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First: My proof of the triangle inequality: If $a,b \in \mathbb{R}$, then $|a+b| \leq |a| + |b|$

Proof: Consider the 4 cases:

1) $a<0$ and $b<0$ 2) $a>0$ and $b<0$ 3) $a>0$ and $b>0$ 4) $a<0$ and $b>0$

$$1. |-a - b| \leq |-a| + |-b| = |a| + |b|$$ $$2. |a - b| \leq |a| + |-b| = |a| + |b|$$ $$3. |a + b| \leq |a| + |b| = |a| + |b|$$ $$4. |b - a| \leq |b| + |-a| = |a| + |b|$$

Hence $|a+b| \leq |a| + |b|$

I believe that is a sufficient proof. My lecturer did a different proof, which didn't really seem to make any sense to me. This is the proof exactly as was written:

Proof: $|a+b|^2 = (a+b)^2 = a^2 + b^2 + 2ab \leq |a|^2 + |b|^2 + 2|a||b| = (|a|+|b|)^2$ From one of our theorems of ordered fields, we know $|a+b| \leq |a| + |b|$

That is where he accepted that it was proven, however it seems to me that he used the theorem to prove the theorem. Is there something that I missed?

Display Name
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  • Note he had the exact same thing written at the top as I did: Theorem : If $a,b \in \mathbb{R}$, then $|a+b| \leq |a| + |b|$ – Display Name Mar 18 '14 at 04:24
  • The last sentence is not an assumption; it is the conclusion. The last step of the proof comes from the fact that $x^2\le y^2$ implies $x\le y$ when $x,y\ge 0$. – user11977 Mar 18 '14 at 04:28
  • @user11977 That makes sense. He is not native to English, I assume that is why it was phrased as such. Thank you. – Display Name Mar 18 '14 at 04:30
  • Note also that in what you did, the extreme right expressions should be $|a|+|b|$. Otherwise equality will not hold, e.g. in case $1$. – Macavity Mar 18 '14 at 04:42
  • @Macavity Yes this is also true. Good point – Display Name Mar 18 '14 at 05:45

1 Answers1

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The proof which you have completed should have utilised $\leq$ rather than $<$ to justify the inequality.

Your instructor was using the fact that $\forall\ a,b$, $|ab|\leq |a||b|$ and the remainder of the proof is just manipulation of different inequalities. He is not using the statement of the theorem at any point.

user2345215
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wanderer
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