[tex]\forall q \in \textbf{Q}, \exists r \in\textbf{Q}[/tex] so that [tex]q + r\in \textbf{Z}[/tex] (Q is set of all rational numbers, and Z is set of all Integers)(adsbygoogle = window.adsbygoogle || []).push({});

Proof:

let q be an arbitrary rational number

thus, [itex]q=\frac{a}{b}[/itex] for some integers a and b, and b is not 0

let [itex]r = \frac{b-a}{b}[/itex] where [itex]b-a,b\in\textbf{Z}[/itex], b is still not 0

[tex]

q + r = \frac{a}{b} + \frac{b-a}{b}[/tex]

[tex] = \frac{a+b-a}{b}[/tex]

[tex] = \frac{b}{b}[/tex]

[tex]=1[/tex] and 1 is an integer

End of proof

I'm not sure if I was redundant with anything, or if I forgot to say anything. I think I only need to find one example since the second quantifier says there exists, which I think means I only need to show one algebraically for an arbitrary rational number. Also, I can take advantage of the fact that an integer is an integer, so I don't have to define it I guess...

I'm in a first year discreet mathematics course. If there's anything wrong with my proof, please let me know.

Thanks, appreciate it!

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# Can someone check my proof?

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