i have a seriouse question here

[Money]

lol it may not seem seriouse to you guys

but here it is

 

if u multiply two consecutive numbers will that

number always be divisible by 2

ex. 2x3=6 6/2=3

ex. 5x6=30 30/2=15

 

plz give me sum numbers that will prove this wrong lol

i argued my ass off wit my math teacher saying

i'll find sum numbers to prove him wrong (help a brotha out lol)

[timo]

You might argue whether 0*1=0 is divisible by 2. Apart from that, every 2nd number is an even number, meaning one of its prime factors is 2. Instead of multiplying the numbers you could multiply their prime factors (in the sense of 8*9 = (2*2*2)*(3*3) = 2*2*2*3*3) => the product will contain at least one prime factor of 2 => the product is divisible by two.

[the tree]

Atheists example of 0 and 1 is the only one I can think of (apart from 0 and -1, but that's the same thing really), I wonder when this was given to you was it phrased as any two consecutive natural numbers or integers? Obviously if you're playing in completely different sets then it doesn't hold true at all.

[Money]

i think it was integers but im not sure

ima find out 2day

[the tree]

If it was then 0 and 1, or 0 and -1 are your best bet. If it was the naturals then Atheist seems to have proved your teacher right.

 

If you're the type to just irritate your teacher for the sake of it then you could move out of the reals: (1+i) and (1+2i) are sort of consecutive, I think.

[Money]

wats 1+i mean ???

lol

[Money]

wait arent there prime even numbers if u take a prime

even number and multiply it by a odd consecutive wont

that product not be divisible by two

 

arent prime not divisible by 2 or anything as a matter of fact

i forgot??? lol

[Klaynos]

wait arent there prime even numbers if u take a prime

even number and multiply it by a odd consecutive wont

that product not be divisible by two

 

arent prime not divisible by 2 or anything as a matter of fact

i forgot??? lol

 

All even numbers have to be divisible by 2, there's probably a nice maths proof for this but they're just the 2 times table So you can't have any even primes other than 2.

 

1+1i means that the number is a complex number with 2 parts to it, a real part, the 1, and an imaginary part, 1i where i is the square root of -1.

[the tree]

wait arent there prime even numbers if u take a prime even number...

There is only one prime even number, that is two.

 

All even numbers have to be divisible by 2, there's probably a nice maths proof for this but they're just the 2 times table

Haha, I'm pretty certain that it just follows straight from the definition.

 

wats 1+i mean ???

I wouldn't worry about it, you'll learn eventually and it'll completely change the way that you think about numbers and possibly the whole universe, but it's not really relevant here.

[Money]

kome on man lol i wanna learn what that means and how to use it

i like learning new stuff im addicted to it but i hate school lol

can one of you guys at least get me started on learning that 1i stuff

dont even have to teach me alot just enough to get me started

i'll learn the rest on my own using the internet

[the tree]

No you wont, complex number theory is tricky and quite boring in parts and needs a teacher or at least enough time to be teaching yourself and a good textbook.

 

The basic principle is the wonderful number known as [math]i[/math] where [math]i=\sqrt{-1}[/math], it is the "imaginary unit" and it can be applied to express the square roots of any negative number, eg [math]\sqrt{-49}=7i[/math], to be able to work with it you need to be competent at algebra particularly in manipulating surds and exponents.

 

As a quick example, find the possible values for [math]z[/math] in the quadratic equation: [math]z^2 - 5 z +5 = 0[/math].

 

(honestly, don't worry about it, concentrate on your school work, which will prove a useful tool when you take maths, or anything for that matter to a higher level)

[Money]

z2 - 5z +5 = 0

 

z= -b+sqrtb^2-4ac

---------------

2a

 

z= 5+sqrt-5^2-4x1x5

------------------

2(1)

 

i tried putting that in to my calc. but it sed domain error

so i just tried taking that apart and solving it and i got

 

Z= -10

 

lol hope im right

[Klaynos]

Nope you're not right and your calculator will have problems with it because,

 

sqrt(25-40) = sqrt(-15)

 

Which your calculator wont like, but you can do i*sqrt(15)

[the tree]

Actually just ignore this post, forget it, I'm tired. I'm gonna get a cup of tea and then rewrite this.

