Let $S$ be the set of all ordered pairs $(x_1, x_2)$ of real numbers. In each case determine whether or not $S$ is a linear space with the operations of addition and multiplication by scalars defined as indicated. If the set is not a linear space, indicate which axioms are violated.
(a) $(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)$, $\quad a(x_1, x_2) = (ax_1, 0)$.
(b) $(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, 0)$, $\quad a(x_1, x_2) = (ax_1, ax_2)$.
(c) $(x_1, x_2) + (y_1, y_2) = (x_1, x_2 + y_2)$, $\quad a(x_1, x_2) = (ax_1, ax_2)$.
(d) $(x_1, x_2) + (y_1, y_2) = (|x_1 + x_2|, |y_1 + y_2|)$, $\quad a(x_1, x_2) = (|ax_1|, |ax_2|)$.
(a) $$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)\quad a(x_1, x_2) = (ax_1, 0)$$
Axiom 1. Closure Under Addition. As we can see, for the ordered pairs $x = (x_1, x_1)$ and $y = (y_1, y_2)$ in $S$ we have a unique element $(x_1 + y_1, x_2 + y_2)$ corresponding to the addition of $x$ and $y.$
Axiom 2. Closure Under Multiplication by Real Numbers. Let $x = (x_1, x_2)$ and $a$ be a real scalar. Then, there is an element $(ax_1, 0)$ in $S$ corresponding to the product $ax.$
Axiom 3. Commutative Law of Addition. Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S.$ Then,
\begin{align*}
x + y &= (x_1 + y_1, x_2 + y_2)
\\
&= (y_1 + x_1, y_2 + x_2)
\\
&= y + x
\end{align*}
Axiom 4. Associative Law of Addition. Let $x = (x_1, x_2),$ $y = (y_1, y_2),$ and $z = (z_1, z_2)$ be elements in $S.$ Then,
\begin{align*}
(x + y) + z &= (x_1 + y_1, x_2 + y_2) + (z_1, z_2)
\\
&= (x_1 + y_1 + z_1, x_2 + y_2 + z_2)
\\
&= (x_1, x_2) + (y_1 + z_1, y_2 + z_2)
\\
&= x + (y + z)
\end{align*}
Axiom 5. Existence of the Zero Element. The ordered pair $O = (0, 0)$ is an element in $S$ such that for any $x = (x_1, x_2)$ in $S$ we have $$x + O = (x_1 + 0, x_2 + 0) = x.$$
Axiom 6*. Existence of Negative Elements. Let $x = (x_1, x_2)$ be an element in $S.$ Then, $(-1)x = (-x_1, 0)$ and $x + (-1)x$ is
\begin{align*}
x + (-1)x &= (x_1, x_2) + (-x_1, 0)
\\
&= (x_1 - x_1, x_2)
\\
&= (0, x_2)
\\
&\neq O
\end{align*}
*The set violates this axiom.
Axiom 7. Associative Law of Multiplication. Let $x = (x_1, x_2)$ and let $a$ and $b$ be real scalars. Then, $bx = (bx_1, 0)$ and we have
\begin{align*}
a(bx) &= (abx_1, 0)
\\
&= (ab)(x_1, x_2)
\\
&= (ab)x
\end{align*}
Axiom 8. Distributive Law for Addition in $V.$ Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S$ and let $a$ be a real scalar. Then we have:
\begin{align*}
a(x + y) &= a(x_1 + y_1, x_2 + y_2)
\\
&= (ax_1 + ay_1, 0)
\\
&= (ax_1, 0) + (ay_1, 0)
\\
&= a(x_1, x_2) + a(y_1, y_2)
\\
&= ax + ay
\end{align*}
Axiom 9. Distributive Law for Addition of Numbers. Let $x = (x_1, x_2)$ be an element of $S$ and let $a$ and $b$ be real scalars. Then we have
\begin{align*}
(a + b)x &= [(a + b)x_1, 0]
\\
&= (ax_1 + bx_1, 0)
\\
&= (ax_1, 0) + (bx_1, 0)
\\
&= a(x_1, x_2) + b(x_1, x_2)
\\
&= ax + bx
\end{align*}
Axiom 10*. Existence of Identity. Let $x = (x_1, x_2)$ be an element of $S$ and let $a = 1.$ Then we have
\begin{align*}
1x &= 1(x_1, x_2)
\\
&= (x_1, 0)
\\
&\neq x
\end{align*}
*The set violates this axiom.
