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

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The discussion centers on proving that the absolute value of the difference between two real numbers, |a - b|, is equal to |b - a|. Participants reference a property from a precalculus textbook that states distance is always positive, leading to the conclusion that |a - b| = |b - a|. A specific example using a = 4 and b = 3 demonstrates that both expressions yield the same result, but some participants caution that using fixed values does not constitute a general proof. The conversation emphasizes the need for a mathematical proof using definitions, specifically the properties of square roots and squares. Overall, the consensus is that the equality holds true for all real numbers based on mathematical principles.
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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.
 
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Given that:

$$|x|\equiv\sqrt{x^2}$$

Can you prove that:

$$|a-b|=|b-a|$$ ?
 
MarkFL said:
Given that:

$$|x|\equiv\sqrt{x^2}$$

Can you prove that:

$$|a-b|=|b-a|$$ ?

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?
 
RTCNTC said:
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:

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

Can you show this must be true?
 
MarkFL said:
Generally, using fixed values doesn't constitute a good proof that will hold for all values. Using the definition I gave, we can write:

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

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.
 
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