Prove that a relation is a function

[Heute]

Homework Statement

Let S ={(a,(b,c)) element R x (Z x [0,1)) : a = b + c}
where R is the real numbers, Z is the integers, and [0,) is the interval from 0 to one.

Prove that S is a function (that is, for every a element of R there exist at most one (b,c) so that (a,(b,c)) is an element of S. Note, then, that b is an integer and c is non-negative)

Homework Equations

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The Attempt at a Solution

Let (a,(b,c)) and (a,(d,e)) be elements of S. If S is a function, then (b,c)=(d,e)

(a,(b,c)) is an element of S iff a = b + c; i.e. a-b = c > 0
(a,(d,e)) is an element of S iff a = d + e; i.e. a-d = e > 0
Since b and d are integers, a - b and a - d have the same digits following the decimal. Since c and e cannot be greater than or equal to one, they must account for all of the "decimal portion" of a-b or a-d respectively.

Another approach: b-d+c-e = 0
in which case I'd have to prove that b=d and c=e, but I'm not sure how.

 

[Dick]

Your first approach shows you have the right idea about what the function must be but let's try the second. If c and e are in [0,1), what interval must c-e belong to? Now b-d must be an integer. How many integers are in that interval?

 

[Heute]

c-e must be in the interval [0,1) and b-d is an integer.

Since b-d (an integer) added to a value c-e on the interval [0,1) gives zero, then c-e must = 0 (since no integer plus (0,1) gives zero). Since b-d = 0, b=d.

I think that's it. Thank you!

 

[Dick]

c-e must be in the interval [0,1) and b-d is an integer.

Since b-d (an integer) added to a value c-e on the interval [0,1) gives zero, then c-e must = 0 (since no integer plus (0,1) gives zero). Since b-d = 0, b=d.

I think that's it. Thank you!

It don't think that's quite it. If all you know is that c and e are in [0,1), I would say c-e must be in (-1,1). Can you show that?

 

[Heute]

Ah yes. I see how that works; c could be zero and e could be slightly less than one. But that doesn't change the form of the proof does it? c-e is still a decimal value or zero.

 

[Dick]

Ah yes. I see how that works; c could be zero and e could be slightly less than one. But that doesn't change the form of the proof does it? c-e is still a decimal value or zero.

Right. d-b=c-e and d-b must be an integer. The only integer in (-1,1) is 0.