- #1
kudzie adore
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prove by Induction that n! ≥ 2^(n-1) n ≥ 1
Prove by Induction that n ≥ 2^(n-1) n ≥ 1
- Thread starter kudzie adore
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In summary, the conversation discusses the process of proving the inequality n! ≥ 2^(n-1) using induction. The individual has successfully proven the base case for n=1 and has assumed the statement to be true for n=k. The next step is to prove the statement for n=k+1, which involves multiplying both sides of the equation by (k+1). The challenge is to determine how to reach the proof for (k+1).
- #2
pwsnafu
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Okay, so show us what you have attempted.
- #3
kudzie adore
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I managed to prove for n= 1 and is true
for n=k k!≥ 2^ (k-1) and I assumed that n=k to be true
then for n= k+1 its (k+1)! ≥ 2^[(k+1)-1]
proof for n=(k+1)
(k!)(k+1) ≥ _____?
the problem is that how do we reach the proof for (k+1)
for n=k k!≥ 2^ (k-1) and I assumed that n=k to be true
then for n= k+1 its (k+1)! ≥ 2^[(k+1)-1]
proof for n=(k+1)
(k!)(k+1) ≥ _____?
the problem is that how do we reach the proof for (k+1)
- #4
tiny-tim
Science Advisor
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hi kudzie adore! welcome to pf! 
(try using the X2 button just above the Reply box)
hint: if an equation is true, then multiplying both sides by the same factor will still be true
(try using the X2 button just above the Reply box
)hint: if an equation is true, then multiplying both sides by the same factor will still be true
FAQ: Prove by Induction that n ≥ 2^(n-1) n ≥ 1
What is induction and how does it relate to proving mathematical statements?
Induction is a mathematical proof technique that is used to show that a statement is true for all possible cases. It involves proving a base case and then showing that if the statement is true for one case, it is also true for the next case. This process continues until the statement is proven to be true for all cases.
What is the statement being proved using induction in this case?
The statement being proved is that n ≥ 2^(n-1) for all values of n ≥ 1.
How do you prove this statement using induction?
First, we prove the base case n = 1, which states that 1 ≥ 2^(1-1) or 1 ≥ 1, which is true. Next, we assume that the statement is true for some arbitrary value k, which means that k ≥ 2^(k-1). Then, we must prove that the statement is also true for k+1, which is (k+1) ≥ 2^((k+1)-1). By substituting in our assumption for k, we get (k+1) ≥ 2^(k-1) * 2, which simplifies to (k+1) ≥ 2^k. Since this is true, our assumption is also true, and the statement is proven by induction.
Can this statement be proved using other methods besides induction?
Yes, this statement can also be proved using other methods such as direct proof or contradiction. However, induction is the most commonly used method for proving statements that involve a recursive or repeating pattern.
Can the statement be generalized to include values of n less than 1?
No, the statement only applies to values of n ≥ 1. For values of n less than 1, the statement does not hold true. For example, when n = 0, the statement would be 0 ≥ 2^(0-1) or 0 ≥ 1, which is not true.