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February 8, 2016
Posted by **Francesca ** on Friday, March 25, 2011 at 9:22pm.

Basic Step: Suppose that n = 1. If max(x, y) = 1 and x and y are positive integers, we have x = 1 and y = 1.

Inductive Step: Let k be a positive integer. Assume that whenever max(x, y) = k and x and y are positive integers, then x = y.

Now let max(x, y) = k + 1, where x and y are positive integers. Then max(x – 1, y – 1) = k, so by the inductive hypothesis, x – 1 = y – 1. It follows that x = y, completing the inductive step.

What's wrong with this proof?

- Discrete Math -
**MathMate**, Friday, March 25, 2011 at 9:52pmmax(x,y) has not been previously defined, but I assume it to be:

max(x,y)

=x if x≥y

=y if x<y.

If that is the case, then we can find numerous counter-examples to the given theorem, such as:

x=4, y=6,

n=max(x,y)=6

and x≠y !

which is sufficient proof that the theorem is false.

Notwithstanding the fallacy of the theorem, there are flaws in the proof.

Typically the inductive step is built on the case k and go on to prove that the case k+1 is true.

The assumption part is correctly made:

namely for an integer n=k, and positive integers x and y, if max(x,y)=k, then x=y. (the assumption is correctly made based on the theorem).

So for the case of n=k+1, and max(x,y)=k+1, we only know that at least one of x or y has to be k+1, not both. So there is no logical reason to conclude that both x and y equal k+1. That's where the conclusion that the proof cannot be made.

Could you check that the theorem has been transcribed correctly in the post?

- Discrete Math -
**Francesca**, Friday, March 25, 2011 at 10:07pmThank you for responding. Yes everything is typed correctly. I want to find what is wrong with proof.

- Discrete Math -
**MathMate**, Friday, March 25, 2011 at 10:43pmAre you convinced that the counter example is valid? If it is, then there is no correct proof, since the theorem is not valid.

If the counter example is not valid, explain why it is not valid.

- Discrete Math -
**MathMate**, Friday, March 25, 2011 at 10:52pmAre you sure the question is not:

"Theorem: For every integer*2n*. If x and y are positive integers*such that n=x+y*and with max(x, y) = n, then x = y."

or did the question say:

"Prove or find a counter example"?

- Discrete Math -
**Francesca**, Friday, March 25, 2011 at 11:04pmNo, the question verbatim is "What is wrong with this proof?"

- Discrete Math -
**MathMate**, Saturday, March 26, 2011 at 8:01amFinally, so "What is wrong with the proof" is part of the question. Sorry that I didn't understand it as such before. It seemed to me it was your proof, and I was explaining to you why you should not even try to prove a conjecture full of holes.

Assuming there was a proof, what's wrong had been stated (at 9:52 pm):

"The assumption part is correctly made:

namely for an integer n=k, and positive integers x and y, if max(x,y)=k, then x=y. (the assumption is correctly made based on the theorem).

So for the case of n=k+1, and max(x,y)=k+1, we only know that at least one of x or y has to be k+1, not both. So there is no logical reason to conclude that both x and y equal k+1. That's where the conclusion that the proof cannot be made. "

In addition, by induction, the proof should have been made for the case n+1, and not n-1.

- Discrete Math -
**Francesca**, Saturday, March 26, 2011 at 9:17amThank you! So, going back to your counterexample in post 9:52:

x=4, y=6,

n=max(x,y)=6

Why does it =6? Sorry if this seems like a silly question. . .

- Discrete Math -
**MathMate**, Saturday, March 26, 2011 at 9:35am"max(x,y) has not been previously defined, but I assume it to be:

max(x,y)

=x if x≥y

=y if x<y. "

So if the above definition max(x,y) corresponds with your teacher's,

then

x=4, and y=6,

then x<y, therefore max(x,y)=y,

so max(4,6)=y.

- Discrete Math -
**Francesca**, Saturday, March 26, 2011 at 9:43amOh okay. . .I get it. . .Thank you so much for your help :)

- Discrete Math -
**Francesca**, Saturday, March 26, 2011 at 10:09amHere is the solution: The mistake is in applying the inductive hypothesis to look at max(x −1, y −1) . Notice that we are doing induction on n not on x or y. Even though x and y are positive integers, x −1 and y −1 need not be (one or both could be 0). In fact, that is what happens if we let x = 1 and y = 2 when k = 1.

I have to admit I do not quite understand this explanation. . .

- Discrete Math -
**MathMate**, Saturday, March 26, 2011 at 12:01pmThe inductive process is

1. to establish for a base case, for example n=1. Then,

2. assuming the hypothesis is true for case n (in general),

3. we have to prove that it follows that it is true for the case n+1.

If we are able to do all that, then the hypothesis is true for all n.

Generally, steps 1 and 2 are easy to establish. It is in step 3 that some algebraic or logical operations are required to establish for the case of "n+1".

In the given erroneous proof, the first two steps were correct, but the third step had something to do with x-1 and y-1, which relates to the case n-1 instead of the case n+1. Worse, neither the case n-1 nor the case n+1 had been mentioned, so step 3 had established nothing.

- Discrete Math -
**Eric**, Tuesday, April 10, 2012 at 8:39pmIt was proven during the previous lecture that

1+2+3+....+n=[n(n+1)]/2.

Using that result, find the sum of all 3-digit palindromes: a palindrome is a number that can be written as ABA, where 1=<A<=9 and 0<=B<=9. Hint: decompose ABA = 100*A + 10*B + A and use above proven result.