But since distance is always positive, can we say that | a - b | = | b - a |?

[mathdad]

On page 9 of my David Cohen Precalculus textbook (3rd Edition), the following property is given:

The distance between a and b is | a - b | = | b - a |.

Question:

Can we say that | a - b | = | b - a | because distance is always positive?

Note: a and b are real numbers.

 

[MarkFL]

Given that:

\(\displaystyle |x|\equiv\sqrt{x^2}\)

Can you prove that:

\(\displaystyle |a-b|=|b-a|\) ?

 

[mathdad]

Given that:

\(\displaystyle |x|\equiv\sqrt{x^2}\)

Can you prove that:

\(\displaystyle |a-b|=|b-a|\) ?

If a and b are real numbers, I can plug any for a and b.

Let a = 4 and b = 3.

| 4 - 3| = | 3 - 4 |

| 1 | = | -1 |

1 = 1

True?

 

[MarkFL]

If a and b are real numbers, I can plug any for a and b.

Let a = 4 and b = 3.

| 4 - 3| = | 3 - 4 |

| 1 | = | -1 |

1 = 1

True?

Generally, using fixed values doesn't constitute a good proof that will hold for all values. Using the definition I gave, we can write:

\(\displaystyle \sqrt{(a-b)^2}=\sqrt{(b-a)^2}\)

Can you show this must be true?

 

[mathdad]

Generally, using fixed values doesn't constitute a good proof that will hold for all values. Using the definition I gave, we can write:

\(\displaystyle \sqrt{(a-b)^2}=\sqrt{(b-a)^2}\)

Can you show this must be true?

I cannot prove anything outside of fixed values at my level of math.

I know that the square root of a square yields the radicand.

I like to look at it this way:

sqrt{(cars - road)^2} = cars - road.

sqrt{(road - cars)^2} = road - cars.