Quick question about surjective functions

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SUMMARY

The function f(a/b) = 2^a * 3^b, where (a/b) is in lowest terms, is not surjective when mapping positive integers to natural numbers. The discussion highlights that while one participant claims it is onto, they fail to demonstrate that for every natural number y, there exists a corresponding (a/b) such that f(a/b) = y. Additionally, the confusion arises from the interpretation of the function's domain, suggesting that it may be more appropriate to analyze its injectivity from rational numbers to natural numbers.

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  • Understanding of surjective functions and their definitions
  • Knowledge of rational numbers and their representation in lowest terms
  • Familiarity with exponential functions, specifically 2^a and 3^b
  • Basic concepts of injective functions and their properties
NEXT STEPS
  • Research the properties of surjective functions in detail
  • Explore examples of injective and surjective functions in mathematical literature
  • Study the representation of rational numbers and their implications in function mapping
  • Investigate the implications of mapping from rational numbers to natural numbers
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Mathematicians, educators, and students studying advanced functions, particularly those interested in the properties of surjective and injective functions in number theory.

jaejoon89
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Question Details:
f(a/b) = 2^a * 3^b where (a/b) is in lowest terms.

Show f is surjective (onto).

Note: f maps positive integers to natural numbers


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Is it sufficient to say that...

It is onto because for every natural number y there is (a/b) s.t. f(x) = y.
 
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I think you have misunderstood the question.
f(a/b) suggests that f maps from rational numbers and not positive integers. If it mapped from positive integers we would have b=1 always. However even then f isn't surjective. Can you find an a/b in lowest terms such that 2^a * 3^b = 5?

It is onto because for every natural number y there is (a/b) s.t. f(x) = y.
No. Here you just state the definition of a surjective function. You never show that y actually exists.

Perhaps you're really supposed to show it's injective as a function from the rational numbers to the natural numbers?
 

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