Explain why rational exponents are not defined when the denominator of the exponent in lowest terms is even and the base is negative
assume x > 0 then (-x)^(1/2) = sqrt(-x) and we cant take sqaure roots of negative nnumbers.............yet
so (-3)^(1/4) is \[\sqrt[4]{-3}\] which is not ok but \[\sqrt[3]{-3}\] is fine
such square roots are possible in complex number system.....if u'll probably not know about it if u r learning elementary number theory......but as for real number system such square root s are not possible..
make sense?
still confused sorry
what is the square root of -2?
doesnt exist
what is the 3rd root of -8?
2
-2 right?
-2
so you cant take the squared but you can take the qubed root of a negative number right?
right
in other words (-1)^(1/2) does not have an answer but (-1)^(1/3) does
yes
extend that and (-1)^(1/(even number)) has no answer but (-1)^(1/(odd number)) does
when I say extend that I just mean extend the idea
(-1)^(1/2) = not ok (-1)^(1/3) = ok (-1)^(1/4) = not ok (-1)^(1/5) = ok (-1)^(1/6) = not ok (-1)^(1/7) = ok
ok how could you explain this in words?
we cant take the nth root of a number unless n is odd
err we cant take the nth root of a negative number unless n is odd
because why?
because nth roots of a negative number when n is even are undefined such as sqrt(-2)
because nothing times itself is negative
a*a is always positive when a is a real number
and thats the explanation?
we cant take the nth root of a negative number unless n is odd because nth roots of a negative number when n is even are undefined such as sqrt(-2)
understand it and say it however you want.
your teacher is not looking for a proof, just wants you to understand the language and the know how.
ok this was very helpful thank you!
x^(1/2) the 1/2 part is the rational exponent so the denominator is the number we are concerned with.
aight man i got to sleep goodluck
thank you!!!!
for sure
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