Thread: rigorous proof

I was told to write a rigorous proof for the following theorem:

for all ,x : (-1)x = -x and
come up with the following proof :

1) (-1)x = (-1)x + 0................................................. .................................................. ..............by using the axiom : for all ,a: a + 0 = a

2) (-1)x + 0 = (-1)x + ( x + (-x))............................................... ................................................by using the axiom: for all ,a : a + (-a) = 0

3) (-1)x + (x + (-x)) = ((-1)x + x) + (-x)................................................ ....................................by using the axiom : for all a,b.c : a + ( b + c) = ( a + b) + c

4) ((-1)x + x) + (-x) = ( x + (-1)x) + (-x)................................................ ....................................by using the axiom : for all a,b : a+ b = b + a

5) ( x + (-1)x) + (-x) = ( 1x + (-1)x) + (-x)................................................ .................................by using the axiom: for all ,a : 1a = a

6) ( 1 + (-1))x + (-x) = ( 1x + (-1)x) + (-x)................................................ .................................by using the axiom : for all a,b,c : (a + b)c = ac + bc

7) ( 1 + (-1))x + (-x) = 0x + (-x)................................................ ...............................................by using the axiom : for all ,a : a + (-a) = 0

8) 0x + (-x) = 0 + (-x)................................................ .................................................. ...........by using the theorem : for all, a: 0a = 0

9) 0 + (-x) = (-x) + 0................................................. .................................................. ............by using the axiom : for all a,b : a + b = b + a

10) (-x) + 0 = -x................................................. .................................................. ..................by using the axiom : for all ,a : a + 0 = a

11) HENCE..FOR ALL ,X : (-1)X= -X....................................


Is that correct ??

Any help will be appreciated

Thanx

you definitely have all the ideas there, and you would probably receive full credit for that, but just a tip, you only have to manipulate one side cuz you start with (-1)x= something

So you want to use axioms and theorems to show that "something"=-x

you do that along the way, and even take it further

basically all of the left sides of the equation all the way down could be left as (-1)x (and probably should be)

Originally Posted by artvandalay11

you definitely have all the ideas there, and you would probably receive full credit for that, but just a tip, you only have to manipulate one side cuz you start with (-1)x= something

So you want to use axioms and theorems to show that "something"=-x

you do that along the way, and even take it further

basically all of the left sides of the equation all the way down could be left as (-1)x (and probably should be)

Thanx ,but can you write a few lines of the proof to show me what you mean??
Thanx again

for all ,x : (-1)x = -x and
come up with the following proof :

1) (-1)x = (-1)x + 0................................................. .................................................. ..............by using the axiom : for all ,a: a + 0 = a

2) (-1)x = (-1)x + ( x + (-x))............................................... ................................................by using the axiom: for all ,a : a + (-a) = 0

3) (-1)x = ((-1)x + x) + (-x)................................................ ....................................by using the axiom : for all a,b.c : a + ( b + c) = ( a + b) + c


That's what I mean, you want (-1)x on one side of the equation and you want to end up with -x on the other side

I caught a mistake that I didn't catch before though in your proof, on line 6 and then line 7, you must first factor the x out, then conclude 1+(-1)=0 and 0x=0, so you skipped a step in there

You may say to yourself, but I did that on the left side, but on line 5 to line 6 you don't make mention that x=1x, but again we're changing all the lefts to (-1)x, so we can ignore that as long as you correct the right side

Originally Posted by artvandalay11

for all ,x : (-1)x = -x and
come up with the following proof :

1) (-1)x = (-1)x + 0................................................. .................................................. ..............by using the axiom : for all ,a: a + 0 = a

2) (-1)x = (-1)x + ( x + (-x))............................................... ................................................by using the axiom: for all ,a : a + (-a) = 0

3) (-1)x = ((-1)x + x) + (-x)................................................ ....................................by using the axiom : for all a,b.c : a + ( b + c) = ( a + b) + c


That's what I mean, you want (-1)x on one side of the equation and you want to end up with -x on the other side

I caught a mistake that I didn't catch before though in your proof, on line 6 and then line 7, you must first factor the x out, then conclude 1+(-1)=0 and 0x=0, so you skipped a step in there

You may say to yourself, but I did that on the left side, but on line 5 to line 6 you don't make mention that x=1x, but again we're changing all the lefts to (-1)x, so we can ignore that as long as you correct the right side

Thanx i see what you mean now .

So we apply the axiom : for all ,a: a + (-a) =0 , in line (2), to line (1) and get the result :

(-1)x = (-1)x + (x + (-x)) is that correct??

And another point : you said we must end up with the equation:

(-1)x = -x .But surely if we end up with only -x can we not use the equality axiom: a= b and b =c and c =d and d = e e.t.c e.t.c and conclude (-1)x = -x ??
Thanx again

Yes that's what I mean.

And you are correct with the transitive property of the equal sign. Idk how much of a stickler your professor is but its generally adviseable to only try to manipulate one side of an equation in a proof like this. It doesn't make the math any less invalid

THANX again ,but the worst part is when i was asked to show which way the laws of logic are involved in the proof .

I suppose i must show that in another thread .

Be so kind to check my reasoning.