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# Prove: 1 + 1 = 2 watch

1. (Original post by JamesF)
Click
2. It just follows from the definitions, what a let down eh...
3. I was pretty sure some professor somewhere proved that 1+1 = 2 using like hundreds of sheets of paper. And only a select few people actually understood it.

It was a few years ago.
4. (Original post by Widowmaker)
I did this in yr 10

let a = b

a² = ab Multiply both sides by a

a² + a² - 2ab = ab + a² - 2ab Add (a² - 2ab) to both sides

2(a² - ab) = a² - ab Factor the left, and collect like terms on the right

2 = 1 Divide both sides by (a² - ab)
You can't divide by zero
5. It practically is an axiom though.
6. (Original post by Zapsta)
You can't divide by zero
xactly wot i woz gonna say!
therfore 1 does not equal 2

and yeh...i think 1+1=2 by definition!
i dunno how could prove it or why!
7. (Original post by 2776)
I was pretty sure some professor somewhere proved that 1+1 = 2 using like hundreds of sheets of paper. And only a select few people actually understood it.

It was a few years ago.
Actually I'm pretty sure i read that too, which is why the above proof has to use Peano's axioms, not the field axioms.
8. is the principia an axiom!?

when i mean axiom i mean basic properties which are assumed in order to analyse and describe other bodies. 2 being 1 greater than 1 is itself pretty much an axiom, and so you cant prove it, unless you decide to invent even simpler axioms and prove 1+1=2 on those basis.

then again, i might be wrong
9. (Original post by Willa)
when i mean axiom i mean basic properties which are assumed in order to analyse and describe other bodies.
That's what Peano's axioms are for.

There is a natural number 0.
Every natural number a has a successor, denoted by S(a).
There is no natural number whose successor is 0.
Distinct natural numbers have distinct successors: if a ≠ b, then S(a) ≠ S(b).
If a property is possessed by 0 and also by the successor of every natural number which possesses it, then it is possessed by all natural numbers. (This axiom ensures that the proof technique of mathematical induction is valid.)
10. (Original post by Widowmaker)
I did this in yr 10

let a = b

a² = ab Multiply both sides by a

a² + a² - 2ab = ab + a² - 2ab Add (a² - 2ab) to both sides

2(a² - ab) = a² - ab Factor the left, and collect like terms on the right

2 = 1 Divide both sides by (a² - ab)
Your first line clearly ststes that a=b. therefore, as previously mentioned, when you try to divide by a²-ab this is the same as a²-a² which is zero. This is undefined.
11. I have a problem with the proof linked to above. When he defines addition, doesn't he use the concept of addition at the end: (a + c)'? How can a definition hold if it uses the concept it is intended to define?
12. (Original post by wanderer)
I have a problem with the proof linked to above. When he defines addition, doesn't he use the concept of addition at the end: (a + c)'? How can a definition hold if it uses the concept it is intended to define?
I think he defines it recursively, and so it holds up.

Define 1 logically as 'a variable representing any single object'

Define the + sign as 'the logical opreator AND'

Define any number greater than one as 'a variable representing a series of single objects'. 2 can then be defined as 1 AND 1, 3 as 1 AND 1 AND 1, etc.

Hence:

1 + 1 = 2

is equivalent to:

1 AND 1 = 1 AND 1

which is a tautology. There you go.
14. So can anyone point me to a definition of a recursive definition then?
15. (Original post by wanderer)
So can anyone point me to a definition of a recursive definition then?

http://www.huminf.aau.dk/cg/Module_II/1072.html

or try,

16. Thanks, I now understand recursive definition. And the proof of 1 +1 = 2. *wanders off feeling smugly clever*.

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