Science:Math Exam Resources/Courses/MATH220/December 2009/Question 06 (a)

MATH220 December 2009

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Question 06 (a)
Let $\mathbb {I}$ be the set of irrational numbers. That is $\mathbb {I} =\mathbb {R} -\mathbb {Q}$. Prove the following statements: (i) Let $x\in \mathbb {R}$. If $x\in \mathbb {I}$ then ${\frac {1}{x}}\in \mathbb {I}$. (ii) Let $n\in \mathbb {Z}$ and $x\in \mathbb {R}$. If $x\in \mathbb {I}$ then $n+x\in \mathbb {I}$.

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
These are straightforward proofs by contradiction. Assume the given elements are rational and then show that this implies that the element x itself must also be rational.

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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. (i) Assume towards a contradiction that $\displaystyle 1/x\in \mathbb {Q}$. This means that there exists nonzero integers a and b such that $\displaystyle {\frac {1}{x}}={\frac {a}{b}}.$ Inverting gives $\displaystyle x={\frac {b}{a}}$ which is a contradiction since x was given to be irrational. Hence $\displaystyle 1/x\in \mathbb {I}$. (ii) Assume towards a contradiction that $\displaystyle n+x\in \mathbb {Q}$. This means that there exists integers a and b with b nonzero such that $\displaystyle n+x={\frac {a}{b}}$ Isolating for x gives $\displaystyle x={\frac {a-nb}{b}}\in \mathbb {Q}$ and this is a contradiction. Hence $\displaystyle n+x\in \mathbb {I}$.

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