Real Analysis/Section 1 Exercises/Answers

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General[edit]

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  6. With the view of getting a contradiction, assume \sqrt{p} is rational. Then \sqrt{p}=r/s for some integers s and r such that they have no common factor other than one, (that is, s and r are in lowest terms). Squaring both sides and rearranging terms gives s^2p=r^2. Since p is prime, r must be divisible by p, say r=pt for some integer t. By substitution, s^2p=p^2t^2, so that s^2=pt^2, and thus s must also be divisible by p, a contradiction of r and s being coprime.
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Inequalities[edit]

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    1. As a side note, this is why you do the "inequality sign flip" when you "multiply by -1"

      \begin{align}a &< b \\ a - a &< b - a \\ 0 &< b - a \\ -b &< b - a - b \\ -b &< (b - b) - a \\ -b &< 0 - a \\ -b &< -a \\ \square \end{align}

    2. Intuition solved!

      \begin{align}a &< b \\ a + c &< b + c \\ a + c &< b + c < b + d \\ a + c &< b + d \\ \square\end{align}

    3. Part 1: Let's get a new property from the given statement by applying (II)

      \begin{align}c &> d \\ -c &< -d\end{align}

      Part 2: Let's solve it now

      \begin{align}a &< b \\ a + (-c) &< b + (-c) \\ a - c &< b + (-d) \\ a - c &< b - d \\ \square\end{align}

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Absolutes[edit]

  1. The following are answers.
    1. As a side note, you should generally not use squares in your proof! However, it works here just because we're dealing with absolute values, the resulting number when you reverse the squaring operation.

      \begin{align}|a + b| &\le |a| + |b| \\ |a + b|^2 &\le (|a| + |b|)^2 \\ a^2 + 2ab + b^2 &\le |a|^2 + 2|a||b| + |b|^2 \\ a^2 + 2ab + b^2 &\le a^2 + 2|ab| + b^2 \\ 2ab &\le 2|ab| \\ ab &\le |ab| \\ \square \end{align}