Logic-Based Comparison: Determining the Larger Number between 2^1000 and 500!

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To determine which is larger between 2^1000 and 500!, one can analyze the number of factors involved. The hint suggests breaking down 2^1000 into 500 factors of 2, represented as (2^2)^500. By comparing these factors to the 500 factors in 500!, it becomes clear that 500! contains larger values overall. The logic lies in recognizing that while 2^1000 has many factors of 2, 500! includes a broader range of integers, leading to its greater size. Thus, 500! is indeed larger than 2^1000.
barneygumble742
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hi...how can i tell which number is larger just from using logic?

for example a question says "which is larger 2^1000 or 500!?" the answer is 500! and the hint is: just compare the 500 factors appearing in each expression. i don't understand the hint but more importantly, I'm just not seeing the logic part of it.

thanks,
barneygumble742
 
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Are you sure you copied this right? Because 2^1000 > 1000^100 > 500.
 
that's not 500 exclamation point. it's 500 factorial. should i have used a different notation?
 
make 2^1000 into 500 factors and compare the factors is what the hint is suggesting
2^1000=(2^2)^500. and then compare it to 500 factors of 500!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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