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Prove consequences of the field axioms

Prove the following consequences of the field axioms using only the six field axioms and the first four theorems of the chapter (the axioms and theorems can be found on p. 18 of Apostol, or anywhere else that lists axioms for a field should give roughly the same information).

  1. Prove that a(b-c) = ab - ac.
  2. Prove that 0 \cdot a = a \cdot 0 = 0.
  3. Prove that if ab = ac and a \neq 0, then b = c. Show also that the multiplicative identity 1 is unique.
  4. Prove that given a, b with a \neq 0 there is exactly one x such that ax = b.
  5. Prove that if a \neq 0, then b/a = b \cdot a^{-1}.
  6. Prove that if a \neq 0, then \left(a^{-1}\right)^{-1} = a.
  7. Prove that if ab = 0 then a = 0 or b = 0.
  8. Prove that (-a)b = -(ab) and (-a)(-b) = ab.
  9. Prove that if b \neq 0 and d \neq 0, then (a/b) + (c/d)= (ad + bc)/(bd).
  10. Prove that if b \neq 0 and d \neq 0, then (a/b)(c/d) = (ac)/(bd).
  11. Prove that if b,c,d \neq 0, then (a/b)/(c/d) = (ad)/(bc).

Note: For the first few of these I was super explicit, listing each axiom and theorem. After maybe part (c) I was a bit more terse. Each step is still included, and hopefully it is clear which axiom is being used. If not, leave a comment and I can add in a few more details.

  1. Proof. We compute, with justifications on the right,

        \begin{align*}    a(b-c) &= a(b+(-c))                  && (\text{Theorem I.3})\\           &= ab + a(-c)                 && (\text{Axiom 3}) \\           &= ab + a(-c) + ac +(-(ac)) \qquad \qquad  && (\text{Adding 0}) \\           &= ab + a(-c+c) +(-(ac))      && (\text{Axiom 3}) \\           &= ab + a(0) + (-(ac))        && (\text{Theorem I.2, } -c+c = 0)\\           &= a(b+0)+(-(ac))             && (\text{Axiom 3}) \\           &= ab + (-(ac))               && (b+0=b \text{ by Axiom 4})\\           &= ab - ac                    && (\text{Theorem I.3}). \qquad \blacksquare \end{align*}

  2. Proof. First, a \cdot 0 = 0 \cdot a by Axiom 1 (commutativity). Then,

        \begin{align*}    a \cdot 0 &= a \cdot (a + (-a)) && (a+(-a) = 0 \text{ by Theorem I.2})\\              &= a \cdot a + a \cdot (-a) \qquad \qquad && (\text{Axiom 3})\\              &= a \cdot a - a \cdot a && (\text{part (a)}) \\              &= 0. && (\text{Theorem I.2}).\qquad \blacksquare \end{align*}

  3. Proof. Let a,b,c \in \mathbb{R} with a \neq 0 and ab = ac. Since a \in \mathbb{R} \smallsetminus \{ 0 \}, we know (Axiom 6) it has an inverse y, i.e., there is a y \in \mathbb{R} such that ay = ya = 1. Then,

        \begin{align*}    ab = ac &\implies yab = yac &&(\text{since products are uniquely determined})\\            &\implies 1 \cdot b= 1 \cdot c \qquad \qquad &&(\text{definition of } y)\\            &\implies b = c && (\text{Axiom 4}). \end{align*}

    This shows that the multiplicative identity is unique since if 1' is also a multiplicative identity, then b = 1 \cdot b = 1' \cdot b \implies 1 = 1'. \qquad \blacksquare

  4. Proof. Since a \in \mathbb{R} and a \neq 0, we know there exists a y \in \mathbb{R} such that a y= 1 (Axiom 4). Let x = yb. Then,

        \[   ax = a(yb) = (ay)b = 1b = b.   \]

    Further, this solution is unique since if x' is another solution, then ax = b and ax' = b implies ax = ax' implies x = x' by part (c). \qquad \blacksquare

  5. Proof. By definition of b/a, we have

        \begin{alignat*}{2}       a \cdot \left( \frac{b}{a} \right) = b &\implies& \ a^{-1} \cdot \left(a \cdot \left( \frac{b}{a} \right) \right) &= a^{-1} b \\       &\implies& (a^{-1} \cdot a) \left( \frac{b}{a} \right) &= b \cdot a^{-1} \\       &\implies& 1 \cdot \left( \frac{b}{a} \right) &= b \cdot a^{-1} \\       &\implies& \frac{b}{a} &= b \cdot a^{-1}. \qquad \blacksquare \end{alignat*}\end{align*}

