# "If equals be added to equals" in Spivak?

• Sep 23rd 2012, 06:43 AM
Porcupine
"If equals be added to equals" in Spivak?
Sorry if this was already posted. I couldn't find any answer anywhere.
Also, sorry if this is the wrong forum.

In just the 2nd page of Chapter 1 of Spivak's Calculus 3ed edition,
Spivak describes to us 3 properties of numbers and then he uses these properties to prove that: a+x=a.
Right on the start he adds (-a) to both sides. Hence (-a)+(a+x)=(-a)+a.

My question is, how do you prove that it is a valid thing to do, to add the same number to equal sides of the equation?

Thanks!
And I'm sorry if this is a silly question!

BTW, the properties P(x) are these:

P(1) : If a, b and c are any numbers, then a+(b+c)=(a+b)+c.

P(2) : If a is any number, then a+0=0+a=0

P(3) : For any number a, there is a number -a such that a+(-a)=(-a)+a=0
• Sep 23rd 2012, 07:57 AM
ione
Re: "If equals be added to equals" in Spivak?
I do not have Spivak's book, but if he were trying to prove that a+x=a, he would not start the proof by assuming what he is trying to prove. He would not start with a+x=a then add (-a) to both sides and get (-a)+(a+x)=(-a)+a.
The properties you listed are taken to be true without proof. So is the property about adding the same number both sides of an equation.
• Sep 23rd 2012, 11:05 AM
Porcupine
Re: "If equals be added to equals" in Spivak?
Yes, you are right.
This is what's written:
"Indeed, if a number x satisfies
a+x=a
for any number a, then x=0 (and consequently this equation also holds for all numbers a).

The thing is that Spivak does not mention this property:
If a=b, then a+c=b+c.

Spivak mentions all the other properties, just not this one.

Euclid, on the other hand, does state this property as an axiom in the Elements:
"If equals are added to equals, then the wholes are equal."
• Sep 23rd 2012, 11:59 AM
emakarov
Re: "If equals be added to equals" in Spivak?
In logic, axioms are divided into three classes. The first class includes logical axioms such as "A /\ B -> A." They are accepted in all theories solely in virtue of the logic we are using. The second class are axioms of equality. It is enough to have two equality axioms (more precisely, the second one is an axiom schema that consists of an infinite number of concrete axioms):

x = x (1)

and

P(x) /\ x = y -> P(y) for all formulas P (2)

The third class consists of axioms specific to a given theory, such as order axioms, ring axioms, axioms of a vector space and so on. Since the vast majority of theories include equality, axioms of equality are often joined with logical axioms.

We can derive a = b -> a + c = b + c from the equality axioms. Indeed, let P(x) be a + c = x + c. Then

P(a) /\ a = b -> P(b)

is an instance of (2). Further, P(a) is an instance of (1), so we can derive

a = b -> P(b)

which is

a = b -> a + c = b + c.

Spivak apparently lists axioms that are specific to real numbers and leaves out logical and equality axioms that are accepted everywhere.
• Sep 23rd 2012, 05:14 PM
HallsofIvy
Re: "If equals be added to equals" in Spivak?
The fact that, for any x and y, x+ a= y+ a, that is, that adding the same thing to both sides of the equation is due to the fact that addition is "well defined"- if you add the same two numbers together, you always get the same thing. Saying that "x= y" is just saying that x and y are (different) names for the same number. x+ a and y+ a are just two examples of adding the same two numbers together and must give the same result: x+ a= y+ a.