Proving the Existence of a Large n for (2^n) > K

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SUMMARY

The discussion focuses on proving that for any natural number K, there exists a natural number n such that \(2^n > K\). The participants establish that the smallest K is 0, where \(2^0 = 1\) satisfies the inequality. The approach involves identifying values of n that satisfy the condition \(2^n > K\) by solving for n, confirming that as n increases, \(2^n\) will surpass any given K due to the exponential growth of the function.

PREREQUISITES
  • Understanding of exponential functions, specifically \(2^n\)
  • Basic knowledge of natural numbers and inequalities
  • Familiarity with mathematical proof techniques
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the properties of exponential growth and its implications in mathematics
  • Learn about mathematical induction as a proof technique
  • Explore the concept of limits in calculus to understand behavior as n approaches infinity
  • Investigate the relationship between logarithms and exponential functions
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Students in mathematics, particularly those studying algebra and proofs, educators teaching mathematical concepts, and anyone interested in understanding the fundamentals of exponential functions.

iwonde
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Homework Statement


Show that for any natural number K, there is an n large enough so that (2^n) > K.


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The Attempt at a Solution


K is a natural number -> the smallest possible K would be 0 (lower bound?) and the smallest 2^n is 1 when n = 0, and the upper bound for both sides are infinite. So if I set n=0 and K =0 I get 2^n > K. I'm not sure if this is the right approach.
 
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Take some number K, find the possible n's that satisfy 2^n>K. Ie. solve for n.
 

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