Hello, so just as an introduction, I am a novice level mathematician. I am a computer software engineer so I have been trying to learn group theory on my own but have been struggling. However, my path to self-learning this field made me realize there are times where I have to compromise my own knowledge as it is vastly overwhelmed by all the special methods, functions, and results that are in the toolbox of advanced mathematicians. This made me excited for ChatGPT but after this incidence I am realizing how detrimental beginners relying on ChatGPT will get.
I saw on Instagram a comparison question between two values: 2^100! and (2^100)! I wasn’t so sure at first but here is my opinion on it (not enough to be a proof):
We know for a fact that when we divide a=(2^100)!/2^100! we should get a less than 1 to show that the denominator is bigger and if a more than 1, the numerator is bigger.
I tried to show that we can express the numerator as (2^{100})^99! which shows 2^100 being multiplied by itself 99! times.
The denominator is (1)(2)(3)…(2^{100}-1)(2^100) which can be seen as values <= 2^100 being multiplied 2^{100}-1 (1 is identity) times.
Then I said for large numbers of n, (2^n)-1<(n-1)! which just shows 99! > 2^{100}-1
So because the denominator 2^100! has more terms and terms that are at least greater than or equal to the terms in (2^100)! and every single term left over will only increase the denominator, our a goes very close to 0, which is less than 1. So therefore, 2^100! is bigger.
This is a little bit of a messy demonstration but I think it is valid. However, someone told me there was a better way we could make this rigorous by applying log2. I responded back saying that one can apply log2 on both sides. I told them that this claim is quite bold as log2(2^100!) is trivial but log2((2^100)!) seems very difficult to simplify. I suggested maybe rewriting it as a sum then taking the integral then applying some boundaries but that seems really hard to do in this situation due to the shear magnitude of (2^100)!. Then, I received a comment that made me distrust ChatGPT forever. You see, I asked this question to ChatGPT previously and they gave me the following response:
“The factorial operation grows much faster than exponential operations as the input value increases. For 2^100, you’re looking at a very large number. However, when you take the factorial of that number, (2^100)!, you’re multiplying 2^100 by every integer less than it down to 1. This operation grows the number astronomically compared to the exponential growth of 2^100!, which is 2 raised to the power of the factorial of 100 (a much smaller number than 2^100 itself).
Therefore, without needing to calculate the exact values, we can deduce that (2^100)! is significantly larger than 2^100!.”
This response genuinely made me question my existence as this reasoning was so weird. This was done by ChatGPT4, probably the best LLM available to the public right now.
So, I already had suspicions that the reason why so many people in the comments were getting this question wrong was because of ChatGPT.
Back to the man who answered me. When questioning how he will evaluate log2((2^100)!), he has said:
“Not it's pretty straightforward. I wrote a comment for this method but i'll rewrite it here. You rewrite log2(2^100 !) as the sum of log2 (i), you majorate by the number of terms time the greatest term, that gives 100 × 2^100. In the end you have to compare 99! With 2^100, which is easy. Same as comparing 297with 99! / 8. End”
The reason why I blame this on ChatGPT is that if you actually tell it to apply log2, it tells you the image as I have posted.
I thought I was about to learn some kind of new method but no, this guy on Instagram just searched up how to do it on ChatGPT then tried to lecture me on what the retard AI suggested. Idk man. This type of misinformation is not at all trivial to a math layman and even for myself tbh. I had been questioning myself over this for so long since ChatGPT keeps giving responses like:
“The exponentiation rule a^mn = (a^m)^n applies here, but it’s essential to apply it correctly. Correctly, if you have 2^100!, it does not simplify or equal (2^100)^99! directly due to the nature of factorial growth and exponential operations.”
I think ChatGPT is starting to make the math area of the internet, once a place where only people with somewhat built-in prerequisite knowledge of foundational mathematical operations and techniques, to be full of people who use ChatGPT and just straight up give misinformation that is very hard to detect.
When I confronted the guy he said:
“I did it in my head in 2 min. It's an actual proof without "we observe" . if you couldn't understd what i said before it's just skill issue”
The man didn’t even bother to EXPLAIN the maths and his conclusion, he is just jamming operations together and seeing if ChatGPT will find a way to justify his methods. I think unless ChatGPT gets better at math, we are going to see a lot of math disinformation on the internet which really sucks.
Edit: (2n)-1<(n-1)! should be (2^n)-1<(n-1)!
Videos
What would be the factorial of zero?
Where can the concept of factorial be used?
What do we understand from the term factorial and what is the formula for the calculation of a factorial.
2 raised to (100 factorial )or (2 raised to 100 ) factorial, i believe its one on the right because i heard somewhere when terms are larger factorial beats exponents but then again im not sure , is there a way to solve it
Me and my friends figured 2100! was bigger after trying it with smaller numbers than 100! but we couldn't prove it or are %100 sure that it is correct. I am really curious as to how it could be proved. Thank you in advance :)
Your list of multiples is correct. But then the total is 97, not 2^97. But good to see some work.
How to solve this in the easiest way possible...
keep dividing 100 by 2 till you get a value < 2
100 / 2 = 50
50 / 2 = 25
25 / 2 = 12 (forget about the remainder)
12 / 2 = 6
6 / 2 = 3
3 / 2 = 1
now just add up all the quotients 50 + 25 + 12 + 6 + 3 + 1 = 97
the same logic can be applied to find the highest power of any number x that divides n! completely or evenly.
If you need to find the exponent of a prime number $p$ in $N!$, you have to look how many times it appears (it will appear every $p$ numbers, and every $p^2$ numbers it will appear twice, and every $p^3$ numbers it will appear three times, etc).
So you are looking for the number : $$\Bigl\lfloor\dfrac{100}{2} \Bigr\rfloor+\Bigl\lfloor\dfrac{100}{2^2} \Bigr\rfloor+\Bigl\lfloor\dfrac{100}{2^3}\Bigr\rfloor+\cdots$$
which also is Legendre's Formula
An easier way to do this:
$$\lfloor\dfrac{100}{2}\rfloor + \lfloor\dfrac{100}{2^2}\rfloor + \lfloor\dfrac{100}{2^3}\rfloor +...$$ $$=50+25+12+6+3+1$$ $$=97$$
Hey guys,
I have a problem to solve this on my own. Is there a trick I don't know for solving 100! ?
With simple ineqalities we have:
$100!\geq 90^{10}\cdot 80^{10}\cdots 20^{10}\cdot 10^{10}$
$100!\geq (9\cdot 8 \cdots 2 \cdot 1)^{10}\cdot 10^{90}$
$100!\geq (9!)^{10}\cdot 10^{90}>10^{100}$
Before there was an error on the algebra, as pointed out in the comments. I try to fix the error following the same approach:
$100!=(1\times..\times 10)\times(11\times..\times 20)\times...\times(91\times..\times 100)=A_1...A_{10}$
so we estimate $A_i \ge 10^{10}$ for $i=2,..9$.
Instead we write $A_1A_{10}=(1\times 100)\times(2\times 99)\times(3 \times 97)\times...\times(10 \times 91)\ge (10^2)^{10}$.
Combining: $100!\ge (10^{10})^8 \times (10^2)^{10}=(10^{10})^{10}=10^{100}$.