Math Is Fun Forum

Yes, I used logarithm to the base 10 for convenience.
The idea is to show that
Infinity + 1 = Infinity
Infinity + Infinity = Infinity
Infinity x Infinity = Infinity
Infinity ^ Infinity = Infinity
Infinity ^ Infinity ^ Infinity ^ Infinity ........Infinite times = Infinity
This is true because because we can never say for what value of n
log(n)infinity would be undefined.
And this is only the fourth stage of iteration (also referred to as 'tetration').
No wonder, Infinity is beyond comprehension for the human mind!

If we are talking of infinite number of zeros after the decimal, before the 1, in b,
the proof is correct!
Because, a²-b² would be equal to a-b.
This is because we have defined 0.0000000000.......1 as 1 ater an infinite number of zeros after the decimal followed by 1,
as kylekatarn denotes 0.(0)1.

However, if the number of zeros is finite, (Don't worry, it is not)
a²-b² would contain more 9s than a-b.

Oops, I made a mistake.

If, as I had said earlier, x∈ Real numbers,
my question would have no answer as
(-1 x 10^n) -1 would give a negative odd number for a
large value of n (n ∈ Natural Numbers).  And this number raised to itself would give a negative number which would be much lesser than zero!
Therefore, the smallest value would be -∞

I forgot while posting that one that a^-n = 1/a^n
Therefore, the minmum value of x^x is not -∞
I am sorry I wasn't concentrating 100%

0^0^0 is not defined.
But for 1, as you pointed out,  2^Mx > x^x^x.
For this problem, let us take value of x ≥ 2, x ∈ N

0.(0)1 is 1/(10^ ∞)
When I had to deal with numbers like 9.9999999999999999999....8,
I put a horizontal arrow pointing towards the right above the 9 after the decimal.Then I put the number 8. This meant that 9 keeps on recurring but the last digit is 8. But your method seems better, more convenient. Whether it would be accepted by Mathematicians worldover is another thing!

But the point I wanted to make is 0.999999999999...................... is a devilish number and would never be found in any mathematical problem or solution. It cannot be represented in a graph, It cannot be said to be irrational, yet it cannot be expressed as a rational number. It is not a transcendental number. As a rule, it should be possible to represent all decimals as a rational number. But this number defies the rule, although it is a recurring decimal. Sin 89.99999999999999999999999999999........ degrees can be said to be equal to this number. 

I give up..........

Problem # n+7

If x ∈ Real Numbers, What is the minimum value of x^x?

Reminded me of one of Murphy's Love Laws:-

Brains x Beauty x Availability = Constant

Thanks polyanin.
The website seems to be GREAT!

And you got it RIGHT!
The Girls can take the odd places and the boys can take the even places.
This is possible in 5! x 5! different ways.
Then, the Boys can take the odd places and the girls can take the even places, which is possible in 5! x 5! different ways.
Hence, the number of arrangements possible is
2 x 5! x 5! = 2 x 120 x 120 = 28,800

I live in Chennai, India, very close to the Equator, and 5 1/2 hours ahead of GMT! Chennai is a port city in the Bay of Bengal and was formerly known as Madras.

Digital roots of 14^n

The digital roots always go in cycles of 6.
That is there are six possible digital roots (sometimes, one or more of them may be the same)Since the digital root of 14 is 5,
we get 6 values of digital roots for
5^6n, 5^6n+1, 5^6n+2,....5^6n+5
They are 1, 5, 7, 8, 4, 2.
For even powers, we see that the digital roots are 1, 7, 4.
This would be true for 5, 14, 23, 32, 41 etc, i.e. all numbers having digital root of 5.

From the work of Mathsy, it is seen that we needed to check only for even powers (as odd powers are divisible by 5).
Since the even powers are divisible by 3, it can be concluded that
14^n + 11 would never be a prime number!
I don't think there are any loopholes in the proof.

It was a little easier after I got 24 zeros for 100! Thereafter, one additional zero each for 125, 250, 375, 500, 625, 750, 875 and 1000 have to be provided. 249 should be the right answer! I didn't know the solution when I posted that question.