Fake Proofs

We have all seen fake proofs. They purposely try to trick you into believing something that is false. The proofs go from easy to hard, so start at the beginning and work your way to the hardest problem, trying to find the mistakes along the way. Good luck!

Find the mistake:

1. 1!=1

2. 0!=1

3. 0!=1!

4. Dividing by !, we see that

5. 0=1

What mistake was made? The mistake occurs between line 3 and line 4, when you assume that you are taking 0 ! and 1 !, while 1! is something that cannot be separated into 1 and !. Therefore, this proof is false.

Find the mistake:

1. 1/∞=0

2. Rotating both sides of the equation 90 degrees counter-clockwise, we see that (note that “1” rotated by 90 degrees is “-”)

3. -18=0

4. -10=8

5. Rotating both sides of the equation 90 degrees clockwise, we see that

6. 1/0=∞

What could possibly be the mistake that we made? If we remember back to math class, while we can do some things to both sides of the equation, rotation on each side does not fall into that category. Therefore, this proof is false since you cannot rotate both sides of an equation.

Find the mistake:

All positive natural numbers can be expressed in fewer than 11 words. Using a proof by contradiction (assume that this statement is false, then find a contradiction), we first assume that this is false. Therefore, there is some integer n that you cannot express in fewer than 11 words. Suppose we call this number “the smallest positive integer not definable in under 11 words”. But this name is only 10 words long, which is a contradiction since the number was written in 10 words. Therefore, the statement must be true.

What went wrong? The mistake we made was only considering the smallest number. If we give that number the name “the smallest positive integer not definable in under 11 words”, what will we give the next number that is that long? Therefore, we cannot name every number in that sense, disproving our proof.

Find the mistake:

1. 2=2

2. 2=1+1

3. 2=1+√1

4. 2=1+√(-1*-1)

5. 2=1+√-1*√-1

6. 2=1+i*i, where i^2=-1

7. 2=1-1

8. 2=0

Where does this proof go wrong? At first, it might seem hard to find a mistake in this since each step is a logical step that you can take, but take a closer look going from step 4 to step 5. The property √(mn)=√m√n only works when m and n are greater than 0. Therefore, this is the step that disproves this fake proof that 2=0. Otherwise, the other logic in this proof is correct.

Find the mistake:

1. x^2+x+1=0

2. x^2=-x-1

3. x=-1-1/x for x not equal to 0

4. Substituting our value of x into the original equation:

5. x^2+(-1-1/x)+1=0

6. x^2-1/x=0

7. x^3=1

8. x=1

9. Substituting our value of x into the original equation:

10. 1^2+1+1=0

11. 3=0

What is the mistake made? Try checking what happens from step 3 to step 5. By the fundamental theorem of algebra, which is just a fancy way to say that a polynomial with degree n (the degree of a polynomial is the largest exponent x is raised to) is equal to 0 at n values, not necessarily distinct. Therefore, the first equation has only 2 roots, while step 7 has 3 roots! Somehow, we must have added an extra root to this equation when we increased the degree by 1. This root happens to be 1, and we added it to the equation in step 5 when the degree of the equation was increased by 1. Therefore, 1 is an extraneous (just a fancy way of saying it doesn’t work) root of this equation. We found the mistake!

Find the mistake:

Lakeside is going to have a surprise fire drill next week, at a time that no one would expect. One of the students at Lakeside realizes that if there hadn’t been a drill on Thursday, everyone would know that the drill would be on Friday, meaning it would break the condition that it would be at a time that no one would expect. Since the drill can’t be on Friday, that means that the last possible day for the drill is on Thursday. But, if everyone went home on Wednesday and there was not a drill, the drill would be on Thursday, breaking the condition that it would be a surprise. Therefore, the drill can’t be on Friday or Thursday. By the same logic, the student deduced that the drill can’t be on Monday, Tuesday, Wednesday, Thursday, or Friday. Therefore, the student was sure that there would be no drill.

On Wednesday, the fire alarm went off, and the student did not expect it.

What was the mistake that the Lakeside student made? The Lakeside student tries to use cases to understand each day. The first case is on Friday, the second is on Thursday, etc. But, when he proves that the drill can’t be on Monday, he adds an extra case to explain how the drill can’t be on any day. However, to add this case, we must have been able to consider it as we started. If we include it in the beginning, we can’t eliminate Friday, since if everybody went home on Thursday, there is still the possibility that there is no drill, meaning that no one can deduce what day the drill is. It follows that the Lakeside student had a flaw in his casework.