# Proof Verification - Spivak's Calculus - Chapter 1 Question 1.3

In this post, I asked for help getting started on a proof in Spivak's Calculus Fourth Edition, Chapter 1 Question 1.3

Prove that if $x^2 = y^2$, then $x = y$ or $x = −y$ using only the following properties of numbers:

A lot of awesome people responded with hints to get me started:

P7 implies that if $ab = 0$ then $a = 0$ or $b = 0$. Various of the others show that if $x^2 = y^2$ then $\left( x + y \right) \left( x − y \right) = 0$.

Based on that hint, I came up with the following proof:

Note 1 $$\begin{array} { c c } x \cdot 0 + x \cdot 0 = x \left( 0 + 0 \right) & By\ P9 \\ x \cdot 0 + x \cdot 0 = x \cdot 0 & By\ P2 \\ x \cdot 0 + x \cdot 0 + \left( -x \cdot 0 \right) = x \cdot 0 + \left( -x \cdot 0 \right) & By\ Addition \\ x \cdot 0 + 0 = 0 & By\ P3 \\ \mathbf{ x \cdot 0 = 0} & By\ P2 \\ \end{array}$$

Note 2 $$\begin{array} { c c } \left( -a \right) \cdot b + a \cdot b = \left[ \left( -a \right) + a \right] \cdot b & By\ P9 \\ \left( -a \right) \cdot b + a \cdot b = 0 \cdot b & By\ P3 \\ \left( -a \right) \cdot b + a \cdot b = 0 & By\ Note\ 1 \\ \left( -a \right) \cdot b + a \cdot b + - \left( a \cdot b \right) = 0 + - \left( a \cdot b \right) & By\ Addition \\ \left( -a \right) \cdot b + 0 = 0 + - \left( a \cdot b \right) & By\ P3 \\ \mathbf{ \left( -a \right) \cdot b = - \left( a \cdot b \right)} & By\ P2 \\ \end{array}$$

Note 3

$$\begin{array} { c c } \left[ x + y \right] \left[ x + (-y) \right] = 0 & \text{By assumption} \\ x \left( x + y \right) + \left( -y \right) \left( x + y \right) = 0 & \text{By P9} \\ x^2 + xy + x \left( -y \right) + y \left( -y \right) = 0 & \text{By P9} \\ x^2 + x \left[ y + \left( -y \right) \right] + y \left( -y \right) = 0 & \text{By P9} \\ x^2 + x \cdot 0 + y \left( -y \right) = 0 & \text{By P3} \\ x^2 + 0 + y \left( -y \right) = 0 & \text{By Note 1} \\ x^2 + y \left( -y \right) = 0 & \text{By P2} \\ x^2 + \left( -y^2 \right) = 0 & \text{By Note 2 for all a = y and b = y} \\ x^2 + \left( -y^2 \right) + y^2 = 0 + y^2 & \text{By Addition} \\ x^2 + 0 = 0 + y^2 & \text{By P3} \\ x^2 = y^2 & \text{By P2} \\ \end{array}$$

Finally, note that the last line of Note 3 implies the first line of Note 3, and the first line of Note 3 implies $x + y = 0$ or $x - y = 0$. Additionally, P2 and P3 can then be used to show that $x = y$ or $x = -y$, QED.

I feel like I've jumped to a conclusion without proof--namely that the first line of Note 3 is true if and only if the last line of Note 3 is true.

Should I have started at the last line of Note 3 and worked backward to the first Line of Note 3?

• Yes. For a proof, you should (try to) do Note 3 in the reverse order. If it works, you are done! – GEdgar Jul 29 '16 at 18:30

## 1 Answer

In logic, the statement $p \to q$ does not necessarily mean $q \to p$. ( That is, if $p$ implies $q$, $q$ does not necessarily imply $p$. In other words, if $p$ is true, $q$ is also true--but there are some cases in the universe of discourse when $q$ is true but $p$ is not. )

In the case of this particular question from Spivak's book, the proof above does say that if $\left[ x + y \right]\left[ x + \left( -y \right) \right] = 0$, then $x^2 + y^2 = 0$ is true. And you're right to suspect that--until we prove otherwise--it may be possible there are sets of numbers for which $x^2 + y^2 = 0$ while $\left[ x + y \right]\left[ x + \left( -y \right) \right] \neq 0$.

So, it is important to show that one can get from $x^2 + y^2 = 0$ to $\left[ x + y \right]\left[ x + \left( -y \right) \right] = 0$ using only the number properties given.

$$\begin{array} { c c } x^2 = y^2 & Given \\ x^2 + (-y^2) = y^2 + (-y^2) & \text{By Addition} \\ x^2 + (-y^2) = 0 & \text{By P3} \\ x^2 + (-y^2) + 0 = 0 + 0 & \text{By Addition} \\ x^2 + (-y^2) + x \cdot 0 = 0 + 0 & \text{By Note 1} \\ x^2 + (-y^2) + x \cdot 0 = 0 & \text{By P2} \\ x^2 + (-y^2) + x \left[ y + (-y) \right] = 0 & \text{By P3 Substitution} \\ x^2 + (-y^2) + y \cdot x + (-y) \cdot x = 0 & \text{By P9} \\ x \cdot x + y \cdot (-y) + x \cdot y + x \cdot (-y) = 0 & \text{By Note 2} \\ x \cdot x + x \cdot y + x \cdot (-y) + y \cdot (-y) = 0 & \text{By P4} \\ x ( x + y ) + (-y)( x + y ) = 0 & \text{By P9} \\ \left[ x + (-y) \right] ( x + y ) = 0 & \text{By P9} \\ \left[ x + (-y) = 0 \right] \lor \left[ x + y = 0 \right] & \text{By Note 1} \\ \left[ x + (-y) + y = 0 + y \right] \lor \left[ x + y + (-y) = 0 + (-y) \right] & \text{By Addition} \\ \left[ x + 0 = 0 + y \right] \lor \left[ x + 0 = 0 + (-y) \right] & \text{By P3} \\ \left[ x = y \right] \lor \left[ x = (-y) \right] & \text{By P2} \\ QED \end{array}$$

• I marked my own answer correct. Feel free to correct it if it's not! :D – StudentsTea Jul 31 '16 at 18:45