  6. Proof. By definition of a^{-1} we know a \cdot a^{-1} = 1. By the same reasoning, \left(a^{-1} \right)\left(a^{-1} \right)^{-1} = 1. Hence, by part (c) (Theorem I.7) we have a = \left(a^{-1}\right)^{-1}. \qquad \blacksquare
  7. Proof. Let a,b \in \mathbb{R} with ab = 0.
    Then, if a \neq 0, we know there exists a^{-1} \in \mathbb{R} such that a\cdot a^{-1}= 1. Thus,

        \[   ab = 0 \implies a^{-1} (ab) = a^{-1} 0 = 0 \qquad \qquad (\text{Theorem I.6})\]

    But,

        \begin{align*}          a^{-1} (ab) = 0 &\implies (a^{-1}a)b = 0 \\       &\implies 1b = 0\\       &\implies b = 0. \qquad \blacksquare \end{align*}

  8. Proof. Using the axioms and theorems we have,

        \begin{align*}       (-a)b &= (-a)b + ab + (-(ab)) \\       &= b(-a+a) + (-(ab)) \\       &= b \cdot 0 + -(ab) \\       &= -(ab).    \end{align*}

    And then, from this result (with -b in place of b) we have (-a)(-b) = -(a(-b)), and so,

        \begin{align*}    (-a)(-b) &= -(a(-b)) \\    &= -((-b)a) \\    &= -(-(ba)) & (\text{Again, using the first part})\\    &= ba = ab.\qquad \blacksquare    \end{align*}

  9. Proof. We know from (d) that for b \neq 0 and d \neq 0 we have,

        \begin{align*}       \frac{a}{b} + \frac{c}{d} &= ab^{-1} + cd^{-1} \\       &= 1 \cdot (ab^{-1} + cd^{-1}) & (\text{Multiplying by 1 }, bd \neq 0 by (g))\\       &= bd \cdot (bd)^{-1} \cdot (ab^{-1} + cd^{-1}) & (\text{Substituting } 1 = (bd)(bd)^{-1}) \\       &= (adbb^{-1} + bcdd^{-1})(bd)^{-1} & (\text{Distributing } bd \text{ and using commutativity})\\       &= (ad + bc)(bd)^{-1} \\       &= \frac{ad + bc}{bd}. \qquad \blacksquare    \end{align*}

  10. Proof. Again, since b \neq 0 and d \neq 0 we know by part (g) that bd \neq 0, so (bd)^{-1} exists and we have,

        \begin{align*}       \left( \frac{a}{b} \right) \left( \frac{c}{d} \right) &= (ab^{-1})(cd^{-1}) \\       &= (bd)(bd)^{-1} (ab^{-1})(cd^{-1}) \\       &= (ac)(bb^{-1}) (dd^{-1})(bd)^{-1} \\       &= (ac)(bd)^{-1} \\       &= \frac{ac}{bd}.\qquad \blacksquare    \end{align*}

  11. Proof. Since b,c,d \neq 0 we know cd^{-1} \neq 0 so (cd^{-1})^{-1} exists. Similarly, we know (bc)^{-1} exists. So,

        \begin{align*}       \frac{a/c}{b/d} = \frac{ab^{-1}}{cd^{-1}} &= (ab^{-1})(cd^{-1})^{-1} \\       &= (ab^{-1})(cd^{-1})^{-1} \cdot (bc)(bc)^{-1} \\       &= (ab^{-1} bc)(cd^{-1})^{-1} (bc)^{-1} \\       &= (ac)(cd^{-1})^{-1} (bc)^{-1} \\       &= (ac)(cc^{-1} d^{-1} d (c^{-1}d)^{-1})^{-1} (bc)^{-1} \\       &= (ac)((c^{-1}d)^{-1})(bc)^{-1}\\       &= (ac)(c^{-1} d)(bc)^{-1} \\       &= (ad)(bc)^{-1} \\       &= \frac{ad}{bc}. \qquad \blacksquare    \end{align*}

6 comments

  1. Ketan says:

    for part (i) I had a different solution:

    Let bx = a and dy = c. Multiply the first equation by d and the second equation by b. This yields: bdx = ad and bdy = bc. Adding up the first and second equations yields: bdx + bdy = bd(x + y) = ad + bc. This implies a/b + c/d = (ad + bc)/(bd)

    • Ketan says:

      For a follow up and in a similar vein, for part (j):

      Let bx = a and dy = c. If we multiply the two equations together we get: bdxy = bd(xy) = ac. This implies (a/b) (c/d) = (ac/bd).

    • Ketan says:

      as a follow up, for part (j):

      Let bx = a and dy = c. Multiply the two equations together and this yields bdxy = bd(xy) = ac. This implies (a/b)(c/d) = ac/bd.

      You could try something similar for division, as in part k, but admittedly this way of proving it may not be as intuitive.

  2. Jorge says:

    Hello,
    On my Apostol Copy (which is admittedly old) it also asks to prove the first 4 theorems, including the cancellation law for addition. Do you have that anywhere?

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