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My question is how do you prove that given an integer $x$ and a number $y$, the only way for $x + y$ to be an integer is if $y$ is also an integer.

I can see how to prove by induction that an integer + an integer will always be an integer, but I'm interested in how to prove that if $y$ isn't an integer (given that $x$ is an integer), then $x+y$ is never an integer, and I'm not sure how to do that.

Thanks, Jeremy

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    $\begingroup$ If you accept that adding two integers gives an integer (or subtracting since subtraction and addition are essentially the same) then it's an extremely easy proof by contradiction: $a + b = x$, $x$ and $a$ are integers therefore $x - a$ is an integer, therefore $b = x - a$ is an integer which contradicts our assumption that $b$ is not an integer therefore $x$ could not possibly be an integer. $\endgroup$ – Jared Apr 15 '14 at 23:28
  • $\begingroup$ "... or subtracting since subtraction and addition are essentially the same" Your intuition here is equivalent to how most axiomatic schema for the integers work: with an axiom for the existence of an additive inverse for every integer. Then, the closure of $\mathbb{Z}$ under addition can be used to add the inverse of an integer, to "subtract" it. This may seem too formal, but it makes it so that we can describe the integers with just two binary operations: addition and multiplication. $\endgroup$ – Zubin Mukerjee Apr 15 '14 at 23:33
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    $\begingroup$ @ZubinMukerjee I understand your more formal proof, but I think it would be difficult for most to grasp your proof. Most (rational) people will accept a priori, that if you add or subtract two integers that this will give some other integer. Now I fully understand that this needs to be proved (or assumed and stated) and thus I appreciate your proof, but it seems to me that a less formal proof that normal people can understand would help...then explain the assumptions made in the informal proof and justify them (formally). $\endgroup$ – Jared Apr 16 '14 at 0:11
  • $\begingroup$ The bottom line is that the proof is simple: assume $x + y$ is an integer and $x$ is an integer, then show that this requires $y$ to be an integer therefore if $y$ is not an integer, then $x + y$ could not possibly be an integer. This is the proof regardless of how you go about proving each step. $\endgroup$ – Jared Apr 16 '14 at 0:17
  • $\begingroup$ WOW is that a question! $\endgroup$ – the hunter Apr 16 '14 at 0:37
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Suppose that $x + y$ is an integer, but that $y$ is not an integer, while $x$ is.

Then $y = \left( {x + y} \right) + \left( { - x} \right)$, and since the sum of integers is an integer (you've shown that a sum of integers is an integer, and $x$ is an integer implies $ - x$ is an integer), and ${x + y}$ and $x$ are both integers, we conclude that $y$ is also an integer, a contradiction with our assumption.

So, $x + y$ is not an integer.

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Questions like these are fundamental and intuitive enough that we need a set of axioms which we can assume.


Suppose, for contradiction, that there exists an integer $x \in \mathbb{Z}$ and a $y \not\in \mathbb{Z}$ such that $x+y \in \mathbb{Z}$.

Then, since integers have additive inverses, $\exists -x \in \mathbb{Z}$ such that $x+(-x)=(-x)+x=0$. Since the set of integers, $\mathbb{Z}$, is closed under addition, we may add this element $-x$ to $x+y$, to get another integer $$(-x) + (x+y)$$

By associativity, this can be rewritten $$((-x)+x)+y$$

But we know $(-x)+x=0$, so

$$0+y=y$$

must be an integer. This contradicts our assumption that $y \not\in \mathbb{Z}$.

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It in part depends on what proof methods are available to you -- what axioms and inference rules. However, this might be a start:

Assume x + y is an integer and that x is also an integer Let z = x + y. Since x is an integer, -x should also be an integer. (Depends on axioms you have available).
You state that you can prove that if x and y are both integers you can prove that the sum is an integer. By that same logic, you can prove that (-x) + z is an integer (since both -x and z are integers).
Therefore, (-x) + (x + y) is an integer.
This implies that ((-x) + x) + y) is an integer. (Associative property of addition)
This implies that 0 + y is an integer (Definition of additive inverse)
This implies that y is an integer (Definition of additive identity)
q.e.d.

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Your proof of $x \in \mathbb{Z} \land y \in \mathbb{Z} \Rightarrow x + y \in \mathbb{Z}$ makes it pretty easy. You just need to use the contrapositive, $x + y \notin \mathbb{Z} \Rightarrow \left( x \notin \mathbb{Z} \lor y \notin \mathbb{Z} \right)$

You're given $a \notin \mathbb{Z}$, and $b \in \mathbb{Z}$. Let $c = a + b$.

$c + (-b) = a + b - b = a \notin \mathbb{Z}$. Using the contrapositive, $c + (-b) \notin \mathbb{Z} \Rightarrow \left(c \notin \mathbb{Z} \lor (-b) \notin \mathbb{Z}\right)$. Of course, $-b \in \mathbb{Z}$. So it must be that $c \notin \mathbb{Z}$.

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