Prove that for any integer $x$, there is an integer $k$ such as $k + 5.3 x < 90$

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.

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

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.

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

• can you explain what you did ? I'm sorry im new to these things. Oct 9, 2020 at 0:54

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 $$x < 0$$.