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So, i'm stuck on trying to understand what happens on the following:

Show that $2^n < n!$ for every natural number $n \ge 4.$

Step 1: substitute n = 4 on the equation $2^4 < 4!$ that gives 16 < 24, so it's true for every $n>=4$

Step 2: in the inductive hypothesis, which is basically substituting $n$ by $k$, so $2^k < k!$

Step 3: i want to show that if its true for k so its true for $(k+1)$, so $2^{k+1} < (k+1)!$

Step 4: to get into step 3 I need to multiply by 2 both sides of inductive hypothesis, so $(2^k) \times 2 < (k!) \times 2$ from here on I cannot understand.

Could you please provide an answer in a way that each operation or comparison or property is contained in a single step?

md2perpe
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Leonardo Fonseca
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    Since $n\geq 4$ is it true that $2<k+1$? Since we are talking about specifically $k$ who are $k\geq n$ while we are in this induction step... aren't you done? – JMoravitz Jan 03 '24 at 17:57
  • Recall that if $0<a<b$ and $0<c$ then we have that $ac<bc$... by definition or basic properties of the $<$ symbol. – JMoravitz Jan 03 '24 at 17:59
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    Obligatory suggested reading: How to write a clear proof by induction. Also, you have typos... in step 1 for instance you end with saying "so it's true for every n>=4" when this step only showed that it was true for exactly n=4 and has not yet implied anything about any other cases. – JMoravitz Jan 03 '24 at 18:01
  • In step 4, on the RHS, with $~2 < 4 \leq k,~$ how does $~2 \times k!~$ compare with $~(k+1) \times k! = (k+1)! ~?$ – user2661923 Jan 03 '24 at 18:20
  • in step 4, if i have a number, say k, which is equal or greater than 4, if i get this same number k and add 1, it will still be greater than 4. if k added by 1 is still greater than 4 then it is also greater than 2. This property is known as transitivity. So i could substitute the 2 on the rhs by k + 1 just because of that property? – Leonardo Fonseca Jan 05 '24 at 02:29

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