Does a fraction that can be simplified as an integer belong to the set of integer numbers?Consider the question: 12 2 ∈ Z Would you consider the above statement mathematically correct? and why?I'm a bit on the side of calling it a true statement. However, since 12 2 is equivalent to 6 but not exactly identical, I am quite skeptical.Also, how about the case: 12 1 ∈ Z I checked this question Are all integers fractions?. it has some insight addressing a similar issue, yet some answers make it even more confusing for me.Thank you!

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Does a fraction that can be simplified as an integer belong to the set of integer numbers?Consider the question:      12    2    ∈      Z  Would you consider the above statement mathematically correct? and why?I&#039;m a bit on the side of calling it a true statement. However, since       12    2   is equivalent to 6 but not exactly identical, I am quite skeptical.Also, how about the case:      12    1    ∈      Z  I checked this question Are all integers fractions?. it has some insight addressing a similar issue, yet some answers make it even more confusing for me.Thank you!

Laura Fletcher · Accepted Answer

TL;DR version: Yes, if you define       Q   properly and imbed       Z   into it.Long version:That depends on definitions.To understand such cases, let&#039;s discuss the formal definition of the rational field       Q  Let   R  :=  {  (  a  ,  b  )  ∣  a  ,  b  ,  ∈      Z    ,  b  ≠  0  } be the set of all fractions. On   R, we define a relation   ∼ by saying  (  a  ,  b  )  ∼  (  c  ,  d  )  :⇔  a  d  =  b  cwhere we evaluate the last equation over the integers (as   a  ,  b  ,  c  ,  d are all integers). Now one can show that   ∼ is an equivalence relation, so we can look at the equivalence classes   R      /    ∼. On these classes, which I will denote by   [  a  ,  b  ], we can define an addition and a multiplication as  [  a  ,  b  ]  +  [  c  ,  d  ]  :=  [  a  d  +  c  b  ,  b  d  ]  [  a  ,  b  ]  ⋅  [  c  ,  d  ]  :=  [  a  c  ,  b  d  ]  .One now shows that1.Addition and multiplication are well defined (i.e. do not depend on the representative of the equivalence class we chose).2.  (  R      /    ∼  ,  +  ,  ⋅  ) forms a field, which we will call       Q  3.The map   f  :      Z    →      Q    ,  x  ↦  [  x  ,  1  ] is an injective ring homomorphism.As   f is injective, we get that       Z   is isomorphic to its image in       Q  , and thus we can identify it.Returning to your first question: As   (  12  ,  2  ) and   (  6  ,  1  ) are in the same equivalence class, we have that       12    2   lies in the image of       Z   under   f - and, if you allow to identify these, in       Z  Note that the above construction does not only work for       Z   but for many other rings. There are even similar constructions for rings that are no integral domains (in this case the map   f is often not injective).

drzwiczkih5a · Accepted Answer

12    2    =  6  ∈      Z   and       12    1    =  12  ∈      Z  . As long as you can simplify the expression to an integer number the number is   ∈      Z

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