}0=1}
a
↑
n
b
=
a
↑
n
−
1
(
a
↑
n
(
b
−
1
)
)
{\displaystyle a\uparrow ^{n}b=a\uparrow ^{n-1}(a\uparrow ^{n}(b-1))}
Therefore
a
↑
n
b
=
a
↑
n
−
1
a
↑
n
−
1
⋯
↑
n
−
1
a
{\displaystyle a\uparrow ^{n}b=a\uparrow ^{n-1}a\uparrow ^{n-1}\cdots \uparrow ^{n-1}a}
(with b copies of a, where
↑
{\displaystyle \uparrow }
is right associative) and so it is seen as an extension of the series of operations
+
,
×
,
↑
,
↑
2
,
…
{\displaystyle +,\times ,\uparrow ,\uparrow ^{2},\dots }
where
↑
{\displaystyle \uparrow }
is basic exponentiation
a
↑
n
1
=
a
{\displaystyle a\uparrow ^{n}1=a}
2
↑
n
2
=
2
2
=
2
×
2
=
2
+
2
=
4
{\displaystyle 2\uparrow ^{n}2=2^{2}=2\times 2=2+2=4}
We can extend the uparrows to include multiplication and addition as the hyper operator .
This system may be consistently expanded to include multiplication, addition and incrementing:
a
↑
−
2
b
=
b
+
1
{\displaystyle a\uparrow ^{-2}b=b+1}
a
↑
−
1
b
=
a
+
b
{\displaystyle a\uparrow ^{-1}b=a+b}
a
↑
0
b
=
a
b
{\displaystyle a\uparrow ^{0}b=ab}
a
↑
1
b
=
a
b
{\displaystyle a\uparrow ^{1}b=a^{b}}
a
↑
n
0
=
1
{\displaystyle a\uparrow ^{n}0=1}
(for
n
≥
1
{\displaystyle n\geq 1}
)
a
↑
n
b
=
a
↑
n
−
1
(
a
↑
n
(
b
−
1
)
)
{\displaystyle a\uparrow ^{n}b=a\uparrow ^{n-1}(a\uparrow ^{n}(b-1))}
Proof of consistency by induction.
We will show that Rules 3, 5 and 6 imply rule 4
Assume that
a
↑
1
β
=
a
β
{\displaystyle a\uparrow ^{1}\beta =a^{\beta }}
for any
β
<
b
{\displaystyle \beta <b}
, then
a
↑
1
b
=
a
↑
0
(
a
↑
1
(
b
−
1
)
)
=
a
⋅
a
b
−
1
=
a
b
{\displaystyle a\uparrow ^{1}b=a\uparrow ^{0}(a\uparrow ^{1}(b-1))=a\cdot a^{b-1}=a^{b}}
by rule 6, rule 3 and assumption
Furthermore,
a
↑
1
0
=
1
=
a
0
{\displaystyle a\uparrow ^{1}0=1=a^{0}}
by rule 5
Thus the assumption is true for all
β
≥
0
{\displaystyle \beta \geq 0}
Likewise we can show that Rules 2, 5, 6 imply Rule 3 and that Rules 1, 5, 6 imply Rule 2.
Therefore, Rules 1, 5, 6 imply Rules 4, 5, 6 and so consistently extend the system.
QED
Clearly some of the properties do not extend.
Todo: How do you change bases.
Example:
3
↑
k
n
≈
10
↑
k
n
′
{\displaystyle 3\uparrow ^{k}n\approx 10\uparrow ^{k}n^{\prime }}
what is n' ?
For k = 1:
a
↑
n
=
a
n
=
b
n
log
b
(
a
)
=
b
↑
(
log
b
(
a
)
n
)
{\displaystyle a\uparrow n=a^{n}=b^{n\log _{b}(a)}=b\uparrow (\log _{b}(a)n)}
For k = 2. For all
a
<
b
{\displaystyle a<b}
there is a unique
δ
{\displaystyle \delta }
such that
b
↑↑
(
n
−
δ
−
1
)
<
a
↑↑
n
<
b
↑↑
(
n
−
δ
)
{\displaystyle b\uparrow \uparrow (n-\delta -1)<a\uparrow \uparrow n<b\uparrow \uparrow (n-\delta )}
for all sufficiently large n
Examples:
10
↑↑
(
n
−
1
)
<
3
↑↑
n
<
10
↑↑
n
{\displaystyle 10\uparrow \uparrow (n-1)<3\uparrow \uparrow n<10\uparrow \uparrow n}
for all
n
≥
3
{\displaystyle n\geq 3}
10
↑↑
(
n
−
3
)
<
2
↑↑
n
<
10
↑↑
(
n
−
2
)
{\displaystyle 10\uparrow \uparrow (n-3)<2\uparrow \uparrow n<10\uparrow \uparrow (n-2)}
for all
n
≥
4
{\displaystyle n\geq 4}
Thus the base of a tetration is not very important, they all grow at approximately the same rate eventually.[ note 1]
In fact these numbers
δ
{\displaystyle \delta }
grow very slowly.
Claim:
a
↑↑
(
n
+
k
−
1
)
<
(
a
↑↑
k
)
↑↑
n
<
a
↑↑
(
n
+
k
)
{\displaystyle a\uparrow \uparrow (n+k-1)<(a\uparrow \uparrow k)\uparrow \uparrow n<a\uparrow \uparrow (n+k)\,}
Note, the left inequality is easy to prove:
(
a
↑↑
k
)
↑↑
n
=
(
(
a
↑↑
k
)
↑
)
n
−
1
(
a
↑↑
k
)
>
(
a
↑
)
n
−
1
(
a
↑↑
k
)
=
a
↑↑
(
n
+
k
−
1
)
{\displaystyle (a\uparrow \uparrow k)\uparrow \uparrow n=((a\uparrow \uparrow k)\uparrow )^{n-1}(a\uparrow \uparrow k)>(a\uparrow )^{n-1}(a\uparrow \uparrow k)=a\uparrow \uparrow (n+k-1)\,}
Claim:
n
↑
n
n
<
(
3
↑
n
3
)
↑
n
(
3
↑
n
3
−
3
)
<
3
↑
n
(
3
↑
n
3
)
=
3
↑
n
+
1
3
{\displaystyle n\uparrow ^{n}n<(3\uparrow ^{n}3)\uparrow ^{n}(3\uparrow ^{n}3-3)<3\uparrow ^{n}(3\uparrow ^{n}3)=3\uparrow ^{n+1}3\,}