Why is $x-(x-1)=1$ instead of $x-x-1=-1$ as in $a*(b*c)=a*b*c$?

Question

As in the title of the question, why is $x-(x-1)=1$ instead of $x-x-1=-1$ as in $a*(b*c)=a*b*c$? Is there any intuitive example to explain it?

Answer 1

If you want an intuitive example, this might help you.

You currently have $x$ dollars (or whatever currency you prefer) in your wallet and you have to pay $x-1$ dollars, then what you have after the transaction is $x-(x-1)$. But since you pay one less than what you initially had, you certainly end up with $1$, i.e. $x-(x-1) = 1$.

We can apply similar reasoning to $x-x-1$. You start with $x$ dollars, then you pay $x$ dollars (you now have $0$ dollars), then you pay another $1$ dollar, and you end up in debt.

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Now, what can you do completely algebraically? Since you maybe cannot always interpret the numbers to something like this. Well, whenever you have a minus in front of a parenthesis, like $$-(x+y) $$ what you may do is that you can change the sign in front if you also change the sign of all terms within the parenthesis, i.e.

$$-(x+y) = (-x - y) = -x - y, $$

or in your example,

$$x-(x-1) = x+(-x+1) = x - x + 1 = 0 + 1 = 1. $$

The reason for this is well explained in the other answers :)

Answer 2

Subtraction is not associative, as you discovered. Multiplication is.

http://www.mathwarehouse.com/dictionary/A-words/definition-of-associative-property.php

Answer 3

The negative number to a number $a$ is a number $b$ such that $a+b=0$. We denote that number $b$ as $-a$.

Now, the negative to $x-1$ is a $-x+1$, since $$(x-1)+(-x+1)=x-1+(-x)+1=0$$

Therefore $$x-(x-1)=x+(-x+1)=x+(-x)+1=1$$

Answer 4

I suppose that we are in the filed of real numbers $\mathbb{R}$ ( but in any field it's the same). In a field there are two operations $+$ ( the sum) and $\cdot$ ( the product) that have to satisfies some axioms ( see here), and a term $-x$ is defined as the element of the field such that $x+(-x)=0$ , called the opposed of $x$.

So when we write $x-(x-1)$ , using the axioms of a field, we interpret this as $x+[-(x-1)]$ where $-(x-1)$ is the opposite of $x-1$, i.e the element $a$ that, added to $x-1$ gives $0$: $a+(x-1)=0 $ but, from the definition of opposite and associativity of addition that means: $a+x+(-1)=0$ and we see that we must have $a=-x+1$.

Since we can prove that the opposite is unique, this gives the result : $ x-(x-1)=x+(-x)+1= 0+1=1$.

Answer 5

The subtlety here is that addition that involves negative numbers is still commutative, whereas subtraction is not commutative.

So with $x + (-x) + 1$ we can shuffle the summands around all we want and result is the same: $-x$ zeroes out $x$ and we're left with 1. Thus $x + 1 + (-x) = 1 + (-x) + x = \ldots = 1$.

But with $x - (x - 1)$, a different result can be had just by shifting the parentheses: $(x - x) - 1 = -1$, as you already noticed. If someone just gave $x - x$ you would say that's 0. When that $x - x$ is put into the parentheses, it still evaluates to 0, regardless of what $x$ is to begin with.

By contrast, $x - 1$ evaluates to 0 only when $x = 1$. The expression $x - (x - 1)$ then consists of two numbers: $x$ and the number 1 less than $x$. The difference between these two numbers is 1, regardless of what $x$ is to begin with.

These distinctions are reflected in the terminology and grammar we use. When adding up two numbers, they're both summands. But when we subtract one number from another, the first number is the minuend and the second number is the subtrahend.

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Similarly, multiplication that involves fractions is still commutative, but division is not.