Science:Math Exam Resources/Courses/MATH220/December 2010/Question 03 (b)

MATH220 December 2010

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Question 03 (b)
Let $\mathbb {I}$ be the set of irrational numbers. That is $\mathbb {I} =\mathbb {R} -\mathbb {Q}$. Prove that if $x\in \mathbb {I}$ then $2x\in \mathbb {I}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Try a proof by contradiction.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Suppose that $\displaystyle x\in \mathbb {I}$ and assume towards a contradiction that $\displaystyle 2x\notin \mathbb {I}$. This means that $\displaystyle 2x\in \mathbb {Q}$ and so we find integers a and b with b nonzero such that $\displaystyle 2x={\frac {a}{b}}$. Dividing by 2 shows us that $\displaystyle x={\frac {a}{2b}}\in \mathbb {Q}$ and this contradicts the fact that $\displaystyle x\in \mathbb {I} =\mathbb {R} -\mathbb {Q}$. Hence $\displaystyle 2x\in \mathbb {I}$.

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