Question

We prove Property 5 of absolute values, the Triangle Inequality:

$|x+y|\le \left|x\right|+\left|y\right|$

(a) Verify that the Triangle Inequality holds for $x=2,y=3$, for$x=-2,y=-3$ , and for$x=-2,y=3$ .

(b) Prove that the Triangle Inequality is true for all real numbers x and y.

Short answer

a.Verified that the Triangle Inequality holds for$x=2,y=3$, for$x=-2,y=-3$, and for $x=-2,y=3$

b.Proved that the Triangle Inequality is true for all real numbers x and y.

Step-by-step solution

. Apply the concept of absolute value

If a is a real number, then the absolute value of a is given by:

$\left|a\right|=\{\begin{array}{l}a\text{;if}a\ge 0\\ -a\text{;if}a<0\end{array}$

. Apply the concept of Triangle Inequality

$|x+y|\le \left|x\right|+\left|y\right|$

. Verify Triangle Inequality

To verify Triangle Inequality for$x=2,y=3$ , for $x=-2,y=-3$, and for$x=-2,y=3$ .

• Substitute$x=2,y=3$in LHS and RHS of Triangle Inequality:

LHS:

$\begin{array}{l}|x+y|=|2+3|\\ \Rightarrow |x+y|=\left|5\right|\\ \Rightarrow |x+y|=5\end{array}$

RHS:

$\begin{array}{l}\left|x\right|+\left|y\right|=\left|2\right|+\left|3\right|\\ \Rightarrow \left|x\right|+\left|y\right|=2+3\\ \Rightarrow \left|x\right|+\left|y\right|=5\end{array}$

So, for$x=2,y=3$ , LHS=RHS

So,$|x+y|\le \left|x\right|+\left|y\right|$for$x=2,y=3$

Therefore, Triangle Inequality is verified for$x=2,y=3$

• Substitute$x=-2,y=-3$in LHS and RHS of Triangle Inequality:

LHS:

$\begin{array}{l}|x+y|=\left|-2+(-3)\right|\\ \Rightarrow |x+y|=|-5|\\ \Rightarrow |x+y|=-(-5)\\ \Rightarrow |x+y|=5\end{array}$

RHS:

$\begin{array}{l}\left|x\right|+\left|y\right|=|-2|+|-3|\\ \Rightarrow \left|x\right|+\left|y\right|=-(-2)+-(-3)\\ \Rightarrow \left|x\right|+\left|y\right|=2+3\\ \Rightarrow \left|x\right|+\left|y\right|=5\end{array}$

So,$x=-2,y=-3$ for, LHS=RHS

So,$|x+y|\le \left|x\right|+\left|y\right|$for$x=-2,y=-3$

Therefore, Triangle Inequality is verified for$x=-2,y=-3$

• Substitute in$x=-2,y=3$ LHS and RHS of Triangle Inequality:

LHS:

$\begin{array}{l}|x+y|=|-2+3|\\ \Rightarrow |x+y|=\left|1\right|\\ \Rightarrow |x+y|=1\end{array}$

RHS:

$\begin{array}{l}\left|x\right|+\left|y\right|=|-2|+\left|3\right|\\ \Rightarrow \left|x\right|+\left|y\right|=-(-2)+3\\ \Rightarrow \left|x\right|+\left|y\right|=2+3\\ \Rightarrow \left|x\right|+\left|y\right|=5\end{array}$

So, for,$x=-2,y=3$$LHS<RHS$

So,$|x+y|\le \left|x\right|+\left|y\right|$for$x=-2,y=3$

Therefore, Triangle Inequality is verified for$x=-2,y=3$

. Apply the concept of absolute value

If a is a real number, then the absolute value of a is given by:

$\left|a\right|=\{\begin{array}{l}a\text{;if}a\ge 0\\ -a\text{;if}a<0\end{array}$

. Apply the concept of Triangle Inequality

$|x+y|\le \left|x\right|+\left|y\right|$

. Prove Triangle Inequality

We need to prove$|x+y|\le \left|x\right|+\left|y\right|$

With loss of generality, let x and y be real numbers.

Case 1:$x=0$

LHS:

$\begin{array}{l}|x+y|=|0+y|\\ \Rightarrow |x+y|=\left|y\right|\end{array}$

RHS:

$\begin{array}{l}\left|x\right|+\left|y\right|=\left|0\right|+\left|y\right|\\ \Rightarrow \left|x\right|+\left|y\right|=\left|y\right|\end{array}$

So, for$x=0$,$LHS=RHS$

So,$|x+y|\le \left|x\right|+\left|y\right|$for x=0

Case 2:$x>0,y>0$

$\begin{array}{l}x>0,y>0\\ \Rightarrow x+y>0\end{array}$

LHS:

$|x+y|=x+y$

RHS:

$\left|x\right|+\left|y\right|=x+y$

So, for,$x>0,y>0$$LHS=RHS$

So,$|x+y|\le \left|x\right|+\left|y\right|$ for $x>0,y>0$

Case 3:$x<0,y<0$

$\begin{array}{l}x<0,y<0\\ \Rightarrow x+y<0\end{array}$

LHS:

$|x+y|=-(x+y)$

RHS:

$\begin{array}{l}\left|x\right|+\left|y\right|=-x+(-y)\\ \Rightarrow \left|x\right|+\left|y\right|=-(x+y)\end{array}$

So, for,$x<0,y<0$$LHS=RHS$

So,$|x+y|\le \left|x\right|+\left|y\right|$for$x<0,y<0$

Case 4:$x>0,y<0$

RHS:

$\left|x\right|+\left|y\right|=x+(-y)$

Subcase 4a

$x+y=0$

LHS:

$|x+y|=0$

Since

$\begin{array}{l}y<0\\ \Rightarrow -y>0\end{array}$

So,

$\begin{array}{l}x+(-y)>\mathrm{0...........}[x>0,-y>0]\\ \Rightarrow \left|x\right|+\left|y\right|>|x+y|\\ \Rightarrow RHS>LHS\end{array}$

Subcase 4b

$x+y>0$

LHS:

$|x+y|=x+y$

Since

$\begin{array}{l}y<0\\ \Rightarrow -y>0\\ \Rightarrow -y>0>y\\ \Rightarrow x+(-y)>x+0>x+y\mathrm{..............}[x>0]\\ \Rightarrow \Rightarrow x+(-y)>x+y\\ \Rightarrow \left|x\right|+\left|y\right|>|x+y|\\ \Rightarrow RHS>LHS\end{array}$

Subcase 4c

$x+y<0$

LHS:

$|x+y|=-(x+y)$

Since

$\begin{array}{l}y<0\\ \Rightarrow -y>0\\ x>0\\ \Rightarrow -x<0<x\\ \Rightarrow x+(-y)>-x+(-y)\\ \Rightarrow x+(-y)>-(x+y)\\ \Rightarrow \left|x\right|+\left|y\right|>|x+y|\\ \Rightarrow RHS>LHS\end{array}$

So,$|x+y|\le \left|x\right|+\left|y\right|$for$x>0,y<0$

Since x and y were arbitrary real numbers, the Triangle Inequality is true for all real numbers x and y.