Math Problem Solved #2 (Number Theory)

[newtonisaacma]

i remember when problem like this i was solving in junior class.how old are you?

[vikingwendel]

3^3 is not 9

[budsmokersonly2008]

One counterexample disproves it nuff nuff my stuff.

[soulessnessD]

if n^3 is odd, then n is odd. therefore a larger odd minus a smaller odd equals an even number, then even number plus one is odd.

if n^3 is even, then n is even. therefore an even minus an even number equals an even number, then plus one is odd.

no n exists such that n^3-n+1 is even.

[ienjoyapples]

little bit easier proof, takes about 2 seconds. just factor n^3-n = n(n+1)(n-1), we need to prove this is always even because we add 1 to it in the original problem. if n is even n(n+1)(n-1) is obviously even and n^3-n +1 odd. if n is odd then n(n+1)(n-1) is even since both n+1 and n-1 are both even. therefore n^3-n is always even and n^3-n+1 is always odd, hence never even.

[PoorMansNews]

wow...this is really cool stuff!!!

[lost4468yt]

I found a easier way, just test every integer, worked for me.

[theneonfire]

If n is even, n cubed is even because any number times an even number is even.

n^3 - n would also be even, as an even number subtracted from an even number is even.

Answer is odd because even number + 1 is odd.

If n is odd, n cubed is odd because an odd number multiplied by successive odd numbers is still odd.

However, n^3 - n is even because an odd number subtracted from an odd number is an even number.

Again, answer is odd because even + 1 is odd.

Q.E.D.

[GuyMcpersonLol]

bro...

bro...

just use induction to prove that it's odd

[fredboy6]

ahhaha basic knowledge of math and you can do 3 cube = 27 ahha

[penam7300]

Wtf!! 3 cubed is not 9.....its 27 for god's sake....am i am the one getting a 50 in algebra.....jeezz

[mwyee]

fOR THE SECOND PART(FOR PROVING ALL ODD NUMBERS FOR N), EASIER WAY IS (2M+1) IS ODD. THEREFORE (2M+1)(2m+1)(2m+1) IS ODD. 2M IS EVEN.

ODD - EVEN IS ALLWAYS ODD.

[orangegold1]

What about a complex number? Would you consider (2i+1) even or odd?

[theboombody]

@orangegold1 Interesting thought. I usually think of imaginary numbers as numbers that aren't positive and aren't negative. I wonder if there are degrees of positivity that would tell how positive or how negative a number is. So far we just have solidly positive and solidly negative numbers.

[momoispink2]

three cubed is 27 not 9

[broulys1729]

there is an easier way. n=odd/even. if n=even then even^3 - even=even and adding 1 gives odd. If n=odd then odd^3 - odd= even and adding 1 gives odd. it follows simple logic even^3 =even, even - even=even, odd - odd=even and odd^3=odd.just this information is sufficient.

[skinom23]

@broulys1729 i totally agree..in fact for all integer n,n^3 - n + 1 is always odd.

[cjbatesii]

Slightly alternate algebra, but the same result:

(2m+1)(2m+1)(2m+1)-(2m+1)+1

(2m+1)((2m+1)^2 - 1) + 1

(2m+1)((2m+1+1)(2m+1-1)+1

(2m+1)((2m+2)2m)+1

(2m+1)(4m^2+4m)+1

8m^3+8m^2+4m^2+4m+1

8m^3+12m^2+4m+1

This statement, divided by 2, obviously always yields an odd number. QED

[lelyn1234]

I know you want to be at the level of a typical youtube audience, but for the most part people who want an example of Mathematical Induction are at a level to appreciate a quick hand with the binomial equation. You could memorize Pascal's triangle if your going to put up a video showing off your math skills.

I still saved it to my playlist, because this is a solid example. Thanks!

[IHelpNexon]

Math Theory:i have 3 apples i multiply them by 0 and that = me having 0 apples....wait wait wait WTF where did my apples go.How did it disappar??

[broulys1729]

@IHelpNexon liked ur question. multiply them by 2 and they increase. from where did they increase. first only 3 friends cud eat them now 6 can?????

[YFLOInternational]

Begin with any fallacy and you have disproved your proposition.

[rutovytskyy]

So how is this related to a number theory? Please change the name of this video.

