# How do I show that there are no other solution

1. Sep 20, 2011

### flyingpig

1. The problem statement, all variables and given/known data

[PLAIN]http://img90.imageshack.us/img90/834/unledml.png [Broken]

3. The attempt at a solution

OKay first of all I showed that x is indeed a solution. But I need to state the Axioms of Fields first.

[PLAIN]http://img194.imageshack.us/img194/2548/unledtao.png [Broken]

i.e.

$$a + (b - a) = (b - a) + a$$ by A2

$$= b + (-a + a)$$ by A1

$$= b + 0$$ by A4

Hence b = b

Showing that it is the only solution is tough for me because I don't know what that means. I am guessing at the end of the proof I would show that x = b - a?

Here goes

$$a + x = b$$
$$a + (-a) + x = b + (-a)$$
$$0 + x = b + (-a)$$ By A4
$$x = b + (-a)$$ By A3

$$x = b - a$$ and say it is the only solution?

Thanks

Last edited by a moderator: May 5, 2017
2. Sep 21, 2011

### PAllen

One way to do it is to let x=b-a+h, and show contradiction unless h=0.

3. Sep 21, 2011

### flyingpig

But we haven't learned proof by contradiction yet. So I can't use that.

4. Sep 22, 2011

### flyingpig

Any comments...? I'll take it even if i am wrong.

5. Sep 22, 2011

### Staff: Mentor

Unfortunately, contradiction is the standard way of proving statements about the uniqueness of something. This technique is not related to the axioms you have to use, so it should be fine to use.

6. Sep 22, 2011

### flyingpig

BUt what was wrong with what I did? I posted this because I wasn't confident

7. Sep 22, 2011

### flyingpig

Without wasting everyone's time...

Proof

Let x = b - a + h

a + x = a + (b - a + h)

= (a + (-a)) + (b + h) by A2

= 0 + b + h By A4

= b + h

Now we have a + x = b + h

Stuck...

8. Sep 22, 2011

### Staff: Mentor

Given a + x = b, you showed that there was a solution, x = b - a.

Now suppose that y $\neq$ x, is also a solution. Can you take it from there?

9. Sep 22, 2011

### flyingpig

Are you referring to Pallen's method? Where did y come from?

10. Sep 22, 2011

### HallsofIvy

What you did was show that b- a $is$ a solution. You did not show that it was the only solution, which was the point of the exercise.

11. Sep 22, 2011

### flyingpig

No I meant this part

$$a + x = b$$
$$a + (-a) + x = b + (-a)$$
$$0 + x = b + (-a)$$ By A4
$$x = b + (-a)$$ By A3

Doens't that prove that x can only be x = b + (-a)?

12. Sep 22, 2011

### Punkyc7

I dont think so, it shows x is a solution but not the only one. Now assume y is a solution

13. Sep 22, 2011

### flyingpig

Assume y is NOT a solution
a + y = b

a + (-a) + y = b + (-a)

0 + y = b + (-a)

y = b + (-a)

y = b - a

But this contradicts x $$\neq$$ y

Woah!

14. Sep 22, 2011

### Punkyc7

Assume y is a solution where y does not equal x
.
.
.
.

15. Sep 22, 2011

### flyingpig

Isn't that what I just showed? If I assume y is a solution I would just go back to my original post and redo that proof again?

16. Sep 22, 2011

### Punkyc7

you said assume y is not a solution, yes and no because y is not x so its different but it gets you the same thing so y=x contradiction

17. Sep 22, 2011

### flyingpig

Oh, I should have said y is solution, but my math follows is correct right?

18. Sep 22, 2011

### Punkyc7

if y is not a solution does it make any sense to say for example 0=2?

19. Sep 22, 2011

### flyingpig

No I meant like

20. Sep 22, 2011

### Punkyc7

yeah that works