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Absolute values are needed when n is even and mn is odd.
sqrt(a)=a^(1/n)
(a^m)^n=a^(m* n)
a* 1/b= a/b
A radical expression with an odd index is always defined and the outcome will be either positive, negative, or zero. Therefore, the value of the exponent mn is not relevant, and the expressions are equivalent in this case. sqrt(x^m) ⇔ x^(mn) forn odd Here, the absolute value is not needed. We can visualize our statement by assigning some values for m and n. For simplicity, and without losing generality, we will consider x=- 1.
n= 3, m= 6 | n= 5, m= 15 |
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sqrt((- 1)^6) ? = (- 1)^(6 3) | sqrt((- 1)^(15)) ? = (- 1)^(15 5) |
sqrt(1) ? = (- 1)^2 | sqrt(- 1) ? = (- 1)^3 |
1=1 âś“ | - 1=- 1 âś“ |
When defined, the outcome of a root with an even index is never negative. sqrt(x^m)≥0 for x^m ≥ 0 Conversely, for negative values of x, the sign of x^(mn) depends on the value of the exponent mn.
If the exponent mn is even, then x^(mn) is positive and, when the root is defined, the expressions will be equivalent. sqrt(x^m) ⇔ x^(mn) forx^m ≥ 0,and n and mn even Again, the absolute value is not needed. We can visualize our statement by assigning some values for m and n. For simplicity, and without losing generality, we will consider x=- 1.
n= 2, m= 4 | n= 6, m= 24 |
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sqrt((- 1)^4) ? = (- 1)^(4 2) | sqrt((- 1)^(24)) ? = (- 1)^(24 6) |
sqrt(1) ? = (- 1)^2 | sqrt(1) ? = (- 1)^4 |
1=1 âś“ | 1=1 âś“ |
If the exponent mn is odd, then x^(mn) will be negative if x is negative. Here, when the root is defined, the expressions will not be equivalent unless we take the absolute value of x. n even and mn odd sqrt(x^m) ⇔ |x|^(mn), forx^m≥ 0 In this case, the absolute value is needed. We can visualize our statement by assigning some values for m and n. We will see how the statement is false without absolute value, and how it is true if we take the absolute value of x. For simplicity, and without losing generality, we will consider x=- 1.
Without absolute value n= 2, m= 6 |
With absolute value n= 2, m= 6 |
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sqrt((- 1)^6) ? = (- 1)^(6 2) | sqrt((- 1)^6) ? = |- 1|^(6 2) |
sqrt(1) ? = (- 1)^3 | sqrt(1) ? = 1^3 |
1≠- 1 * | 1=1 ✓ |
We conclude that absolute values are needed when n is even and mn is odd.