1. ## Proof by counter-example

How can i prove using a counter example that:

The number n^3 − n^2 + n + 4 is divisible by 5, for all positive integers n is false ?

2. ## Re: Proof by counter-example

you could evaluate it for n=4....

3. ## Re: Proof by counter-example

It appears that numbers of the form $n = 5i+4$ and $n = 5i+5, i = 0, 1, 2, \cdots$ don't work.

4. ## Re: Proof by counter-example

Thanks guys but I never did it before so I would appreciate if I could have an example followed by the information of how it is done and what is happening during the process.

5. ## Re: Proof by counter-example

You were told exactly what to do! Set n= 4 in the given formula and do the calculation. What do you get?

Do you know what 'counter example' means?

6. ## Re: Proof by counter-example

Ok what I have done is:

if n = 5

then

5^3 - 5^2 + 5 + 4 = 109

Thus 109 is not divisible by 5

Is that the right way of present my answer or there is any specific notation to do this?

7. ## Re: Proof by counter-example

Originally Posted by michele
Ok what I have done is:

if n = 5

then

5^3 - 5^2 + 5 + 4 = 109

Thus 109 is not divisible by 5

Is that the right way of present my answer or there is any specific notation to do this?
Your logic as written isn't quite right. Just remove the "thus" before "109".

"Thus $n^3-n^2+n+4$ is not divisible by $5$ for all $n$"

8. ## Re: Proof by counter-example

Originally Posted by michele
Ok what I have done is:

if n = 5

then

5^3 - 5^2 + 5 + 4 = 109

Thus 109 is not divisible by 5
As romsek said, "thus" is the wrong word. It implies that the fact that 109 is not divisible by 5 follows from what went before. In fact, "what went before" shows that 109 is of the from "n^3- n^2+ n+ 4" for n= 5. The fact that it is not divisible by 5 follows from the fact that 109= 5(21)+ 4.

Is that the right way of present my answer or there is any specific notation to do this?