For real numbers a and b, is it true that a = b?
(1) a^4 = b^4
(2) a^5 = b^5
OA:B
Please explain
For real numbers a and b, is it true that a = b?
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First consider statement (1).
It can be true even if a and b are of opposite signs. For example:
a = 2 and b = -2 gives a^4 = b^4
a = 2 and b = 2. gives a^4 = b^4
You cannot be sure whether or not a and b have equal signs because raising to an even power will always be positive. Hence (1) is not sufficient.
Consider statement (2).
If true, then a and b must have equal signs because any real number raised to an odd number will maintain its sign. For example:
(-2)^5 = -32. So in the case that a or b are negative then raising them to an odd number (5) will maintain their sign.
(2) is sufficient and the answer is B.
Feel free to ask again if not clear.
It can be true even if a and b are of opposite signs. For example:
a = 2 and b = -2 gives a^4 = b^4
a = 2 and b = 2. gives a^4 = b^4
You cannot be sure whether or not a and b have equal signs because raising to an even power will always be positive. Hence (1) is not sufficient.
Consider statement (2).
If true, then a and b must have equal signs because any real number raised to an odd number will maintain its sign. For example:
(-2)^5 = -32. So in the case that a or b are negative then raising them to an odd number (5) will maintain their sign.
(2) is sufficient and the answer is B.
Feel free to ask again if not clear.
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Statement 1:NandishSS wrote:For real numbers a and b, is it true that a = b?
(1) a^4 = b^4
(2) a^5 = b^5
It's possible that a=1 and b=1, in which case a=b.
It's possible that a=1 and b=-1, in which case a≠b.
INSUFFICIENT.
Statement 2:
It's possible that a=0 and b=0, in which case a=b.
If ab≠0, then we can divide each side by b�:
a�/b� = b�/b�
(a/b)� = 1
a/b = 1
a = b.
Thus, in every case, a=b.
SUFFICIENT.
The correct answer is B.
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S1:
a� = b�
a� - b� = 0
(a² - b²) * (a² + b²) = 0
(a + b) * (a - b) * (a² + b²) = 0
So we could have (a + b) = 0, in which case a = -b, or we could have (a - b) = 0, in which case a = b. Not sufficient!
S2:
a� = b�
a� - b� = 0
(a - b) * (a� + a³b + a²b² + ab³ + b�) = 0
If (a - b) = 0, then a = b. If a ≠ b, then (a� + a³b + a²b² + ab³ + b�) = 0.
But if (a� + a³b + a²b² + ab³ + b�) = 0, then (a� + a²b² + b�) = -(a³b + ab³) = -ab * (a² + b²). Since the left side is positive, the right side must be positive, so -ab must be positive. But then ab is negative, and a and b have different signs, which is impossible! We know they're equal when raised to odd powers, so they're both even or both odd.
Hence we must have a = b, and this is also SUFFICIENT.
a� = b�
a� - b� = 0
(a² - b²) * (a² + b²) = 0
(a + b) * (a - b) * (a² + b²) = 0
So we could have (a + b) = 0, in which case a = -b, or we could have (a - b) = 0, in which case a = b. Not sufficient!
S2:
a� = b�
a� - b� = 0
(a - b) * (a� + a³b + a²b² + ab³ + b�) = 0
If (a - b) = 0, then a = b. If a ≠ b, then (a� + a³b + a²b² + ab³ + b�) = 0.
But if (a� + a³b + a²b² + ab³ + b�) = 0, then (a� + a²b² + b�) = -(a³b + ab³) = -ab * (a² + b²). Since the left side is positive, the right side must be positive, so -ab must be positive. But then ab is negative, and a and b have different signs, which is impossible! We know they're equal when raised to odd powers, so they're both even or both odd.
Hence we must have a = b, and this is also SUFFICIENT.
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The question is plain and simple and can be done within 15 seconds. Just know that even powers cannot be predicted where as odd powers can be predicted.
This confirms why option B is sufficient!
This confirms why option B is sufficient!