I'm struggling to get to grips with the Triangle Inequalities. The problem is I don't really understand what it means. This is what my lecturer has written in the notes: $$ |a+b|≤|a|+|b|. $$ First of all, I don't understand why it's less than or equal to.
If I plug in numbers, for example $a=2$ and $b=5$, I struggle to see how $|2+5|≤|2|+|5|$.
If someone could explain this to me I would be very grateful.
On
$|x|$ means the absolute value of $x$. This means that if the value of $x<0$, then $|x|=-x$ otherwise $|x|=x$. In simple words, $|x|$ is formed by removing the minus sign from $x$, if any. So $|4|=4$ and $|-3|=3$
The equation states an obvious fact: $$|a+b|\leq|a|+|b|$$
If $a$ and $b$ are both positive, we have $|a+b|=|a|+|b|$, else we have $|a+b|<|a|+|b|$
Try putting these values, and you should be able to figuring it out: $(2,5)$, $(2,-5)$, $(-2,-5)$
What this has to do with a triangle is unclear to me, though.
On
I guess you are talking about the triangle equality for real numbers, and there is a hint:
When $ab\ge0$, show that $|a+b|=|a|+|b|$.
When $a<0<b$, show that $|a+b|<|b|<|a|+|b|$.
On
If you want to convince yourself that the statement is true, then you should break the statement part into cases: both $a$ and $b$ are positive, both $a$ and $b$ are negative, one is positive and the other is negative, etc.
If you want to know why it is called the triangle inequality, pick any $x$. Let $y = x - a$ and $z = y - b$. Then we have that $a = x - y$ and $b = y - z$. Thus, the inequality becomes: $$ |x - z| \le |x - y| + |y - z| $$
Now think of $x$, $y$, and $z$ as vertices of a triangle, and the absolute value as a distance function. Then the meaning is clear: the distance between $x$ and $z$ is shorter than the distance between $x$ and $z$ through the intermediate point $y$.
On
OK. I will try to explain why $$ |a+b|\leq |a| + |b| $$ is true in a somewhat intuitive/pictorial way.
First off, do you know what $|x|$ means for some number $x$? You probably know that $|x|$ means the absolute value of $x$, but what does that even mean? The absolute value of a number $x$ may be thought of as its "absolute" distance from zero on the real number line. For example, how far away is $3$ from $0$? This is clearly $3$ units away. What about $-4$? Well, this is just $4$ units away from $0$ but in a different direction (remember, the only thing that matters is absolute distance and distances are always positive). Thus, we get the following definition for the absolute value of a number $x$: $$ |x|= \begin{cases} x &\text{if} & x\geq 0,\\ -x &\text{if} & x<0. \end{cases} $$ This guarantees us that the distance from $x$ to $0$ will always be positive, regardless of the value of $x$.
Phew. Okay. Now consider what $|a+b|\leq|a|+|b|$ means. What if $a$ is positive and $b$ is negative? What number will be larger: $|a+b|$ or $|a|+|b|$? Can you see how this might confirm the truth of the triangle inequality?
If you need more explanation or want a general proof (although I think a general proof may confuse you more than help you at the moment), then let me know and I'll add it.
On
It's all about signed values.
When $a$ and $b$ have the same sign, you always get the strict equality. When the sign differs, you get the inequality.
The given inequation summarizes these two situations.
On
Suppose you supply two arbitrary real numbers $a$ and $b$. Let's construct a "triangle" (on the number line) whose vertices are $0$, $a$, and $-b$. (We're free to choose vertices as we like; pick $-b$ instead of $b$ for reasons to be explained in a moment.)
Since the distance between real number $x$ and $y$ is $|x - y|$, the sides of our "triangle" have lengths $$ |a - 0| = |a|,\qquad |-b - 0| = |-b| = |b|,\qquad |a - (-b)| = |a + b|. $$ (Now you know why we picked $-b$: To get $a + b$ inside the absolute value.:)
Two sides of our triangle have length $|a|$ and $|b|$. It's reasonable to ask: Based on this information alone, how long can the third side be?
It should be fairly clear geometrically that:
The third side cannot exceed $|a| + |b|$, the sum of the lengths of the known sides. In symbols, $$ |a + b| \leq |a| + |b|. $$ This is the triangle inequality.
The third side cannot be shorter than the distance between the real numbers $|a|$ and $|b|$. In symbols $$ |a + b| \geq \bigl||a| - |b|\bigr|. $$ This is the reverse triangle inequality.
In your particular example, if two sides of a triangle have length $2$ and $5$, then the third side cannot be shorter than $|2 - 5| = |-3| = 3$, and cannot be longer than $|2 + 5| = |7| = 7$. (On the number line, the sides of a "triangle" are parallel, so the third side is actually equal either to $3$ or to $7$. However, similar-looking inequalities hold for distance in the plane, or in space, or in higher-dimensional spaces. In these spaces, the sides of a triangle need not be parallel. In the plane, the third side of your triangle could have any length between $3$ and $7$.)
One nice proof of the triangle inequalities (for real numbers) is to show that if $x$ and $c$ are real numbers, then $|x| \leq c$ if and only if $-c \leq x \leq c$. (That is, an upper bound on the absolute value of $x$ can be "traded" for a symmetric pair of upper and lower bounds on $x$.)
Clearly, \begin{align*} -|a| &\leq a \leq |a| &&\text{for all real $a$,} \\ -|b| &\leq b \leq |b| &&\text{for all real $b$.} \\ \text{ Adding,}\quad -\bigl(|a| + |b|\bigr) &\leq a + b \leq |a| + |b|. && \end{align*} The third line has the form $-c \leq x \leq c$ for $x = a + b$ and $c = |a| + |b|$, and so can be "traded" for $|x| \leq c$, namely, for $$ |a + b| \leq |a| + |b|. $$ Note, in addition, that $$ |a - b| = |a + (-b)| \leq |a| + |-b| = |a| + |b|. $$
To prove the reverse triangle inequality, we apply the preceding reasoning to the equations \begin{align*} b &= (a + b) - a & \text{obtaining}\quad |b| &\leq |a + b| + |a|, \\ a &= (a + b) - b & \text{obtaining}\quad |a| &\leq |a + b| + |b|. \end{align*} Rearranging the inequalities on the right, we have $$ -|a + b| \leq |a| - |b| \leq |a + b|. $$ This chain of inequalities is also of the form $-c \leq x \leq c$, this time with $x = |a| - |b|$ and $c = |a + b|$, and so can be "traded" for $$ \bigl||a| - |b|\bigr| \leq |a + b|. $$
$a\le b$ means $a$ is less than or equal to $b$, so we have $3\le 3$ and $3\le 4$ for example, since $3$ is equal to $3$ and $3$ is less than $4$.
For example, $|2+5|=7=|2|+|5|$, so $|2+5|\le |2|+|5|$. And $|-1+5|=4<6=|-1|+|5|$, so $|-1+5|\le |-1|+|5|$.