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Prove that

$$|x-a| \leq \frac{|a|}{2} \Rightarrow |x| \geq \frac{|a|}{2}$$

I've tried by direct proof and contradicion but nothing worked. I would like a hint or a tip of what should I do.

Thanks in advance!

Giiovanna
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3 Answers3

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What kind of proof do you want? Doing this visually on a number line is clear and easy. Even if you do not want to present such a proof, doing this first often gives you an idea on how to do the formal proof. That's what I did to find (one of) my formal proofs for your question.

Now that I am back from church, I can add more detail... but others have already written what I was going to write. The simplest formal proof I came up with was explained by @abel in his comment to the main question: "Use $|a+b|≥||a|−|b||$ with $b=x−a$."

If you cannot use that identity, you can go back to basics. The visual number line proof involves viewing the interval around the point $a$ on the number line with radius $\frac{|a|}2$. The drawing would depend on whether $a$ is positive or negative, so you should do your proof based in these two cases.

The first step , assuming the hypothesis, would be

$$-\frac{|a|}2\le x-a \le \frac{|a|}2$$

Then use the sign of $a$ to show that $\frac a2\le x$ for $a\ge 0$ and $x\le\frac a2$ for $a\le 0$. I see that @Lorence has given the details for this approach. In my opinion, that is the best answer here (and you should have accepted that one... no offense intended to Willie Rosario).

Rory Daulton
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  • I did it but I couldnt contiue. Also, I tried to use the fact that $|x| \geq x $. If a is positive, I could do but if it is negative, I couldnt – Giiovanna Mar 08 '15 at 11:41
  • I did it, thanks. – Giiovanna Mar 08 '15 at 11:49
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    To be brutally honest, I think this solution is not helpful at all! specially for someone who is just learning the material. –  Mar 08 '15 at 13:20
  • @WillieRosario: I had to leave my house for a while before I could answer Giiovanna's response to what I first wrote. I see that others, especially Lorence, have already done a good job in completing the answer. I have now edited my answer to include a summary of what I was planning to write. I believe that my incremental approach, giving a hint and offering to give more if needed, is indeed helpful to a student of the material. Do you disagree? – Rory Daulton Mar 08 '15 at 16:37
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    well you are ofending me with your last line in parenthesis bro no matter what. –  Mar 08 '15 at 16:40
  • Is it necesary to remark the OP that he shouldve accepted the other answer? What is your point? –  Mar 08 '15 at 16:42
  • @WillieRosario: I do not understand why you would denigrate my answer in multiple comments then take offense at me. My remark about accepting Lorence's answer was directed at Giiovanna, to help her and this site in the future in making good choices--it was not about you. You have not yet answered my question about my incremental approach in my answer. – Rory Daulton Mar 08 '15 at 16:51
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    I still think your last comment on the parenthesis suggesting the OP to accept another answer is unreasonable, rude, and unnecessary. Disturbing because my solution is complete, clear and to the point. not much talking. Straight to the point. Your answer is convoluted with so much stuff hard to understand. –  Mar 08 '15 at 16:54
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$i.a\geqslant 0$

$$|x-a|\leqslant\frac a2\Longleftrightarrow\frac a2\leqslant x\leqslant \frac{3a}{2}\Longrightarrow|x|\geqslant\frac{|a|}{2}$$

$ii.a< 0$

$$|x-a|\leqslant-\frac a2\Longleftrightarrow\frac {3a}{2}\leqslant x\leqslant \frac{a}{2}\Longrightarrow|x|\geqslant\frac{|a|}{2}$$

Furthermore, easy to get $$|x|\leqslant{\frac{3|a|}{2}}$$

According to No.$i$ and No.$ii$,we can prove it easily.

Faye Tao
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Solution:

Let $y = \frac{a}{2} $. Then $a = 2y$ and $|y| = \frac{|a|}{2} $. We can show the contrapositive of the statement:

$$ |x| < |y| \implies |x-2y| > |y| $$

To show this, notice

$$ 3|y| = |3y| = |2y +y +x-x | = |(2y-x) + y + x| \leq |2y - x| + |y| + |x| < |2y-x| + 2|y| $$

Substracting $2|y|$ from both sides gives

$$ |2y - x| > |y| $$

But $|2y - x| = |x - 2y| $.

Community
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  • it is unfair and unreasonalbe for you to remark that rory's post as unhelpful for an early learner and then post a solution using contrapositive!. please take a look at lorence's post. it is much more elementary and serves the purpose better than yours. – abel Mar 08 '15 at 13:52
  • you said you were brutally honest. i just tried to be honest. – abel Mar 08 '15 at 14:00
  • My solution is elementary. Rory's solution I cannot understand what he means. Very hard to follow. I did explain every step of mine and everyone knows what a contrapositive is. Lorence's Solution is good, but Rory's solution is complete non-undersantable and cryptic disaster, sorry. –  Mar 08 '15 at 14:01
  • i would not say everyone knows how to use contrapositive; yes it is easy to understand for a mature math student. even many of the discrete math students don't know how to correctly reason contrapositive arguments. – abel Mar 08 '15 at 14:06
  • Adding to Rory's bad solutions is his lack of response to a question the OP formulates to him. I find this behavior extremely unkind. –  Mar 08 '15 at 14:07
  • ok bro, you win. – abel Mar 08 '15 at 14:11