[musselma]

To complete this proof you must show that zero does not work either, as 0 is neither even nor odd.

[Theo21Nickerson]

if you want to prove something wrong all you need is to prove one counterexample.

isn't it the basic math concept?

[Entertainmentwf]

f@Theo21Nickerson Logically he needs to show there cannot exist an n for that statement. It's an existence statement. What you are describing is a universal quantified statement (i.e. if he was disproving for instance that for all n>0, that 2n+3 is even, all he would need to do is give one contradictory n that disproved the statement). That is why disproving existence statements tend to be much more difficult since you need to cover the entire domain of the problem.

[Creolebway]

i dont understand why this is somehow a tough problem?

[Averilli1]

Nice btw you can mathematically abbreviate "therefore" by dotting the corners of an equilateral triangle(so a dotted triangle without its sides) :)

[Entertainmentwf]

@Averilli1 Never use the three dots. It's something mathematicians frown upon. It is much more formal to write out Therefore. The three dots can mean quite a bit more in other cases so best to stick to the formal terms.

[Averilli1]

@Entertainmentwf Ohh okay thanks! I will bare that in mind. :)

[Entertainmentwf]

@Averilli1 anytime :). I had it beaten into my skull by one of my mathematics professors who is an Erdos Number 1 and friends with the late Paul Erdos.

[packer777]

Hey bro, the proof is good but in the beginning when u did 3 ^3 you said it was 9 lol simple mistake but everything else was good. Get a bigger board!

[Henriamaa]

What about if n is negative integer. n cubed positive value of n is positive but n cubed for a negative value is negative.

[timaliceetbouqui]

n belongs to N which is a positive group integer. n can not be negative...

[kanechokeslam84]

Another thing you should be an eighth grade or highschool math teacher and I understand you when you explain math and I cant believe im actually watching a math video and i enjoyed it WOW weird post more math videos

[kanechokeslam84]

you kinda look like the kid from elf....... but your still my hero =D

[Bartkei]

This is, I think, a much easier proof: You can rewrite n^3 - n + 1 as n(n^2 - 1) + 1. The first term in that expression is a multiplication of an even number and an odd number. The first term therefore evaluates to an even number. Adding 1 to an even number yields an odd number.

[SlighBaby]

dud you rite but srry u made a mistake in your multiplication 3*3*3 = 27

[MathManXarcTan]

I went about it this way:

Notice that

n^3 - n = n ( n+1) (n-1)

and notice that this is either

even x odd x odd = even

or odd x even x even = even.

So n^3 - n is even, no matter what. Cool. That means n^3 - n + 1 is always odd. Done.

[1CSeth1]

even if 3*3 is 27 the equation ends up an odd number anyways

[seung87]

not much number theory

just simple analysis

[joygopaldas]

Stupid use method of induction u get the result.

[bothar54]

Jesus ...3*3 is not 9 its 27.

If you're going to try and educate people at least make sure its accurate.

[LogInForPaper]

you totally need to get a bigger whiteboard

[krisdestruction]

@LogInForPaper lol

[zixerqwert]

i was watching this hoping to see something amazing...that was the easiest thing in the world. you could have proved it wasn't even so quickly. i also have a formula for all prime numbers that work - work is still being made on it, it's very close.

[WildChildftw]

You have a formula for all prime numbers? Wow, that's pretty amazing.

[nasko700]

The second part could be solved this way: You have (2m+1)^3 - (2m+1) + 1

So (2m+1) is odd and every odd on third degree is also odd. Now we have odd minus odd plus one. Odd minus odd is allways even number. Even plus 1 is odd and the problem is solved. That solution easier for me.

[luked82]

I've got a Calculus problem for you to solve. It is shown at the end of my video called String Art is Calculus.

[th4n4t0s25]

3^3-3+1...it still came out an odd number.

got your point.

.. so no biggy, come on guys!

[witherspa]

Consider my mind blown!

[Nanumir]

3^3 ain't 9, don't worry I know what you meant lol happens to us all

[TheSuperGodsent]

3*3*3 is 27 not 9

[rohypnol55]

Stupidly long route... n^3 - n + 1 = n(n-1)(n+1) + 1. And n(n-1)(n+1) is a product of 3 consecutive numbers, and at least one of these must be even, thus n(n-1)(n+1) is even, and adding the 1 gives the result.