(b) $$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, 0)\quad a(x_1, x_2) = (ax_1, ax_2)$$
Axiom 1. Closure Under Addition. As we can see, for the ordered pairs $x = (x_1, x_1)$ and $y = (y_1, y_2)$ in $S$ we have a unique element $(x_1 + y_1, 0)$ corresponding to the addition of $x$ and $y.$
Axiom 2. Closure Under Multiplication by Real Numbers. Let $x = (x_1, x_2)$ and $a$ be a real scalar. Then, there is an element $(ax_1, ax_2)$ in $S$ corresponding to the product $ax.$
Axiom 3. Commutative Law of Addition. Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S.$ Then,
\begin{align*}
x + y &= (x_1 + y_1, 0)
\\
&= (y_1 + x_1, 0)
\\
&= y + x
\end{align*}
Axiom 4. Associative Law of Addition. Let $x = (x_1, x_2),$ $y = (y_1, y_2),$ and $z = (z_1, z_2)$ be elements in $S.$ Then,
\begin{align*}
(x + y) + z &= (x_1 + y_1, 0) + (z_1, z_2)
\\
&= (x_1 + y_1 + z_1, 0)
\\
&= (x_1, x_2) + (y_1 + z_1, 0)
\\
&= x + (y + z)
\end{align*}
Axiom 5*. Existence of the Zero Element. For any $x$ in $S$ whose second term is nonzero, there is no ordered pair $O$ in $S$ for which the sum $x + O = x.$
*The set violates this axiom.
Axiom 6*. Existence of Negative Elements. Because there is no defined zero element for all $x$ in $S,$ we cannot prove the existence of negative elements.
*This set violates this axiom.
Axiom 7. Associative Law of Multiplication. Let $x = (x_1, x_2)$ and let $a$ and $b$ be real scalars. Then, $bx = (bx_1, bx_2)$ and we have
\begin{align*}
a(bx) &= (abx_1, abx_2)
\\
&= (ab)(x_1, x_2)
\\
&= (ab)x
\end{align*}
Axiom 8. Distributive Law for Addition in $V.$ Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S$ and let $a$ be a real scalar. Then we have:
\begin{align*}
a(x + y) &= a(x_1 + y_1, 0)
\\
&= (ax_1 + ay_1, 0)
\\
&= (ax_1, ax_2) + (ay_1, ay_2)
\\
&= a(x_1, x_2) + a(y_1, y_2)
\\
&= ax + ay
\end{align*}
Axiom 9*. Distributive Law for Addition of Numbers. Let $x = (x_1, x_2)$ be an element of $S$ and let $a$ and $b$ be real scalars. Then we have
\begin{align*}
ax + bx &= (ax_1, ax_2) + (bx_1, bx_2)
\\
&= (ax_1 + bx_1, 0)
\\
\\
(a + b)x &= [(a + b)x_1, (a + b)x_2]
\\
&= (ax_1 + bx_1, ax_2 + bx_2)
\\
&\neq ax + bx
\end{align*}
*The set violates this axiom.
