Showing two certain sets have no elements in common

[k3k3]

Homework Statement

Let x and y be irrational numbers such that x-y is also irrational.
Let A={x+r|r is in Q} and B={y+r|r is in Q}
Prove that the sets A and B have no elements in common.

Homework Equations

The Attempt at a Solution

Since x and y are in A and B, then x+r$_{1}$-y-r$_{2}$ is irrational.

Using contradiction, assume their sum is z and z is an element of A$\cap$B

x-y=z

Then x+r$_{1}$-y-r$_{2}$=z

Since z is in both A and B, then z can also be said to be x+r$_{1}$+y+r$_{2}$

Then x+r$_{1}$-y-r$_{2}$=x+r$_{1}$+y+r$_{2}$

Solving for the right side,

0=2(y+r$_{2}$)

But since the sum is irrational and 0 is not irrational, A and B cannot contain the same elements.


Am I thinking about this correctly?

 

[HallsofIvy]

Homework Statement

Let x and y be irrational numbers such that x-y is also irrational.
Let A={x+r|r is in Q} and B={y+r|r is in Q}
Prove that the sets A and B have no elements in common.

Homework Equations

The Attempt at a Solution

Since x and y are in A and B, then x+r$_{1}$-y-r$_{2}$ is irrational.

x and y are NOT in A and B, $x+ r_1$ and $y+ r_2$ are. And, from that, it follows that they are rational numbers so the difference is rational, not irrational.

Using contradiction, assume their sum is z and z is an element of A$\cap$B

x-y=z

Then x+r$_{1}$-y-r$_{2}$=z

Since z is in both A and B, then z can also be said to be x+r$_{1}$+y+r$_{2}$

Then x+r$_{1}$-y-r$_{2}$=x+r$_{1}$+y+r$_{2}$

Solving for the right side,

0=2(y+r$_{2}$)

But since the sum is irrational and 0 is not irrational, A and B cannot contain the same elements.


Am I thinking about this correctly?

 

[sunjin09]

Homework Statement

Let x and y be irrational numbers such that x-y is also irrational.
Let A={x+r|r is in Q} and B={y+r|r is in Q}
Prove that the sets A and B have no elements in common.

Homework Equations

The Attempt at a Solution

Since x and y are in A and B, then x+r$_{1}$-y-r$_{2}$ is irrational.

Using contradiction, assume their sum is z and z is an element of A$\cap$B

x-y=z

Then x+r$_{1}$-y-r$_{2}$=z

Since z is in both A and B, then z can also be said to be x+r$_{1}$+y+r$_{2}$

Then x+r$_{1}$-y-r$_{2}$=x+r$_{1}$+y+r$_{2}$

Solving for the right side,

0=2(y+r$_{2}$)

But since the sum is irrational and 0 is not irrational, A and B cannot contain the same elements.


Am I thinking about this correctly?

No, if z is in both A and B, then z=x+r1=y+r2; then you take the subtraction ...

 

[k3k3]

Did I show that x+r1-(y+r2)=x+r1-(y+r2)?