2
$\begingroup$

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:

enter image description here enter image description here

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?

$\endgroup$
  • 1
    $\begingroup$ Yes. For a proof, you should (try to) do Note 3 in the reverse order. If it works, you are done! $\endgroup$ – GEdgar Jul 29 '16 at 18:30
0
$\begingroup$

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.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.