I have been given a statement that I need to prove using the contradiction method and I am just a little unsure of how to go about setting this up and executing. Here is the statement:
If x is any nonzero rational number and y is any irrational number, then $\frac{2y}{x} $ is irrational.

  • $\begingroup$ The statement is true. Do you mean that you need to prove it? $\endgroup$ – ajotatxe Jun 10 '16 at 17:26
  • $\begingroup$ ah yes, edited the question $\endgroup$ – sanch Jun 10 '16 at 17:29

Hint: product and quotient of rationals are rational. So $x/2$ is a rational and if ...


Suppose this was not true. Then, we can express $\frac{2y}x$ in the form $\frac pq$, where $p,q\in\mathbb Z,q\ne0$. Moreover, as $x$ is rational, and x is non-zero, we can express $x$ as $\frac ab$ where $a,b\in\mathbb Z$ and $a\ne0,b\ne0$.

Thus, $$\frac{2y}{x}=\frac{2yb}{a}=\frac{p}{q}$$ $$y=\frac{ap}{2bq}$$ Doing so, we have represented $y$ as a ratio of two integers, where the second integer must be non-zero. Thus, $y$ must be rational, which is a contradiction. Hence, we can conclude that $\frac{2y}{x}$ must be irrational.


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