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How can I prove this is false?

How can I prove this is false? Topic: Case statement x#
April 19, 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.
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Best Answers: How can I prove this is false?

Cora 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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Cora 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.
Cora 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 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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Bertred Originally Answered: Christianity is False, i can prove it with its contradictions?
im an atheist and i am going to comment. i only found one strong point in there and it is C. could go into it further, like how the first real religion was zoroastrianism. and D is a point i use often, but it does not disprove anything. the problem with disproving religion is that it is not possible, because no matter what you say, all a religious person has to say it "God (or choose another deity) made it that way." what they do is work around everything you say by saying "it is meant to be like that." its very annoying. hence, the beginning of pastafarianism. Bobby Henderson eventually got fed up with it and said. "I can do it too." and did.
Bertred Originally Answered: Christianity is False, i can prove it with its contradictions?
A: if jesus died for everyones sins. there would be no hell because nomatter what everybody is pre-forgivin.--- Jesus died for the forgiveness of all sins, yes. But you have to accept that forgiveness to receive it. B: i do not remember the case name but there was a case that went to federal court regarding something that battled between state and religion . the state won by getting the religious rep. (A priest) to admit that the bible was not ment to be taken literally. Guess what? Priests sin too. Just because one priest does not support his faith, does not produce the entire religion or belief wrong. If an athiest converts to Christianity, does that prove atheism wrong? C: what happened to all the religions before christianity? AKA: Mayans, Aztecs, Egyptions......your religion just pops up half way through human existance.....--- Christianity has been around since the beginning of the world- Genesis 1:1: In the beginning, God created the heavens and the earth. I'm not sure where you found the fact that Christianity pops up in the middle of human existence, but obviously you don't know what you're talking about. D: the reasons athiesists and agnostics know more about your dumb religion than you is because they want to believe or want to find the answer, they are not nieve enough to fall for anything--- You aren't a very good athiest if you think this fact is true...the fact that you don't even know Genesis 1:1(which is the first verse in the Bible, in case you've been baffled) shows that you know absolutely nothing about the Bible. And I doubt many atheists just read the Bible trying to prove it right. Those who do can't comprehend the fact that there are things bigger, smarter, and more powerful than they are. Those who don't are trying to take something out of context to prove it false, which they have failed to do. All attempts at proving any religion or belief about religions have failed. Because whether you're an atheist or Christian, or any other religion, it is all based on faith. I see the literary evidence over hundreds of years much more believable than scientific theory based on birds or guesswork that is possible in the right conditions. Sorry, but you haven't proven anything, nor have you swayed me in the slightest. And by the way, you may not care about your grammar, but it's easier to take something seriously when you believe that the person is educated. Better luck next time!

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