Axiom 10. Existence of Identity. Let $x = (x_1, x_2)$ be an element of $S$ and let $a = 1.$ Then we have
\begin{align*}
1x &= 1(x_1, x_2)
\\
&= (x_1, x_2)
\\
&= x
\end{align*}
(c) $$(x_1, x_2) + (y_1, y_2) = (x_1, x_2 + y_2)\quad a(x_1, x_2) = (ax_1, ax_2)$$
Axiom 1. Closure Under Addition. As we can see, for the ordered pairs $x = (x_1, x_1)$ and $y = (y_1, y_2)$ in $S$ we have a unique element $(x_1, x_2 + y_2)$ corresponding to the addition of $x$ and $y.$
Axiom 2. Closure Under Multiplication by Real Numbers. Let $x = (x_1, x_2)$ and $a$ be a real scalar. Then, there is an element $(ax_1, ax_2)$ in $S$ corresponding to the product $ax.$
Axiom 3*. Commutative Law of Addition. Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S.$ Then,
\begin{align*}
x + y &= (x_1, x_2 + y_2)
\\
y + x &= (y_1, y_2 + x_2)
\\
x + y &\neq y + x
\end{align*}*The set violates this axiom.
Axiom 4. Associative Law of Addition. Let $x = (x_1, x_2),$ $y = (y_1, y_2),$ and $z = (z_1, z_2)$ be elements in $S.$ Then,
\begin{align*}
(x + y) + z &= (x_1, x_2 + y_2) + (z_1, z_2)
\\
&= (x_1, x_2 + y_2 + z_2)
\\
&= (x_1, x_2) + (y_1, y_2 + z_2)
\\
&= x + (y + z)
\end{align*}
Axiom 5. Existence of the Zero Element. Let $O = (0, 0).$ Then for any $x = (x_1, x_2)$ in $S$ we have $$x + O = (x_1, x_2 + 0) = (x_1, x_2) = x.$$
Axiom 6*. Existence of Negative Elements. Let $x = (x_1, x_2)$ be an element in $S.$ Then, the sum $x + (-1)x$ is
\begin{align*}
x + (-1)x &= (x_1, 0) \neq O
\end{align*}*The set violates this axiom.
Axiom 7. Associative Law of Multiplication. Let $x = (x_1, x_2)$ and let $a$ and $b$ be real scalars. Then, $bx = (bx_1, bx_2)$ and we have
\begin{align*}
a(bx) &= (abx_1, abx_2)
\\
&= (ab)(x_1, x_2)
\\
&= (ab)x
\end{align*}
Axiom 8. Distributive Law for Addition in $V.$ Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S$ and let $a$ be a real scalar. Then we have:
\begin{align*}
a(x + y) &= a(x_1, x_2 + y_2)
\\
&= (ax_1, ax_2 + ay_2)
\\
&= (ax_1, ax_2) + (ay_1, ay_2)
\\
&= a(x_1, x_2) + a(y_1, y_2)
\\
&= ax + ay
\end{align*}
Axiom 9*. Distributive Law for Addition of Numbers. Let $x = (x_1, x_2)$ be an element of $S$ and let $a$ and $b$ be real scalars. Then we have
\begin{align*}
(a + b)x &= [(a + b)x_1, (a + b)x_2]
\\
&= (ax_1 + bx_1, ax_2 + bx_2)
\\
\\
ax + bx &= (ax_1, ax_2) + (bx_1, bx_2)
\\
&= (ax_1, ax_2 + bx_2)
\\
\\
(a + b)x &\neq ax + bx
\end{align*}*The set violates this axiom.
Axiom 10. Existence of Identity. Let $x = (x_1, x_2)$ be an element of $S$ and let $a = 1.$ Then we have
\begin{align*}
1x &= 1(x_1, x_2)
\\
&= (x_1, x_2)
\\
&= x
\end{align*}
(d) $$(x_1, x_2) + (y_1, y_2) = (|x_1 + x_2|, |y_1 + y_2|),\quad a(x_1, x_2) = (|ax_1|, |ax_2|)$$
Axiom 1. Closure Under Addition. Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S.$ Then, there exists a unique element $x + y = (|x_1 + x_2|, |y_1 + y_2|)$ corresponding to the sum of $x$ and $y.$
Axiom 2. Closure Under Multiplication by Real Numbers. Let $x = (x_1, x_2)$ be an element in $S$ and let $a$ be a real scalar. Then there exists an element $ax = (|ax_1|, |ax_2|)$ corresponding to the product of $a$ and $x.$
Axiom 3*. Commutative Law of Addition. Let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements in $S.$ Then, we have
\begin{align*}
x + y &= (|x_1 + x_2|, |y_1 + y_2|)
\\
y + x &= (|y_1 + y_2|, |x_1 + x_2|)
\\
x + y & \neq y + x
\end{align*}*The set violates this axiom.
