MTH 181
Solution to Problem 9 of Section 1.1
V1
Fall 2015
Problem 9: For integers x, y, and z, consider the open sentence
! R(x, y, z) : ( 2x − 1) ( z − 2 ) + ( 3y + 1) ( 2z − 1) + ( 3x − 2 ) ( 2y + 1) ≤ 8
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For what values of x, y, and z is ! R(x, y, z) a true statement?
Solution: First observe that since x, y, and z are integers, only the factor ! ( z − 2 ) of the first term can be
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zero and this happens when ! z = 2 , and the squared factors of the other terms must be at least 1. This is
because all the squared factors of the three terms except ! ( z − 2 ) are zero only when x, y, and z are not
2
integers. Hence,
! 1 ≤ ( 2x − 1) , 0 ≤ ( z − 2 ) ; ! 1 ≤ ( 3y + 1) , 1 ≤ ( 2z − 1) ; ! 1 ≤ ( 3x − 2 ) , 1 ≤ ( 2y + 1)
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and all three terms on the left side of the inequality are 4 or smaller because they squares of integers.
Claim 1: ! 0 ≤ x ≤ 1 ; hence,! x = 0 or 1 because x is an integer.
Suppose ! x ≤ −1 , then ! 3x − 2 ≤ −5 ⇒ ! ( 3x − 2 ) ≥ 25 ⇒ ( 3x − 2 ) ( 2y + 1) ≥ 25 which is too large.
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Similarly, if ! x ≥ 2 then ! 3x − 2 ≥ 4 ⇒ ! ( 3x − 2 ) ≥ 16 ⇒ ( 3x − 2 ) ( 2y + 1) ≥ 16 which is too large.
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Claim 2: ! −1 ≤ y ≤ 0 ; hence, ! y = −1 or 0 because y is an integer.
Suppose ! y ≤ −2 , then ! 3y + 1 ≤ −5 ⇒ ! ( 3y + 1) ≥ 25 ⇒ ( 3y + 1) ( 2z − 1) ≥ 25 which is too large.
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Similarly, if ! y ≥ 1 , then ! 3y + 1 ≥ 4 ⇒ ! ( 3y + 1) ≥ 16 ⇒ ( 3y + 1) ( 2z − 1) ≥ 16 which is too large.
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Claim 3: ! 0 ≤ z ≤ 1 ; hence, ! z = 0 or 1 because z is an integer.
Suppose ! z ≤ −1 , then ! 2z − 1 ≤ −3 ⇒ ! ( 2z − 1) ≥ 9 ⇒ ( 3y + 1) ( 2z − 1) ≥ 9 which is too large.
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Similarly, if ! z ≥ 2 , then ! 2z − 1 ≥ 3 ⇒ ! ( 2z − 1) ≥ 9 ! ⇒ ( 3y + 1) ( 2z − 1) ≥ 9 which is too large.
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Therefore, since ! x = 0 or 1, y = −1 or 0, and z = 0 or 1, we have exactly eight points to check:
! (0, 0, 0), (1, 0, 0), (0, − 1, 0), (1, − 1, 0), (0, 0, 1), (1, 0, 1), (0, − 1, 1), (1, − 1, 1)
Check:
! R(0, 0, 0) : ( 0 − 1) ( 0 − 2 ) + ( 0 + 1) ( 0 − 1) + ( 0 − 2 ) ( 0 + 1) = 9 ≤ 8
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! R(1, 0, 0) : ( 2 − 1) ( 0 − 2 ) + ( 0 + 1) ( 0 − 1) + ( 3 − 2 ) ( 0 + 1) = 6 ≤ 8
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! R(0, − 1, 0) : ( 0 − 1) ( 0 − 2 ) + ( −3 + 1) ( 0 − 1) + ( 0 − 2 ) ( −2 + 1) = 12 ≤ 8
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! R(1, − 1, 0) : ( 2 − 1) ( 0 − 2 ) + ( −3 + 1) ( 0 − 1) + ( 3 − 2 ) ( −2 + 1) = 9 ≤ 8
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! R(0, 0, 1) : ( 0 − 1) (1− 2 ) + ( 0 + 1) ( 2 − 1) + ( 0 − 2 ) ( 0 + 1) = 6 ≤ 8
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! R(1, 0, 1) : ( 2 − 1) (1− 2 ) + ( 0 + 1) ( 2 − 1) + ( 3 − 2 ) ( 0 + 1) = 3 ≤ 8
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! R(0, − 1, 1) : ( 0 − 1) (1− 2 ) + ( −3 + 1) ( 2 − 1) + ( 0 − 2 ) ( −2 + 1) = 9 ≤ 8
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! R(1, − 1, 1) : ( 2 − 1) (1− 2 ) + ( −3 + 1) ( 2 − 1) + ( 3 − 2 ) ( −2 + 1) = 6 ≤ 8
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So finally, the only solutions of the inequality are the four points
! P1 = (1, 0, 0), P2 = (0, 0, 1), P3 = (1, 0, 1), P4 = (1, − 1, 1) . ! !