High School Does Absolute Value Affect Fraction Equality?

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The discussion confirms that the expression |(x + 1)/(x + 2)| is equal to |x + 1|/|x + 2| for any x not equal to -2. Participants tested various values for x, such as 10 and -10, to demonstrate this equality. The reasoning is based on the properties of absolute values, specifically that |xy| = |x||y|. The conclusion is that the two expressions are identically equal, reinforcing the concept of absolute values in fractions. This understanding is crucial for evaluating expressions involving absolute values in mathematics.
askor
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Is it correct that ##\frac{|x + 1|}{|x + 2|}## equal to ##\left|\frac{x + 1}{x + 2} \right|##?

Please explain, I don't understand.

Thank you
 
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Pick an x but not x=-2 and test it:

I pick x = 10

|x+1| / |x+2| = |11| / |12| = 11 / 12 = | 11/12 | = | (x+1) / (x+2) |

then pick x=-10

| x+1| / |x+2| = |-9| / |-8| = 9 / 8 = | 9/8 | = | -9/-8 | = | (x+1) / (x+2) |

Try other values for x and then decide if it is true or not.
 
askor said:
Is it correct that ##\frac{|x + 1|}{|x + 2|}## equal to ##\left|\frac{x + 1}{x + 2} \right|##?
Yes, the two expressions are identically equal.
askor said:
Please explain, I don't understand.
Think about the expressions x + 1 and x + 2. Each of them is negative, zero, or positive, depending on the value of x. Now, as long as ##x \ne -2##, ##\frac{x +1}{x + 2}## will have some value. Does it matter whether we take the absolute values of the numerator and denominator separately, or evaluate the fraction and then take its absolute value?
 
askor said:
Is it correct that ##\frac{|x + 1|}{|x + 2|}## equal to ##\left|\frac{x + 1}{x + 2} \right|##?
Absolutely!

It's only ##|xy| = |x||y|## in disguise.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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