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How can I prove this is false?

Topic: Case statement x#
June 21, 2019 / By Fanni
Question: How can I prove the below statement is false, using complete sentences and as little symbolism as possible: If a > b (for a and b positive real numbers not equal to 1) then (log base a of x) > (log base b of x) for all x.

Best Answers: How can I prove this is false?

Cora | 1 day ago
How rigorous do you need this to be? There's an easy solution, an interesting solution, and a rigorous solution. I know you didn't want symbols, but they can be helpful, so how bout I try both? I'll explain things in English first, and then provide the symbols. Now, for convenience, let's type "log base n of x" as log[n](x) -- EASY SOLUTION The very easy way to do this is, since the proposition is for ALL values of x, to find a particular value of x that makes it false. Say, for x = 1. Now, log( 1 ) = 0 for any real positive base not equal to 1. That is, log[a]( 1 ) = 0 = log[b]( 1 ) This shows that log[a]( x ) is not greater than log[b]( x ) for all x. -- INTERESTING SOLUTION: There are more interesting proofs, though: To show that this is false, we don't need to prove this is false for all possible values and a, b and x. We just need to show that, for some particular values of a, b and x, such that a > b, the statement is false Take a and b to both be greater than 1, that is: a > b > 1 Now, suppose log[a]( x ) > log[b]( x ) for all x Then, log[a]( x ) > log[b]( x ) for values of x > 1 Use the properties of logarithms to convert both sides of the inequality to natural logs (ln): ln( x ) / ln( a ) > ln( x ) / ln( b ) Divide both sides by ln( x ). Since x > 1, ln( x ) is positive, and 1 / ln( a ) > 1 / ln( b ) *** INEQUALITY 1 Since a and b are both greater than 1, both ln( a ) and ln( b ) are positive. So, multiplying both sides of INEQUALITY 1 by ln( a ) ln( b ), ln( a ) < ln( b ) Raise e to both sides of the equation: e^ln( a ) < e^ln( b ), so a < b, FALSE We are given that a > b, so this is false, and is enough proof to show that the proposal is not true for all values of x. -- RIGOROUS SOLUTION This is to show that there is no way to choose values for a and b to make the proposition true for all values of x. Now, we could go farther, divide it into cases. Case 1 is as above. Then there are two other cases: When a and b are both less than one, and when a is greater than 1 but b is less than 1. CASE 2: 0 < b < a < 1 The proof is similar to the above, up to INEQUALITY 1. Then, it goes: Since a and b are both less than 1, ln( a ) and ln( b ) are both negative, and ln( a ) ln( b ) is positive. Multiply INEQUALITY 1 by ln( a ) ln( b ) to get: ln( a ) < ln( b ) FALSE CASE 3: 0 < b < 1 < a This is the interesting one. Since a > 1, ln( a ) is positive, whereas b < 1, and ln( b ) is negative. Then, ln( a ) ln( b ) is negative, and we get: ln( a ) > ln( b ) a > b, TRUE and the proposal is true subject to the following constraints: x > 1, and 0 < b < 1 < a That is, it holds true wherever a and b are on opposite sides of the 1, and x > 1. -- To be even more rigorous, you could then show what happens when x = 1, and x < 1 Of course, for x = 1, as we showed at the top, log[a]( 1 ) = log[b]( 1 ) = 1, so the proposition is false. For x < 1, the system behaves opposite to the way it behaves for x > 1. That is, it's false in cases the other one is true, and vice versa. Taking it from the natural log conversion step: Since x < 1, ln( x ) is negative. Dividing both sides of the inequality by ln( x ), 1 / ln( a ) < 1 / ln( b ) *** INEQUALITY 2 Now, if a and b are both greater than or less than 1, as in Cases 1 and 2, they have the same sign, and multiplying INEQUALITY 2 by ln( a ) ln( b ) yields: ln( a ) > ln( b ) TRUE However, when b < 1 and a > 1, as in Case 3, multiplying factor is negative, and ln( a ) < ln( b ) FALSE So, for x < 1, the proposition is false for all values of a and b such that a > 1 > b; true otherwise. --- So, we have enough information to describe the entire truth table for the proposition: ................... x < 1 ... x = 1 ... x > 1 0 < b < a < 1 ....T...........F..........F 0 < b < 1 < a ....F...........F.........T 1 < b < a..........T............F.........F Therefore, even excluding the case x = 1, there is no way to choose values for a and b to make the proposition hold true for all values of x.
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We found more questions related to the topic: Case statement x#

