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I have this question in my homework, and I think that either I don't understand or it's too obvious... I found $k < 90 -5.3x$ but I am not sure if I understood the question. Can anyone confirm? I am taking a beginner class in mathematical proofs.

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  • $\begingroup$ Take $x=10$ for some $k\in Z$ $\endgroup$
    – Lion Heart
    Oct 9, 2020 at 0:23
  • $\begingroup$ I am not supposed to prove something by giving a single example $\endgroup$
    – Rewind
    Oct 9, 2020 at 0:29
  • $\begingroup$ I mean you can get a solution by taking, x is a multiple of 10 $\endgroup$
    – Lion Heart
    Oct 9, 2020 at 0:35

3 Answers 3

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Take any $k\leq\varphi(90-5.3x)$ and the inequality holds.Here $\varphi(x)=[x]$ when x is not an integer, $\phi(x)=x-1$ when x is an integer.

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If $x\ge0$, you can take $k=-6x$. If $x<0$, you can take $k=42$.

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  • $\begingroup$ can you explain what you did ? I'm sorry im new to these things. $\endgroup$
    – Rewind
    Oct 9, 2020 at 0:54
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You found $k < 90 - 5.3 x$, which is the range for the possible values of $k$. The question asks you to show that some integer $k$ satisfies the inequality. Hence we now need to find an integer strictly less than $90-5.3x$.

If you are familiar with the floor function/least integer function, taking $k = \lfloor-5.3 x\rfloor$ works, since $\lfloor -5.3 x \rfloor \le -5.3 x < 90-5.3x$, and $\lfloor-5.3x\rfloor$ is an integer.

If you are not, it would be useful to split this into cases:

If $x \ge 0$, we find that $-6x < 90-6x \le 90-5.3x$, and since $x$ is an integer, $-6x$ is an integer as well.

If $x < 0$, we have $0 < -6x$ and thus $90 < 90-6x$ and $90$ is an integer.

In this case, this is a proof by cases, and both cases $x \ge 0$ and $x < 0$ needs to be included in the proof; we take $k = -6x$ for $x \ge 0$ and $k = 90$ for $x < 0$.

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