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Let x ∈ R, n ∈ N then there exist a unique y ∈ R s.t. yn = x if
- a)n is even
- b)n is even and x > 0
- c)n is odd
- d)None
Correct answer is option 'C'. Can you explain this answer?
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Let x ∈ R, n ∈ N then there exist aunique y ∈R s.t. ...
Explanation:
To prove the given statement, we need to show that for every real number x and every positive integer n, there exists a unique real number y such that yn = x.
a) n is even:
If n is even, then we can rewrite the equation yn = x as y2k = x, where k is a positive integer. Taking the k-th root on both sides, we get y = x^(1/2k). Since x^(1/2k) is a real number for any positive integer k, we can conclude that there exists a unique real number y for any even value of n.
b) n is even and x > 0:
In this case, the argument is similar to part (a), but with the restriction that x must be greater than 0. Taking the k-th root on both sides of the equation y2k = x, we get y = x^(1/2k). Since x^(1/2k) is a positive real number for any positive integer k, we can conclude that there exists a unique real number y for any even value of n and x > 0.
c) n is odd:
If n is odd, then we can rewrite the equation yn = x as y(2k+1) = x, where k is a positive integer. Taking the (2k+1)-th root on both sides, we get y = x^(1/(2k+1)). Since x^(1/(2k+1)) is a real number for any positive integer k, we can conclude that there exists a unique real number y for any odd value of n.
d) None:
Option d) is incorrect because we have shown that there exists a unique real number y for both even and odd values of n, with or without the additional condition of x > 0.
Therefore, the correct answer is option c) - n is odd.
To prove the given statement, we need to show that for every real number x and every positive integer n, there exists a unique real number y such that yn = x.
a) n is even:
If n is even, then we can rewrite the equation yn = x as y2k = x, where k is a positive integer. Taking the k-th root on both sides, we get y = x^(1/2k). Since x^(1/2k) is a real number for any positive integer k, we can conclude that there exists a unique real number y for any even value of n.
b) n is even and x > 0:
In this case, the argument is similar to part (a), but with the restriction that x must be greater than 0. Taking the k-th root on both sides of the equation y2k = x, we get y = x^(1/2k). Since x^(1/2k) is a positive real number for any positive integer k, we can conclude that there exists a unique real number y for any even value of n and x > 0.
c) n is odd:
If n is odd, then we can rewrite the equation yn = x as y(2k+1) = x, where k is a positive integer. Taking the (2k+1)-th root on both sides, we get y = x^(1/(2k+1)). Since x^(1/(2k+1)) is a real number for any positive integer k, we can conclude that there exists a unique real number y for any odd value of n.
d) None:
Option d) is incorrect because we have shown that there exists a unique real number y for both even and odd values of n, with or without the additional condition of x > 0.
Therefore, the correct answer is option c) - n is odd.
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Let x ∈ R, n ∈ N then there exist aunique y ∈R s.t. yn = x ifa)n is evenb)n is even and x > 0c)n is oddd)NoneCorrect answer is option 'C'. Can you explain this answer?
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