Axiom 4*. Associative Law of Addition. Let $x = (x_1, x_2),$ $y = (y_1, y_2),$ and $z = (z_1, z_2)$ be elements in $S.$ Then we have
\begin{align*}
(x + y) + z &= (|x_1 + x_2|, |y_1 + y_2|) + (z_1, z_2)
\\
&= (|x_1 + x_2 + y_1 + y_2|, |z_1 + z_2|)
\\
\\
x + (y + z) &= (x_1, x_2) + (|y_1 + y_2|, |z_1 + z_2|)
\\
&= (|x_1 + x_2|, |y_1 + y_2 + z_1 + z_2|)
\\
\\
(x + y) + z &\neq x + (y + z)
\end{align*}*The set violates this axiom.
Axiom 5*. Existence of the Zero Element. Suppose $x = (x_1, x_2)$ is an element in $S$ such that $x_1, x_2 < 0.$ Then, there is no element $O = (o_1, o_2)$ in $S$ such that $x + O = x$ since we have
$$x + O = (|x_1 + x_2|, |o_1 + o_2|)$$
implying that for any $x,$ the components of $x + O$ will always be positive.
*The set violates this axiom.
Axiom 6*. Existence of Negative Elements. Because there is no zero element in $S$ such that $x + O = x$ for all $x$ in $S,$ there is no way to prove the existence of negative elements.
*The set violates this axiom.
Axiom 7. Associative Law of Multiplication. Let $x = (x_1, x_2)$ be an element of $S$ and let $a$ and $b$ be real scalars. Then we have
\begin{align*}
a(bx) &= a(|bx_1|, |bx_2|)
\\
&= (|abx_1|, |abx_2|)
\\
&= ab(x_1, x_2)
\\
&= (ab)x
\end{align*}
Axiom 8. Distributive Law for Addition in $V.$ Let $a$ be a scalar and let $x = (x_1, x_2)$ and $y = (y_1, y_2)$ be elements of $S.$ Then we have
\begin{align*}
a(x + y) &= a(|x_1 + x_2|, |y_1 + y_2|)
\\
&= (|ax_1 + ax_2|, |ay_1 + ay_2|)
\\
&= ax + ay
\end{align*}
Axiom 9. Distributive Law for Addition of Numbers*. Let $a$ and $b$ be real scalars and let $x = (x_1, x_2)$ be an element of $S.$ Then we have
\begin{align*}
(a + b)x &= \left[|(a + b)x_1|, |(a + b)x_2|\right]
\\
&= \left(|ax_1 + bx_1|,|ax_2 + bx_2|\right)
\\
\\
ax + bx &= a(x_1, x_2) + b(x_1, x_2)
\\
&=(|ax_1|, |ax_2|) + (|bx_1|, |bx_2|)
\\
&= \left(|ax_1 + ax_2|, |bx_1 + bx_2|\right)
\\
\\
(a + b)x &\neq ax + bx
\end{align*}*The set violates this axiom.
Axiom 10*. Existence of Identity. Let $x = (x_1, x_2)$ and let $a = 1.$ We then have:
\begin{align*}
ax &= 1x
\\
&= 1(x_1, x_2)
\\
&= (|1 \cdot x_1|, |1 \cdot x_2|)
\\
&= (|x_1|, |x_2|)
\end{align*}
But if $x_1$ or $x_2$ is less than zero, then $1x \neq x,$ violating the axiom.
