Mathematics, 07.10.2020 17:01 lyndamahe0

Determine whether the absolute value of a difference is sometimes, always, or never the same as the difference of the absolute values. Ia-bI = IaI - IbI

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Answer from: quinten3700

|a-b| = |a| - |b| if and only if $a \ge 0\,\land\,b\ge 0$ . That is, it is sometimes the same.

Absolute Value is a mathematic Form which represents either the Magnitude or the Distance respect to Origin of a given Real Number, whose is depicted below:

$|x| = \left \{ {{x,\,x\ge 0} \atop {-x,\,x< 0}} \right.$ , for all $x\in \mathbb{R}$ (1)

In this question we must check if |a|-|b| = |a - b| . Hence, we must check it for the following four cases:

(i) $a \ge 0\,\land\,b\ge 0$

(ii) $a\ge 0\,\land \,b < 0$

(iii) $a < 0\,\land\,b$

(iv) $a< 0\,\land b\,\ge 0$

Case I

1) $|a| \ge 0$ , $|b| \ge 0$ , $|a-b| \ge 0$ Definition of absolute value.

2) $a \ge 0\,\land\,b\ge 0$ Given.

3) a - b = a - b Definition of absolute value/Reflexive Property

4) |a-b| = |a| - |b| Result.

Case II

1) $|a| \ge 0$ , $|b| \ge 0$ , $|a-b| \ge 0$ Definition of absolute value.

2) $a\ge 0\,\land \,b < 0$ Given.

3) $a - b =^{?} a - (-b)$ Definition of absolute value.

4) $a - b =^{?} a + b$ -(-x) = x

5) $-b =^{?} b$ Compatibility with addition/Modulative property/Existence of additive inverse/Absurd found.

6) $|a-b| \ne |a| - |b|$ Reductio ad absurdum.

Case III

1) $|a| \ge 0$ , $|b| \ge 0$ , $|a-b| \ge 0$ Definition of absolute value.

2) $a < 0\,\land\,b$ Given.

3) $a-b =^{?} -a-(-b)$ Definition of absolute value.

4) $a-b =^{?} -(a -b)$ $(-1)\cdot x = -x$ /Distributive property/Absurd found.

5) $|a-b| \ne |a| - |b|$ Reductio ad absurdum.

Case IV

1) $|a| \ge 0$ , $|b| \ge 0$ , $|a-b| \ge 0$ Definition of absolute value.

2) $a< 0\,\land b\,\ge 0$ GIven.

3) $a-b =^{?} -a - b$ Definition of absolute value.

4) $a =^{?} -a$ Compatibility with addition/Modulative property/Existence of additive inverse/Absurd found.

5) $|a-b| \ne |a| - |b|$ Reductio ad absurdum.

In a nutshell, |a-b| = |a| - |b| if and only if $a \ge 0\,\land\,b\ge 0$ . That is, it is sometimes the same.

Please see this question related to Absolute Values: link

Answer from: goly9453

Sometimes True

Explanation:

Let's see what happens when a = b

Let's pick something like a = 2, so b = 2.

|a-b| = |2-2| = |0| = 0

|a|-|b| = |2|-|2| = 2-2 = 0

Both result in 0.

We see that |a-b| = |a|-|b| is true for this example.

The given equation is either "sometimes true" or "always true"

Now let's pick something where a < b. Let a = 1 and b = 2

|a-b| = |1-2| = |-1| = 1

|a|-|b| = |1|-|2| = 1-2 = -1

We get different results, so |a-b| = |a|-|b| is not true for this example.

So this leads us to say the original equation is sometimes true.

Answer from: Quest

Answer for 2: his profit would be $2.88 explanation: you have to first multiply 72 and 0.12 together to get $8.64 once you do that you know how much money the shopkeeper made from selling the candy bars individually. so you have to do 8.64-5.76=2.88 hope that ; )answer for 3: yeah i’m sorry i have no idea what number 3 is. i’m sorry i couldn’t with that.
I'm going to add u to plsss answer questions 2 n 3

I'm going to add u to plsss answer questions 2 n 3

Answer from: Quest

1e+27

step-by-step explanation:

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Determine whether the absolute value of a difference is sometimes, always, or never the same as the...