Originally Answered: How can I prove this is false?
How rigorous do you need this to be? There's an easy solution, an interesting solution, and a rigorous solution. I know you didn't want symbols, but they can be helpful, so how bout I try both? I'll explain things in English first, and then provide the symbols. Now, for convenience, let's type "log base n of x" as log[n](x) -- EASY SOLUTION The very easy way to do this is, since the proposition is for ALL values of x, to find a particular value of x that makes it false. Say, for x = 1. Now, log( 1 ) = 0 for any real positive base not equal to 1. That is, log[a]( 1 ) = 0 = log[b]( 1 ) This shows that log[a]( x ) is not greater than log[b]( x ) for all x. -- INTERESTING SOLUTION: There are more interesting proofs, though: To show that this is false, we don't need to prove this is false for all possible values and a, b and x. We just need to show that, for some particular values of a, b and x, such that a > b, the statement is false Take a and b to both be greater than 1, that is: a > b > 1 Now, suppose log[a]( x ) > log[b]( x ) for all x Then, log[a]( x ) > log[b]( x ) for values of x > 1 Use the properties of logarithms to convert both sides of the inequality to natural logs (ln): ln( x ) / ln( a ) > ln( x ) / ln( b ) Divide both sides by ln( x ). Since x > 1, ln( x ) is positive, and 1 / ln( a ) > 1 / ln( b ) *** INEQUALITY 1 Since a and b are both greater than 1, both ln( a ) and ln( b ) are positive. So, multiplying both sides of INEQUALITY 1 by ln( a ) ln( b ), ln( a ) < ln( b ) Raise e to both sides of the equation: e^ln( a ) < e^ln( b ), so a < b, FALSE We are given that a > b, so this is false, and is enough proof to show that the proposal is not true for all values of x. -- RIGOROUS SOLUTION This is to show that there is no way to choose values for a and b to make the proposition true for all values of x. Now, we could go farther, divide it into cases. Case 1 is as above. Then there are two other cases: When a and b are both less than one, and when a is greater than 1 but b is less than 1. CASE 2: 0 < b < a < 1 The proof is similar to the above, up to INEQUALITY 1. Then, it goes: Since a and b are both less than 1, ln( a ) and ln( b ) are both negative, and ln( a ) ln( b ) is positive. Multiply INEQUALITY 1 by ln( a ) ln( b ) to get: ln( a ) < ln( b ) FALSE CASE 3: 0 < b < 1 < a This is the interesting one. Since a > 1, ln( a ) is positive, whereas b < 1, and ln( b ) is negative. Then, ln( a ) ln( b ) is negative, and we get: ln( a ) > ln( b ) a > b, TRUE and the proposal is true subject to the following constraints: x > 1, and 0 < b < 1 < a That is, it holds true wherever a and b are on opposite sides of the 1, and x > 1. -- To be even more rigorous, you could then show what happens when x = 1, and x < 1 Of course, for x = 1, as we showed at the top, log[a]( 1 ) = log[b]( 1 ) = 1, so the proposition is false. For x < 1, the system behaves opposite to the way it behaves for x > 1. That is, it's false in cases the other one is true, and vice versa. Taking it from the natural log conversion step: Since x < 1, ln( x ) is negative. Dividing both sides of the inequality by ln( x ), 1 / ln( a ) < 1 / ln( b ) *** INEQUALITY 2 Now, if a and b are both greater than or less than 1, as in Cases 1 and 2, they have the same sign, and multiplying INEQUALITY 2 by ln( a ) ln( b ) yields: ln( a ) > ln( b ) TRUE However, when b < 1 and a > 1, as in Case 3, multiplying factor is negative, and ln( a ) < ln( b ) FALSE So, for x < 1, the proposition is false for all values of a and b such that a > 1 > b; true otherwise. --- So, we have enough information to describe the entire truth table for the proposition: ................... x < 1 ... x = 1 ... x > 1 0 < b < a < 1 ....T...........F..........F 0 < b < 1 < a ....F...........F.........T 1 < b < a..........T............F.........F Therefore, even excluding the case x = 1, there is no way to choose values for a and b to make the proposition hold true for all values of x.
Originally Answered: How can I prove this is false?
Try to find a counterexample. a = 10, b = 1/10 and x = 10 gives a counterexample that is easy to evaluate.

Bertred
Try to find a counterexample. a = 10, b = 1/10 and x = 10 gives a counterexample that is easy to evaluate.
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Originally Answered: Christianity is False, i can prove it with its contradictions?
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Originally Answered: Christianity is False, i can prove it with its contradictions?