multiplication=addition

[MathematicalPhysicist]

i think everyone is familiar to this but i wonder if someone has made his research to study it.
what i mean is the namubers that when the result of multiplying them together is the same as adding them:
for example:
2*2=2+2=4 (trivial, i know (-: ).
3*1.5=3+1.5=4.5

my question has someone prooved that the number of these pairs are finite or infinite in the real numbers?

[matt grime]

xy=x+y

rearrange and solve for R.

for N (well, this is the number theory thread)

xy-x-y+1=1

(x-1)(y-1)=1

in N x=y=2

[Muzza]

Perhaps I'm misunderstanding something... You want real solutions to this equation:

x + y = xy
=>
y = x / (x - 1)

I.e, for any x != 1 we can chose a y so that x + y = xy, namely y = x / (x - 1). Any real x (except for x = 1) will give a real y, so there are an infinite number of such pairs...

A more interesting question would be to consider only integers...

[MathematicalPhysicist]

Originally posted by Muzza
Perhaps I'm misunderstanding something... You want real solutions to this equation:

x + y = xy
=>
y = x / (x - 1)

I.e, for any x != 1 we can chose a y so that x + y = xy, namely y = x / (x - 1). Any real x (except for x = 1) will give a real y, so there are an infinite number of such pairs...

A more interesting question would be to consider only integers...
i didnt think about it very much so perhaps from this had risen the misunderstanding.
i guess in the real numbers is really trivial.
so lets say just prooving for integers how do you go around prooving this?
[x,y]=[Z] or [x,x/(x-1)]=[Z]

[Muzza]

so lets say just prooving for integers how do you go around prooving this?
[x,y]=[Z] or [x,x/(x-1)]=[Z]


I have no idea what this means. matt grime provided an answer for what happens if you only consider the natural numbers (i.e, x = y = 2 is the only solution).

[MathematicalPhysicist]

z doesnt represent the integers?

[Muzza]

Aha, yes it does, but you used the equals sign (and non-curly braces), and I've never seen = used to specify membership in a set (http://www.combinatorics.net/weblib/A.3/img1.png seems to be more commonly used).

[NateTG]

Originally posted by loop quantum gravity
i didnt think about it very much so perhaps from this had risen the misunderstanding.
i guess in the real numbers is really trivial.
so lets say just prooving for integers how do you go around prooving this?
[x,y]=[Z] or [x,x/(x-1)]=[Z]

For \frac{x}{x-1} to be an integer, x-1 must dividex. Since the difference between the two is 1, their greatest common factor is 1, so the only solutions are x-1=1 and x-1=-1 (2+2=2*2 and 0+0=0*0).

[matt grime]

NateG, you should state you are only working in N there, by saying integer you implying Z. In Z, there are other answers. All we know is that here (x-1) is a unit, hence x-1=-1 or 1, yielding the other integer answer of x=y=0 as well as x=y=2

[NateTG]

For \frac{x}{x-1} to be an integer |x-1| must divide |x|. The GCF of the two absolute values is 1 (by Euclid's algorithm). Therefore |x-1| \leq 1 since it divides |x| and obviously divides itself. For x \in \mathbb{Z} that leaves three solutions: 0,1, and 2, but 1 leads to division by zero, so the only solutions are x=0 and x=2.

P.S. My apologies for the mixed formatting

[matt grime]

forget that last post if you saw it, i absolutely apoligize, i misread your post.

[NateTG]

Originally posted by matt grime
forget that last post if you saw it, i absolutely apoligize, i misread your post.

Don't sweat it, I've posted some beauties myself. I should be less adverserial in my response though. ;)

[MathNerd]

Integer solutions of xy = x + y where x,y \epsilon Z

xy - x - y = 0

(x-1)(y-1) = 1

let x' = x -1 and y' = y - 1 so x',y' \epsilon Z

So x' = 1 / y'

The only integers whose reciprocals are also integers are 1 and -1

So y' = 1 and x' = 1

So y = 2 and x = 2

AND

So y' = -1 and x' = -1

So y = 0 and x = 0

Therefore there are only two solutions to this diophantine equation x = y = 2 and x = y = 0