Math Induction: Where Does the >2xk Come From?

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SUMMARY

The discussion centers on the mathematical proposition involving induction, specifically addressing the expression 2^(k+1) and its relationship to k+1. Participants clarify that for k > 1, the inequality 2k > k + 1 holds true, confirming the validity of the proposition. The consensus is that 2^(k+1) is indeed greater than or equal to k+1, reinforcing the correctness of the induction step in the proof.

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  • Understanding of mathematical induction
  • Familiarity with exponential functions, specifically 2^(k+1)
  • Basic algebraic manipulation skills
  • Knowledge of inequalities and their properties
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  • Study the principles of mathematical induction in depth
  • Explore the properties of exponential functions and their growth rates
  • Review algebraic techniques for manipulating inequalities
  • Investigate common induction proofs in mathematics
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Students of mathematics, educators teaching mathematical induction, and anyone interested in understanding the foundations of proofs involving inequalities and exponential growth.

coconut62
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Please refer to the image attached.

Where does the > 2 x k come from?

Based on the proposition, shouldn't it be > k+1?
 

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look two lines above the questioned statement.
 
I got it, thanks.
 
But one more thing: 2^(k+1) is equal or bigger than k+1. Then it's not necessarily bigger than k+1. How can we say that the proposition is true for that case?
 
coconut62 said:
But one more thing: 2^(k+1) is equal or bigger than k+1. Then it's not necessarily bigger than k+1. How can we say that the proposition is true for that case?
If k > 1, then 2k > k + 1.

2k = k + 1 only if k = 1.
 
Greater and Greater-Equal

coconut62 said:
But one more thing: 2^(k+1) is equal or bigger than k+1. Then it's not necessarily bigger than k+1. How can we say that the proposition is true for that case?


The propisition is true since it says 2^{k+1} > k + k \ge k+1 and together 2^{k+1} > k+1.
 

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