How do I show that there are no other solution

[flyingpig]

Homework Statement

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

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

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

Now can I just jump to

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

Thanks

 

[PAllen]

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

 

[flyingpig]

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

 

[flyingpig]

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

 

[Mark44]

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.

 

[flyingpig]

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.

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

 

[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...

 

[Mark44]

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?

 

[flyingpig]

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?

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

 

[HallsofIvy]

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

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.

 

[flyingpig]

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.

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)?

 

[Punkyc7]

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

 

[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!

 

[Punkyc7]

Assume y is a solution where y does not equal x
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[flyingpig]

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

 

[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

 

[flyingpig]

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

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

 

[Punkyc7]

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

 

[flyingpig]

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

No I meant like

Assume y is 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 from before

 

[Punkyc7]

yeah that works

 

[flyingpig]

yeah that works

These proofs take A LOT of ingenuity...

 

[Punkyc7]

yeah, most of them seem obvious but to prove them is a different story
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when you have been doing proofs for a while read this
http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Challen/proof/proof.html

 

[flyingpig]

yeah, most of them seem obvious but to prove them is a different story
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when you have been doing proofs for a while read this
http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Challen/proof/proof.html

I had a laugh like a real one.

 

[flyingpig]

This is just out of curiousity, but how would you go about doing it if we take Pallen's suggestion and let x = b-a+h

 

[Punkyc7]

if h is anything other than zero then x would not equal b-a

 

[Mark44]

Assume y is a solution

and that y \neq x

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

Then, since x = b - a, and y = b - a, it follows that x - y = b - a - (b - a) = 0, hence x = y.
This contradicts the assumption that x and y are different, so we conclude that the solution to the equation a + x = b is unique.

 

[deluks917]

a + x = b
a + y = b

a + x = a + y
x = y