[UnitedNetizen]

very nicely done

[tseavamvas]

radical maths dude! 3^3=9 ok...in imaginary set it works someway...

[viharsarok]

Lol, 7 minutes for a trivial elementary school level excercise. Also, 3^3=27, you fucking retard.

[HARUHism20XX]

Work in the ring:Z/2Z

[MADDALENA21]

3^3= 27

[thrashmeta1]

damn your a dumbass 3^3 is not 9... you should go out and get a basic knowledge on math

[rustytrombone111111]

i could be a mistake, you know some people make mistakes sometimes it happens, im sure you've made mistakes before..

[TheSuperGodsent]

the final number would still be odd

[nthnthebsebllplr]

3 cubed is 27...haha

[cwldoc]

Simple summary of proof:

Every integer has the same parity as its cube

(because odd X odd = odd and

even X even = even)

So subtracting any integer from its cube results in an even number.

(because odd - odd =even and

even - even = even)

Then adding one makes the result odd, regardless of the integer with which we started!

[CogitoErgoCogitoSum]

If you put in an even number, you get an odd one. If you put in an odd number, you get an odd one still. Are you saying that there are twice as many odd numbers as even numbers?

[gccpp]

Where/what is "there"? :)

[FightBadIdeas]

It's faster if you use the unique factorization of integres and a few simple "lemmas":

If n is odd, then all powers of n are odd.

If n is even, then all powers of n are even.

(In particular, n^3 and n^1.)

If you add (or subtract) two odd numbers, the result is even; also, if you add (or subtract) two even numbers, the result is still even. (In particular, n^3-n is even.)

So, for any integer n and for any m,k>=0,

n^m +- n^k +- 1 is an odd number.

Your result is a particular case of this.

[FightBadIdeas]

Besides, I think this proof is less frightening because there is less computing here.

[stageoneinnovations]

3 cubed is 27 !!! not 9 !!!

[Heissenburger]

Proof 2:- If n is an integer, n^3 - n = n(n^2-1) = (n-1)n(n+1) which is 3 consecutive numbers multiplied together. They can't all be odd, so there is an even number among the three, that makes the whole product even. Thus n^3 - n + 1 would always be odd, and never even.

Proof 3:- If n is an integer, and n^3 - n + 1 is even, then working modulo 2, n^3 - n + 1 = 0 (in the ring Z/2Z)

In the ring Z/2Z, either n=0 or n=1, both of which won't work (verify by substition).

[unzippedrach]

i hate math

[Cagebreaker21]

We will get all your money!!!

[0149162532]

Your proof while not perfect is correct. However, I will question your basic math skills. I have made that mistake before but never on a video. I would like you to prove that 3^3=3*3 if you can do this i would be very impressed.

[mrjames113083]

Wait. You want me to prove that "3 raised to the 3rd power" equals "3 times 3"?

[0149162532]

no i want you too look at your video when n=3

[mrjames113083]

OH NUTS. That would be my bad.

[mrjames113083]

And I've added an annotation in the video at that part.

[oryxfreeride]

@mrjames113083 I make mistakes like that all the time. Personally, if this were my video, I would make an annotation, just a few seconds before you actually make the mistake, pointing to the guy on the screen with a giant "moron" in the annotation box. And then put up an annotation saying how the guy on the screen didn't even realize his silly mistake, but it doesn't matter.

Yes you'd be calling yourself a moron, or idiot if you prefer, but it would be funny...

[wuptiswupp]

@0149162532

his long route proved that he didn't even basic knowledge in math. :)

the answer can be found in a quick way.

[minhnhut2025]

@wuptiswupp did you read the question? If not, please RE-READ it again, CAREFULLY

[mbabitt]

like n^a-n^b+1, n^a+n^b-1, and n^a-n^b-1.

[mrjames113083]

n^a+n^b+1 is something that I can prove. It's easy, but time consuming. I'll try to find a way to simplify the process without losing the main idea. Stay tuned.

[mbabitt]

In general this is true for n^a+n^b+1 for any a and b. You can also negate signs and this will still be true. Can you prove this?

[mrjames113083]

I can try to prove the general case, but what do you mean by negate signs and it'll be true. I'm unclear as to what that means.

[agentredlum]

n^a+n^0+1 will be even for all even non-zero integer n.

[mbabitt]

If n is not even.