What is the proper way to prove that lx-2l = l2-xl?

[The Student]

By looking at it, it is intuitively obvious. But I can't seem to show the proper way to prove it by using basic properties of real numbers.

 

[daon2]

By looking at it, it is intuitively obvious. But I can't seem to show the proper way to prove it by using basic properties of real numbers.

There's a couple of "easy" ways.

1. |x-2| is "the distance between x and 2 on the real number line". Since the distance between x and 2 is the same as between 2 and x, they are the same.

2. |x-2| = |(-1)(2-x)| = |-1|*|2-x| = 1*|2-x|=|2-x|

 

[The Student]

There's a couple of "easy" ways.

1. |x-2| is "the distance between x and 2 on the real number line". Since the distance between x and 2 is the same as between 2 and x, they are the same.

2. |x-2| = |(-1)(2-x)| = |-1|*|2-x| = 1*|2-x|=|2-x|

Oh yeah! Thanks!

 

[HallsofIvy]

Yet another way (using the basic definition of "absolute value"):
|x|= x if $\displaystyle x\ge 0$, -x if x< 0.

If $\displaystyle x\ge 2$ then $\displaystyle x- 2\ge 0$ and $\displaystyle 2- x\le 0$. |x- 2|= x- 2 and |2- x|= -(2- x)= x- 2.

If $\displaystyle x< 2$ then $\displaystyle x- 2\le 0$ and $\displaystyle 2- x> 0$. |x- 2|= -(x- 2)= 2- x and |2- x|= 2- x.

 

[DrPhil]

By looking at it, it is intuitively obvious. But I can't seem to show the proper way to prove it by using basic properties of real numbers.

Yet another another way - I seen to remember this definition from my youth:

......$\displaystyle \displaystyle |A| = \sqrt{A^2}$

Then $\displaystyle \displaystyle |x-2|= \sqrt{(x - 2)^2} = \sqrt{x^2 - 2x + 4} = \sqrt{4 - 2x + x^2} = \sqrt{(2 - x)^2} = |2 - x|$

 

[HallsofIvy]

Yet another another way - I seen to remember this definition from my youth

......$\displaystyle \displaystyle |A| = \sqrt{A^2}$

Then $\displaystyle \displaystyle |x-2|= \sqrt{(x - 2)^2} = \sqrt{x^2 - 2x + 4} = \sqrt{4 - 2x + x^2} = \sqrt{(2 - x)^2} = |2 - x|$

Did they have "absolute value" and "square root" that long ago?!