Is there any proof that $1-1=0$ and that $(-1)x=x$?
These proofs should be created using only multiplication, addition and several basic rules like $a+x=0$, $a+0=a$, $1a=a$ etc.
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Well, both rely on the fact, that there is a unique additive inverse to any number $x$, called $-x$ with the propertythat $x+(-x)=0$.Then the first statement is a definition and the second one can be shown by evaluating $x+(-1)x$. – Verdruss Oct 04 '17 at 09:17
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What exactly to you mean by $-1$? – drhab Oct 04 '17 at 09:20
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Here's a MathJax tutorial :) – Shaun Oct 04 '17 at 09:20
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2First of all you need to establish what your axioms are. – Shaun Oct 04 '17 at 09:22
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Depends on what you define to be your axioms, but yeah, there are proofs.
The usual way to go about things is to first define addition as a commutative operation with a unit, and to define that unit to be called $0$. Then you define multiplication, and you define the unit of multiplication to be equal to $1$.
Subtraction, $a-b$, is usually defined as $a-b = a+(-b)$, and for each $a$, $-a$ is defined as the unique solution to the equation $a+x=0$.
Using this, you can already see that
- $1-1 = 1+(-1)$ (by definition of $-$), and
- $1+(-1)=0$ (by definition of $-1$).
You also define as your axioms that addition and multiplication are distributive, meaning that $(a+b)c=ac+bc$. Using this, you can show that $0\cdot x + 0\cdot x = (0+0)\cdot x$ and then, using the definition of $0$, get the equation $0\cdot x+0\cdot x = 0\cdot x$ which means (after we add $-(0\cdot x)$ to both sides of the equation) that $0\cdot x = 0$ for all $x$.
Then we can go for $(-1)\cdot x$ like so:
- $(-1)\cdot x + x = (-1)\cdot x + 1\cdot x$ (by the definition of $1$)
- $(-1)\cdot x + 1\cdot x = (-1+1)\cdot x$ (by distributivity)
- $(-1+1)\cdot x = 0\cdot x$ (by the definition of $-1$).
- $0\cdot x=0$ (we already proved that.
So, altogether, we have the equation $$(-1)\cdot x + x = 0$$ which means, by definition, that $(-1)\cdot x = -x$.
5xum
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Hmm. Now that I'm looking at it, I don't quite understand how you infered (-1)x = -x from (-1)x+x=0. Is it that trivial? Because to me it seems like a subtraction – Gillian Oct 04 '17 at 15:31
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@Gillian $-x$, by definition, is the unique number that, when you add it to $x$, gives you $0$. If I add $(-1)\cdot x$ to $x$, I get $0$. Therefore, $(-1)\cdot x$ is a number that, when I add it to $x$, gives me $0$. Therefore, $(-1\cdot x) = -x$. – 5xum Oct 05 '17 at 07:00
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@Gillian Or in equations, $y=-x$ if and only if $y+x=0$. Since $(-1)\cdot x + x = 0$, we can conclude that setting $y=(-1)\cdot x$ causes $y$ to satisfy the equation $y+x=0$, therefore, by definition, $y=-x$. – 5xum Oct 05 '17 at 07:01