Prove algebraically that ##n^3+3n-1## is odd

[chwala]

Homework Statement

Prove algebraically that $n^3+3n-1$ is odd for all positive integers $n$.

Relevant Equations

Algebra

This is a past paper question; Find the solution here; Well understood

Now i was just thinking along this lines;

Let $n=x$,
then $f(x)=x^3+3x-1$
$f(x) =x(x^2+3)-1$

since $x∈ℤ^{+}$ then $x(x^2+3)$ will always be even implying that $x(x^2+3)-1$ is odd.
Would this approach hold or i have to stick with ms? Thanks.

 

[anuttarasammyak]

I prefer transformation of
$$n^3+3n-1=(n-1)^3+3n^2$$
Say n is odd (n-1)^3 is even and 3n^2 is odd so it is odd.
Say n is even (n-1)^3 is odd and 3n^2 is even so it is odd.
So we can say it is odd.

[EDIT]
$$n^3 \equiv n(mod\ 2)$$
$$n^3+3n-1 \equiv 4n -1 \equiv -1 \equiv 1 (\mod 2)$$ So it is odd.

 

[malawi_glenn]

Why not use the parity rules O + O - O = O and E + E - O = O directly?

 

[chwala]

Why not use the parity rules O + O - O = O and E + E - O = O directly?

@malawi_glenn how to show this directly? I just looked at parity...some work though required...you still have to let $n=2k+1$ for odd numbers and let $n=2k$ for even numbers. Note that this was already used on the attached mark scheme guide.

...or you have a more direct way?

 

[malawi_glenn]

if n is odd, then you have
O² + O*O - O = O

if n is even, then you have
E²+ O*E - O = O

 

[Mark44]

Why not use the parity rules O + O - O = O and E + E - O = O directly?

@malawi_glenn how to show this directly? I just looked at parity...some work though required...you still have to let n=2k+1 for odd numbers and let n=2k for even numbers.

The first one above doesn't take much work.
$(2k+1) + (2m+1) - (2n+1) = 2(k + m + n + 1) + 1$, which is an odd number. The second one can be shown just as easily.

The basic ideas are that

• odd + odd = even
• odd + even = odd
• even + even = even
• odd * odd = odd
• even * even = even
• odd * even = even

You can manipulate these equations to get another set that involves subtraction and division.

 

[nuuskur]

@OP Don't confuse the reader with renaming to $x$ for no reason at all. Let $n$ be a natural number and give an argument why $n^3+3n-1$ is odd. For example $n(n^2+1)+(2n-1)$ is the sum of even and odd, because $n(n^2+1)$ is even for all $n$.

 

[malawi_glenn]

Don't confuse the reader with renaming to x for no reason at all.

Which post are you referring to?

 

[nuuskur]

Which post are you referring to?

The very first one. We have $n$ and then it suddenly changes to $x$, it's confusing.

 

[malawi_glenn]

The first one above doesn't take much work.
$(2k+1) + (2m+1) - (2n+1) = 2(k + m + n + 1) + 1$, which is an odd number. The second one can be shown just as easily.

The basic ideas are that

• odd + odd = even
• odd + even = odd
• even + even = even
• odd * odd = odd
• even * even = even
• odd * even = even

You can manipulate these equations to get another set that involves subtraction and division.

Yes, these rules are pretty handy to know.
@op: try to prove these. Then use them basically whenever you can

 

[PeroK]

if n is odd, then you have
O² + O*O - O = O

if n is even, then you have
E²+ O*E - O = O

That all fine, except that notation may not be part of the course. Moreover, the marking scheme suggests to me that an approach from first principles is expected.

 

[chwala]

That all fine, except that notation may not be part of the course. Moreover, the marking scheme suggests to me that an approach from first principles is expected.

Would a student be penalised for using an approach other than what was expected? Like in this case using parity? Or it depends on examining body...

 

[PeroK]

Homework Statement:: Prove algebraically that $n^3+3n-1$ is odd for all positive integers $n$.
Relevant Equations:: Algebra

This is a past paper question; Find the solution here; Well understood

View attachment 318976

Now i was just thinking along this lines;

Let $n=x$,
then $f(x)=x^3+3x-1$
$f(x) =x(x^2+3)-1$

since $x∈ℤ^{+}$ then $x(x^2+3)$ will always be even implying that $x(x^2+3)-1$ is odd.
Would this approach hold or i have to stick with ms? Thanks.

If you were looking for an esoteric solution, try this:

Let $f(n) = n^3 + 3n - 1$. Note that $f(0) = -1$ is odd. We will show by induction that $f(n)$ is odd for all integers $n$.

Now$$f(n+1) - f(n) = (n+1)^3 + 3(n+1) - n^3 - 3n = 3(n^2 + n + 2)$$Let $g(n) = n^2 + n + 2$ and note that $g(0) = 2$ is even. Now:$$g(n+1) - g(n) = 2(n+1)$$The difference is even, hence by induction $g(n)$ is even for all integers $n$. Hence $f(n+1) - f(n)$ is even for all integers $n$. Hence $f(n)$ is odd for all integers $n$.

 

[malawi_glenn]

That all fine, except that notation may not be part of the course. Moreover, the marking scheme suggests to me that an approach from first principles is expected.

It's just schematic.
I do that notation in my math class (but swedish letters U for odd (udda) and J för even (jämn)) :-D

 

[nuuskur]

Would a student be penalised for using an approach other than what was expected? Like in this case using parity? Or it depends on examining body...

NO!!! If the technique is correct, then the correct result will be obtained.

If the problem statement doesn't specify what methods are allowed, then, in my opinion, it's in extremely bad taste to penalise students.

As an example, we do $\varepsilon-\delta$ proofs early in analysis tutorials to get comfortable with the definition. In test assignments we specifically state that prove the limit is so and so by definition of limit (there is no ambiguity here).

 

[Mark44]

Would a student be penalised for using an approach other than what was expected?

NO!!! If the technique is correct, then the correct result will be obtained.

I disagree. If the correct result was obtained, but not by a specific technique that was explicitly stated, then the student almost certainly would be penalized.

I do agree, though, if there isn't a specific method required, then no penalty, which is what the following is saying.

If the problem statement doesn't specify what methods are allowed, then, in my opinion, it's in extremely bad taste to penalise students.

 

[pbuk]

This is a past paper question

Then the only consideration should be getting maximum possible marks. You can do this by using the methods that are taught in the syllabus to solve the problem, and the method that is taught for solving questions of the form "prove something is odd for all $n \in \N$" is to consider odd and even separately.

This appears to be from a public exam similar to a UK 'A'-level. Markers are paid about £3 per script for these, which leaves seconds to decide on each mark: they do this by comparing it to the marking scheme, so make it easy for them to give you full marks.

The instructions to markers will say something like

If a student uses a method which is not explicitly covered by the marking instructions the same principles of marking should be applied. Credit should be given to any valid methods. Examiners should seek advice from their senior examiner if in any doubt.

so if you do go "off-piste" you might be lucky, but if the marker is falling behind on their quota you might not. It is typically this kind of thing that gets picked up in a re-mark, but why go to the time, trouble, cost and risk of a re-mark when it is easier to use the method you were taught first time? There are no marks for being a smart-arse.

Disclaimer: I know what I am talking about: I used to be a real smart-arse.