[the tree]

O.k. now that I'm caffeinated enough to compose at least one decent problem, give me values for [math]z[/math] for which [math]z^{2}-6z+10=0[/math].

 

Remember not to even get your calculator out.

 

[hide]No, I'm not going to give you the answer, just some more information for when you've got there.

 

What you've got is complex conjugates, that is two complex numbers, one in the form (p+qi) and the other in the form (p-qi), the product of complex conjugates is always a real number and you should be able to work out why easily enough.

 

In fact, in any equation with one complex number as it's root, at least one other root is it's complex conjugate. This is quite a handy thing to know when you're solving equations of an order higher than 2. Complex conjugates are also handy for dividing complex numbers and quite a few other things in terms of complex algebra.[/hide]

[John Cuthber]

At least for the positive integers (ie the numbers we usually think of as "numbers" ) the product of any even number and any number is always even. For any 2 consecutive numbers, one of them will be even so their product will be odd.

I think zero is considered to be even after all, if I divide it by 2 I get 0 and (more importantly) no remainder, so 0 and 1 multiplied together give 0 and that's even too.

[the tree]

Damnit you're right, the definition of an even number does include zero, well that's irritating. I guess for all integers his teacher was right then. Not that surprising I guess, Atheists proof was sound enough and it's intuitively correct as you explained it.

 

Although Atheists proof still doesn't cover 0 and 1, it's seems like you just have to work that special case individually.

[Dave]

Although Atheists proof still doesn't cover 0 and 1, it's seems like you just have to work that special case individually.

 

I don't see how it doesn't cover the 0/1 combination at all; 0 is divisible by any non-zero integer just by definition. Even if you don't believe this, you have just agreed that 0 is an even number, and hence it must be divisible by 2 by definition anyway.

 

For a slightly more formal proof, try the following. Let [imath]a \in \mathbb{Z}[/imath] be any integer. Let us assume that a is even. So there exists [imath]k \in \mathbb{Z}[/imath] such that [imath]a=2k[/imath]. Hence, [imath]a(a+1) = 2k(2k+1) = 4k^2 + 2k[/imath]. Clearly 2 divides the right hand side, and so [imath]a(a+1)[/imath] is divisible by 2. A similar argument follows for a odd.

[the tree]

Atheist's proof was centered around the fact that of any two consecutive numbers, one of them will have 2 as a prime factor, which doesn't apply to 0 and 1 (unless 0 has prime factors and I'm missing something here). That's all.

[alan2here]

You could check here.

http://img127.imageshack.us/img127/3816/image1vv3.png

 

They all give integers (whole numbers) when devided by 2

 

Maybe he ment consecutivly diffrently than you imagined.

For example

 

1 * 1 = 1

1 / 2 = 0.5

[the tree]

Maybe he ment consecutivly diffrently than you imagined.

For example

 

1 * 1 = 1

1 / 2 = 0.5

What do you mean? "two consecutive numbers multiplied together" seems fairly clear.

[Dave]

Atheist's proof was centered around the fact that of any two consecutive numbers, one of them will have 2 as a prime factor, which doesn't apply to 0 and 1 (unless 0 has prime factors and I'm missing something here). That's all.

 

Looks like Atheist's proof was slightly more convoluted than I first thought. You don't need to use prime decomposition, only that odd and even numbers are adjacent.

[timo]

Depends: Why would the product of an even and an odd number be even?

In the end, it all boils down to the question what you already accept as being true and what statements you think needs further comments/steps/proofs. For some, the statement that the product of two or more consecutive naturals is even probably wouldn't need any further comment at all. For you, the statement that a product of an odd and an even number is even seems to be enough. Others might want to know why. The next step beyond my little explanation why this is true would be asking why a pair of consecutive numbers is always {odd,even} or {even,odd}.

 

Sidenote: Of couse, the case 0*1 is covered. It's explicitly mentioned.