# B Justification of addition in Spivak, Ch.1

#### Von Neumann

Summary
Based on the properties provided, why does x + 3 = 5 lead to (x + 3) + (-3) = 5 + (-3)?
I am 100% reading too much into this, but I am curious which of the properties provided by Spivak allow one to justify a specific argument. For reference/context, the properties are:

P1: If a, b, and c are any numbers, then
$$a +(b + c) = (a + b) +c$$

P2: If a is any number, then
$$a + 0 = 0 + a = a$$

P3: For every number a, there is a number -a such that
$$a + (-a) = (-a) + a = 0$$

Specifically, I am curious about the following:

$$x + 3 = 5$$

$$x + 3 + (-3) = 5 + (-3)$$

From this point, I understand the argument as the following:

By Property P3,

$$x + 0 = 2$$

By Property P2 ,then

$$x = 2$$

Is the step in question justifiable from any of the listed properties? Or, rather, is this basic property of addition intended to be assumed? Most properties are introduced from first principles, so I don't know if this should be any different. However, directly before the proof Spivak says "It is then possible to find the solution of certain simple equations by a series of steps (each justified by P1, P2, or P3) ...".

Here is the best that I can come up with:

$$x + 3 = 5$$

By P2,

$$x + 3 + 0 = 5$$

By P3,

$$x + 3 + (-3 + 3) = 5$$

By P1,

$$x + (3 + (-3)) + 3 = 5$$

Then, (by basic algebra?)

$$x + (3 + (-3)) = 5 + (-3)$$

By P3,

$$x + 0 = 2$$

By P2,

$$x = 2$$

• Stephen Tashi

#### jedishrfu

Mentor
Here’s a reference where they call that step the additive property of equality.

And this reference says it too:

#### Von Neumann

Here’s a reference where they call that step the additive property of equality.

And this reference says it too:

Yes. Can you be more specific how we can prove this from the three properties listed?

#### Math_QED

Homework Helper
If I understand your question you are trying to prove:

$a=b \implies a+c = b+c$

Is this correct?

It can be justified using someting very elementary (more elementary than the properties you listed) that you probably read over:

Addition is a function $+: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$.

This means, for every $(x,y) \in \mathbb{R}\times \mathbb{R}$, there is a unique number $x+y\in \mathbb{R}$.

So, if $a=b$, then $(a,c)=(b,c)$ and thus by uniqueness it follows that $a+c=b+c$.

Ps: Spivak didn't formally introduce the function notion at this point, so don't worry about this too much. But it is a very good question and shows that you don't take things for granted!

Last edited:
• jim mcnamara and Von Neumann

#### Stephen Tashi

Here is the best that I can come up with:

$$x + 3 = 5$$

By P2,

$$x + 3 + 0 = 5$$
That seems to assume "A quantity may be substituted for it's equal in any algebraic expression". I seem to remember that phrase given as an axiom in a secondary school textbook. But to get to $x+3+0 = 5$ correctly, you'd have to use the transitive property of $=$.

$(x+3)+0 = (x+3)$ by P2
$(x+3) = 5$ given as the initial step
$(x+3)+0 = 5$ transitive property of $=$

• jim mcnamara and Von Neumann

#### Von Neumann

If I understand your question you are trying to prove:

$a=b \implies a+c = b+c$

Is this correct?

It can be justified using someting very elementary (more elementary than the properties you listed) that you probably read over:

Addition is a function $+: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$.

This means, for every $(x,y) \in \mathbb{R}\times \mathbb{R}$, there is a unique number $x+y\in \mathbb{R}$.

So, if $a=b$, then $(a,c)=(b,c)$ and thus by uniqueness it follows that $a+c=b+c$.

Ps: Spivak didn't formally introduce the function notion at this point, so don't worry about this too much. But it is a very good question and shows that you don't take things for granted!
This is exactly my question! I think I need to get better at formulating exactly what I have a problem understanding. Your response makes sense. I was worried about overthinking this aspect.

• jim mcnamara

#### Von Neumann

That seems to assume "A quantity may be substituted for it's equal in any algebraic expression". I seem to remember that phrase given as an axiom in a secondary school textbook. But to get to $x+3+0 = 5$ correctly, you'd have to use the transitive property of $=$.

$(x+3)+0 = (x+3)$ by P2
$(x+3) = 5$ given as the initial step
$(x+3)+0 = 5$ transitive property of $=$
Yes! Those seem to be the steps I was performing implicitly in going directly from

$$x + 3 = 5$$

to

$$x + 3 + 0 = 5$$

I think the prologue of the text will make a lot more sense with this information. Thanks for responding!

"Justification of addition in Spivak, Ch.1"

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