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prove that for every rational number z and every irrational number x, there exists a unique irrational number y such that x+y=z I have seen this solved before however I am not looking to solve this using gcd. The unique portion is messing me up. I'm not sure if I should set this up as a if p then q. im confused on how to set it up. I understand a rational sum is possible if both are rational. Any suggestions would be great as to how to go about the problem. Thank you! I have something like Let X not in Q and z in Q such that X+y=z. Then I get stuck because I don't know if I should verfy the existence of just one y but first picking two different y values.

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  • $\begingroup$ Think about it this way: What would $y$ have to be? What value do we KNOW that $y$ is? $\endgroup$ – Carl Schildkraut Sep 29 '16 at 1:44
  • $\begingroup$ Why not split this into two parts. Start with for every number z and every number x, there exists a unique number y such that x+y=z. Then given z rational and x irrational, the unique y found at the previous step must be ... $\endgroup$ – dxiv Sep 29 '16 at 1:48
  • $\begingroup$ Well y we know is a unique irrational or z-x $\endgroup$ – Sam Sep 29 '16 at 2:28
  • $\begingroup$ I don't get how splitting it up would help. I would just end up the same as the start right? Because I would go back to assuming z rational and X irrational $\endgroup$ – Sam Sep 29 '16 at 2:30
  • $\begingroup$ You know that $y=z-x$ is unique, so what remains to be shown is that it is irrational. By contradiction, if it were rational and given that $z$ is rational, then $x=z-y$ would be... $\endgroup$ – dxiv Sep 29 '16 at 2:44
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prove that for every rational number z

Let $z \in \mathbb{Q}$ be an arbitrary rational number.

and every irrational number x,

Let $x \in \mathbb{R} \setminus \mathbb{Q}$ be an arbitrary irrational number.

there exists a unique irrational number y such that x+y=z

Let $y \in \mathbb{R}$ be the unique real number defined by $y = z - x$.

Then $y$ (like all reals) is either rational or irrational.

Suppose $y$ were rational. Then $x = z - y$ would be the difference of two rational numbers, thus itself a rational number. But that contradicts the premise that $x$ is irrational, so the assumption that $y$ was rational is proven false. Therefore $y$ is irrational.

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  • $\begingroup$ Oh okay I get it now! Could I split the y into two cases of y is rational and y is irrational. $\endgroup$ – Sam Sep 29 '16 at 3:57
  • $\begingroup$ No, because y cannot be rational! $\endgroup$ – user247327 Jan 29 at 19:10

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