Multiplication=addition 2*2=2+2=4

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In summary, the conversation revolved around a mathematical equation that represents the product of two numbers being equal to their sum. The question was whether there are a finite or infinite number of pairs that satisfy this equation in the real numbers. It was proven that there are an infinite number of solutions in the real numbers, but there are only two solutions in the integers.
  • #1
MathematicalPhysicist
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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?
 
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  • #2
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
 
  • #3
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...
 
  • #4
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 let's say just prooving for integers how do you go around prooving this?
[x,y]=[Z] or [x,x/(x-1)]=[Z]
 
  • #5
so let's 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).
 
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  • #6
z doesn't represent the integers?
 
  • #7
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 (
img1.png
seems to be more commonly used).
 
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  • #8
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 let's say just prooving for integers how do you go around prooving this?
[x,y]=[Z] or [x,x/(x-1)]=[Z]

For [tex]\frac{x}{x-1}[/tex] to be an integer, [tex]x-1[/tex] must divide[tex]x[/tex]. Since the difference between the two is 1, their greatest common factor is 1, so the only solutions are [tex]x-1=1[/tex] and [tex]x-1=-1[/tex] (2+2=2*2 and 0+0=0*0).
 
  • #9
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
 
  • #10
For [tex]\frac{x}{x-1}[/tex] to be an integer |x-1| must divide |x|. The GCF of the two absolute values is 1 (by Euclid's algorithm). Therefore [tex]|x-1| \leq 1[/tex] since it divides |x| and obviously divides itself. For [tex]x \in \mathbb{Z}[/tex] 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
 
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  • #11
forget that last post if you saw it, i absolutely apoligize, i misread your post.
 
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  • #12
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. ;)
 
  • #13
Integer solutions of [tex] xy = x + y [/tex] where [tex] x,y \epsilon Z [/tex]

[tex] xy - x - y = 0 [/tex]

[tex] (x-1)(y-1) = 1 [/tex]

let [tex] x' = x -1[/tex] and [tex] y' = y - 1 [/tex] so [tex] x',y' \epsilon Z [/tex]

So [tex] x' = 1 / y' [/tex]

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

So [tex] y' = 1 [/tex] and [tex] x' = 1 [/tex]

So [tex] y = 2 [/tex] and [tex] x = 2 [/tex]

AND

So [tex] y' = -1 [/tex] and [tex] x' = -1 [/tex]

So [tex] y = 0 [/tex] and [tex] x = 0 [/tex]

Therefore there are only two solutions to this diophantine equation [tex] x = y = 2 [/tex] and [tex] x = y = 0 [/tex]
 
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1. What is multiplication?

Multiplication is a mathematical operation that involves adding a number to itself a certain number of times. It is represented by the symbol "x" or "*", and is often used to find the total of equal groups or to calculate the area of a rectangle.

2. How is multiplication related to addition?

Multiplication is related to addition because it is essentially a shortcut for repeated addition. For example, 2 x 3 is the same as 2 + 2 + 2. Both operations result in a total of 6.

3. Why is 2*2 equal to 2+2?

This is because the numbers 2 and 2 are being added together the same number of times, which is 2. So, 2*2 is essentially saying "add 2 two times" and 2+2 is literally adding 2+2.

4. Can multiplication and addition be used interchangeably?

No, multiplication and addition cannot be used interchangeably. While they are related, they are different operations with different rules. For example, the commutative property holds for addition (2+3=3+2), but not for multiplication (2*3 does not equal 3*2).

5. How can I use multiplication in real life situations?

Multiplication can be used in a variety of real life situations, such as calculating the cost of multiple items at a store, finding the total area of a room, or determining the number of days in a certain number of weeks. It is a useful tool for solving problems involving equal groups or repeated addition.

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