Answer: Prove function has even values
Friday, 25 of January , 2008 at 9:48 pm
Question: Prove that the function 
has even values for all natural numbers (
).
* Updated question for clarity.
Answer: Solving this problem is really trivial. The easiest way (by far) to solve this problem is to use Mathematical induction.
Wikipedia describes Mathematical Induction as:
The simplest and most common form of mathematical induction proves that a statement holds for all natural numbers n and consists of two steps:
- The basis: showing that the statement holds when n = 0.
- The inductive step: showing that if the statement holds for n = m, then the same statement also holds for n = m + 1.
This is exactly what we need to do to solve the problem. First, we need to prove that 
is even. This can easily be done:
Therefore, since 0 is an even number, we have satisfied the first step.
The second step is to show that if 
is even than 
is also even.
If we assume is even, we know that 
has to be even because 
.
Therefore, the function 
is even for all 
because:
- is even.
- If is even, it implies that
is even.
Question: Prove that the function has even values for all natural numbers ().
* Updated question for clarity.
Answer: Solving this problem is really trivial. The easiest way (by far) to solve this problem is to use Mathematical induction.
Wikipedia describes Mathematical Induction as:
The simplest and most common form of mathematical induction proves that a statement holds for all natural numbers n and consists of two steps:
- The basis: showing that the statement holds when n = 0.
- The inductive step: showing that if the statement holds for n = m, then the same statement also holds for n = m + 1.
This is exactly what we need to do to solve the problem. First, we need to prove that is even. This can easily be done:
Therefore, since 0 is an even number, we have satisfied the first step.
The second step is to show that if is even than is also even.
If we assume is even, we know that has to be even because .
Therefore, the function is even for all because:
- is even.
- If is even, it implies that is even.
Comment by Nikita Nikolaev
Made Saturday, 26 of January , 2008 at 11:43 am
Actually, the proof is wrong.
First, you rely on the fact that 0 belongs to your set and hence you are not proving the statement for the set of natural numbers, but rather for a set of nonnegative integers, at the very least.
Second, the arguments are not correct. You try to show that is even and then for some given 
, 
is also even. But we need to recognise that those two expressions are numbers and it is meaningless to say whether a number is even or odd in the same sense as for the function.
The definition for an even function is the following
This clearly does not hold for natural numbers, nor does it for integers, unless the function is defined on the singleton 
. This can be seen from the following argument:
Comment by Nikita Nikolaev
Made Saturday, 26 of January , 2008 at 11:45 am
In addition to the fact that 
and 
is not defined.
Comment by eldila
Made Saturday, 26 of January , 2008 at 7:35 pm
Thanks for the comments. I prefer when people point out my mistakes instead of blindly accepting them. First off, I would like to point out that I have no formal education with regards to pure mathematics, so please take my comments with a grain of salt. However, I believe the biggest problem was that the problem was not properly posed.
I got this question from a textbook called Algebra by Saunders MacLane and Garrett Birkhoff. The original problem is stated as follows: By induction, prove that 
is always even. Considering this is in a chapter called natural numbers, I assume that he wants it proven for only natural numbers which he defines as .
I believe the question could be more correctly posed as:
Show 
that there 
where 
.
I believe that my proof would be sufficient (minus a few details). Please let me know if I’m mistaken.
Comment by eldila
Made Saturday, 26 of January , 2008 at 7:41 pm
Also, in order to write latex equations, you need to surround your latex with [ tex ] and [ /tex ]. (Note: I added spaces around tex so that the tags wouldn’t be parsed.)
Comment by eldila
Made Saturday, 26 of January , 2008 at 8:23 pm
I posed the question differently in the original post which should clarify the question.
Comment by Nikita Nikolaev
Made Saturday, 26 of January , 2008 at 9:51 pm
eldila,
yes! That was it is absolutely correct and now makes sense. Moreover, in this case, the proof is brilliant!
In other words you can say that for all natural numbers, the value of that function is an even number. That would the correct proposition.
And thanks for the correction with LaTeX equations — I didn’t realise that at first.
Comment by eldila
Made Saturday, 26 of January , 2008 at 10:17 pm
Well, I’m glad we got everything cleared up. The original question wasn’t posed very well. It gave the impression that I wanted you to prove that f(n) was an even function for natural numbers. This question, as you pointed out, doesn’t even make any sense.
Anyways, if you find any mistakes in any of my future posts, be sure to comment on it. Also, if you come across any interesting questions, you can always email them to me or post the problem on your blog.
Cheers
Comment by Nikita Nikolaev
Made Sunday, 27 of January , 2008 at 7:40 am
I will, I will. I actually have got quite a few, so will need to take some time sometime and post them.
Thanks