Proving a number is irrational.

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SUMMARY

The discussion centers on proving that log2(3) is irrational through a contradiction method. The proof begins by assuming that log2(3) can be expressed as a rational number x = P/Q, leading to the equation 2P = 3Q. The conclusion is reached by demonstrating that 2P is always even while 3Q is always odd, resulting in a contradiction. The proof is confirmed to be valid, with a note on the necessity of P and Q being positive integers to avoid exceptions.

PREREQUISITES
  • Understanding of logarithmic functions and their properties
  • Familiarity with proof by contradiction techniques
  • Basic knowledge of even and odd integers
  • Experience with integer exponentiation
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  • Study the properties of logarithms in depth, focusing on irrational numbers
  • Learn more about proof techniques, particularly proof by contradiction
  • Explore the implications of even and odd integers in mathematical proofs
  • Investigate other examples of irrational numbers and their proofs, such as √2
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Mathematics students, educators, and anyone interested in number theory and proof techniques, particularly those focusing on irrational numbers.

cragar
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Homework Statement


Prove that log_2(3) is irrational.

The Attempt at a Solution



This is also equivalent to 2^x=3 from the definition of logs.
Proof: For the sake of contradiction let's assume that x is rational and that their exists integers P and Q such that x=P/Q .
so now we have 2^{\frac{P}{Q}}=3
now I will take both sides to the Q power .
so now we have 2^P=3^Q
since P and Q are integers, there is no possible way to have 2 raised to an integer to equal 3 raised to an integer, because 2^P will always be even and 3^Q will always be odd. so this is a contradiction and therefore x is irrational.
 
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Looks good :)
 
sweet ok , I'm new to writing proofs so just want some confirmation.
 
I can't imagine that you would lose points for this, but for the sake of pedantry you might want to point out that P and Q would have to both be positive integers. Just because 2^0=3^0 and 2^P, 3^Q aren't even and odd respectively when P and Q are negative.
 

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