Absolute Value (algebraic version)....1

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The discussion focuses on rewriting absolute value expressions without using absolute value notation. For the first question, the expression |1 - sqrt{2}| + 1 simplifies to sqrt{2}. In the second question, given that x < 3, the expression |x - 3| is rewritten as -x + 3. Participants express agreement on the solutions provided. The thread highlights the challenges of learning precalculus while successfully addressing absolute value concepts.
nycmathguy
Homework Statement
Rewrite each expression without using absolute value notation.
Relevant Equations
n/a
Absolute Value (algebraic version)
Rule:

| x | = x when x ≥ 0

| x | = -x when x > 0

Rewrite each expression without using absolute value notation.

Question 1

|1 - sqrt{2} | + 1

The value 1 - sqrt{2} = a negative value.

So, -(1 - sqrt{2}) = - 1 + sqrt{2}.

When I put it all together, I get this:

-1 + sqrt{2} + 1

Answer: sqrt{2}

You say?

Question 2

| x - 3 | given that x < 3.

If x < 3, then x - 3 is a negative value.

So, -(x - 3) becomes -x + 3.

You say?
 
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I agree also for this.
Ssnow
 
Ssnow said:
I agree also for this.
Ssnow
I got it right again. Not bad for a person that has been attacked since joining the site for trying to learn precalculus, a subject that IS WAY OVER MY HEAD.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks

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