Proving 2^n < n without Equality: Why Ask?

  • Thread starter pivoxa15
  • Start date
In summary, the conversation discusses proving the inequality 2^n<=n! for all n>=4, and whether it is necessary to prove the stronger inequality of 2^n<n! in order to prove the original one. The possibility of a misprint in the textbook is also mentioned.
  • #1
pivoxa15
2,255
1
In my maths textbook it asks to prove 2^n<=n! for all n>=4

I could prove it no problems using induction but could show 2^n<n! without the equality inequality.

My question is why would the textbook ask for a weaker condition? Is it a misprint?

If I can show < then it automatically implies <= holds as well dosen't it?
 
Mathematics news on Phys.org
  • #2
If I can show < then it automatically implies <= holds as well dosen't it?
Yes, if x is strictly less than y, then x is certainly less than or equal to y. With regards to your former question, it might just be a typo- either way, you can solve the problem using induction.
 
  • #3
It can't be a missprint. See what happens with n=4.

Daniel.

PS. Apparently, it can be a missprint.
 
Last edited:
  • #4
dextercioby said:
It can't be a missprint. See what happens with n=4.

Daniel.

2^4=16 , 4!= 24, so 2^4<4!
 
  • #5
I was only asking the relationship between 2^n and n!

Looking at the graphs for 2^n and n! it seems that no where is 2^n=n!

n>=4 is definitely correct.
 
  • #6
Look at what are the factors of 2^n and the factors of n!...
 
  • #7
There are two "=" signs in question. The first one is definitely a misprint but n>=4 is correct.
 

1. What is the importance of proving 2^n < n without equality?

Proving 2^n < n without equality has significance in various mathematical and scientific fields, as it helps to establish the upper bound of growth rate for certain functions and algorithms. It also allows for a better understanding of exponential functions and their limitations.

2. How does one go about proving 2^n < n without equality?

To prove 2^n < n without equality, one can use mathematical induction, which involves establishing a base case and then proving the inequality for the next value using the previous one. Another method is to use logarithms to simplify the expression and then compare the growth rates.

3. Can 2^n ever be equal to n?

No, 2^n can never be equal to n. This is because exponential growth occurs at a much faster rate than linear growth, and thus, 2^n will always surpass n at some point.

4. What are the applications of proving 2^n < n without equality?

Proving 2^n < n without equality has applications in computer science, particularly in the analysis of algorithms and data structures. It is also useful in determining the time complexity of algorithms and predicting their efficiency.

5. Are there any exceptions to the inequality 2^n < n?

Yes, there are a few exceptions to the inequality 2^n < n, such as when n is equal to 0 or 1. In these cases, 2^n is equal to n. However, for any other positive integer value of n, 2^n will always be less than n.

Similar threads

I Prove Even # Transpositions in Identity Permutation w/ Induction & Contradiction
    • General Math
Replies
2
Views
1K
[ASK] Mathematical Induction: Prove 7^n-2^n is divisible by 5.
    • General Math
Replies
5
Views
2K
MHB Proving Triangle Area ≤ $\frac{1}{2}$ in a Square with $(n+1)^2$ Points
    • General Math
Replies
1
Views
765
B Wolfram Alpha's Why the 2 i π n?
    • General Math
Replies
3
Views
1K
B Combinatorics and Magic squares
    • General Math
Replies
2
Views
1K
MHB Is There an Easier Method to Prove $n^2>n$ for Negative Integers?
    • General Math
Replies
2
Views
1K
MHB Proving Inequality: \(\frac{1}{n^2}\) Sum < \(\frac{7}{4}\)
    • General Math
Replies
2
Views
1K
MHB Solve Prove by Induction: 3^2n+1 + 2^n-1 Divisible by 7
    • General Math
Replies
2
Views
1K
B Points on a Circle (moving through an obstacle)
    • General Math
Replies
1
Views
777
MHB Probability of misprints on each page
    • Set Theory, Logic, Probability, Statistics
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top