PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

<hint>
  <p>
    Follow the proof in <xref ref="example-integers-binomial-theorem" />.
  </p>
</hint>

<hint>
  <p>
    The base case,
    <m>S(0): (1 + x)^0 - 1 = 0 \geq 0 = 0 \cdot x</m> is true.
    Assume <m>S(k): (1 + x)^k -1 \geq kx</m> is true.
    Then
    <md>
      <mrow>(1 + x)^{k + 1} - 1 &amp; = (1 + x)(1 + x)^k -1</mrow>
      <mrow>&amp; = (1 + x)^k + x(1 + x)^k - 1</mrow>
      <mrow>&amp; \geq kx + x(1 + x)^k</mrow>
      <mrow>&amp; \geq kx + x</mrow>
      <mrow>&amp; = (k + 1)x</mrow>
    </md>,
    so <m>S(k + 1)</m> is true.
    Therefore, <m>S(n)</m> is true for all positive integers <m>n</m>.
  </p>
</hint>

<hint>
  <p>
    For <xref ref="fn_ineq" /> and <xref ref="fn_prod" /> use mathematical induction.
    <xref ref="fn_formula" /> Show that <m>f_1 = 1</m>,
    <m>f_2 = 1</m>, and <m>f_{n + 2} = f_{n + 1} + f_n</m>.
    <xref ref="fn_ratio" /> Use part <xref ref="fn_formula" />.
    <xref ref="fn_coprime" /> Use part <xref ref="fn_prod" /> and <xref ref="exercise-integers-gcd-1" />.
  </p>
</hint>

<hint>
  <p>
    Use the Fundamental Theorem of Arithmetic.
  </p>
</hint>

<hint>
  <p>
    Let <m>S = \{s \in {\mathbb N} : a \mid s</m>, <m>b \mid s \}</m>.
    Then <m>S \neq \emptyset</m>, since <m>|ab| \in S</m>.
    By the Principle of Well-Ordering,
    <m>S</m> contains a least element <m>m</m>.
    To show uniqueness, suppose that <m>a \mid n</m> and
    <m>b \mid n</m> for some <m>n \in {\mathbb N}</m>.
    By the division algorithm,
    there exist unique integers <m>q</m> and <m>r</m> such that <m>n = mq + r</m>,
    where <m>0 \leq r \lt m</m>.
    Since <m>a</m> and <m>b</m> divide both <m>m</m>, and <m>n</m>,
    it must be the case that <m>a</m> and <m>b</m> both divide <m>r</m>.
    Thus, <m>r = 0</m> by the minimality of <m>m</m>.
    Therefore, <m>m \mid n</m>.
  </p>
</hint>

<hint>
  <p>
    Since <m>\gcd(a,b) = 1</m>,
    there exist integers <m>r</m> and <m>s</m> such that <m>ar + bs = 1</m>.
    Thus, <m>acr + bcs = c</m>.
    Since <m>a</m> divides both <m>bc</m> and itself,
    <m>a</m> must divide <m>c</m>.
  </p>
</hint>

<hint>
  <p>
    Every prime must be of the form 2, 3, <m>6n + 1</m>, or <m>6n + 5</m>.
    Suppose there are only finitely many primes of the form <m>6k + 5</m>.
  </p>
</hint>

<hint>
  <p>
    (a) <m>3 + 7 \mathbb Z = \{ \ldots, -4, 3, 10, \ldots \}</m>; (c)
    <m>18 + 26 \mathbb Z</m>; (e) <m>5 + 6 \mathbb Z</m>.
  </p>
</hint>

<hint>
  <p>
    There is a nonabelian group containing six elements.
  </p>
</hint>

<hint>
  <p>
    The are five different groups of order 8.
  </p>
</hint>

<hint>
  <p>
    <md>
      <mrow>(aba^{-1})^n &amp; = (aba^{-1})(aba^{-1}) \cdots (aba^{-1})</mrow>
      <mrow>&amp; = ab(aa^{-1})b(aa^{-1})b \cdots b(aa^{-1})ba^{-1}</mrow>
      <mrow>&amp; = ab^na^{-1}</mrow>
    </md>.
  </p>
</hint>

<hint>
  <p>
    Since <m>abab = (ab)^2 = e = a^2 b^2 = aabb</m>,
    we know that <m>ba = ab</m>.
  </p>
</hint>

<hint>
  <p>
    <m>H_1 = \{ id \}</m>, <m>H_2 = \{ id, \rho_1, \rho_2 \}</m>,
    <m>H_3 = \{ id, \mu_1 \}</m>,
    <m>H_4 = \{ id, \mu_2 \}</m>,
    <m>H_5 = \{ id, \mu_3 \}</m>, <m>S_3</m>.
  </p>
</hint>

<hint>
  <p>
    The identity of <m>G</m> is <m>1 = 1 + 0 \sqrt{2}</m>.
    Since <m>(a + b \sqrt{2}\, )(c + d \sqrt{2}\, ) = (ac + 2bd) + (ad + bc)\sqrt{2}</m>,
    <m>G</m> is closed under multiplication.
    Finally, <m>(a + b \sqrt{2}\, )^{-1} = a/(a^2 - 2b^2) - b\sqrt{2}/(a^2 - 2 b^2)</m>.
  </p>
</hint>

<hint>
  <p>
    Since <m>a^4b = ba</m>,
    it must be the case that <m>b = a^6 b = a^2 b a</m>,
    and we can conclude that <m> ab = a^3 b a = ba</m>.
  </p>
</hint>

<hint>
  <p>
    <m>P_n</m>, the vector space of polynomials with degree at most <m>n</m>,
    has dimension <m>n+1</m> by <xref ref="theorem-exponent-laws" />. [Cross-reference is just a demo,
    content is not relevant.] What happens if we relax the defintion and remove the parameter <m>n</m>?
  </p>
</hint>

<hint>
  <p>
    What did you see last time you went driving?
  </p>
</hint>

<hint>
  <p>
    Maybe go out for a drive?
  </p>
</hint>

<hint>
  <p>
    What did you see last time you went driving?
  </p>
</hint>

<hint>
  <p>
    Maybe go out for a drive?
  </p>
</hint>

<hint>
  <p>
    You can take a derivative on any one of the choices to see if it is correct or not,
    rather than using techniques of integration to find
    <em>a single</em> correct answer.
  </p>
</hint>

<hint>
  <p>
    Dorothy will not be much help with this proof.
  </p>
</hint>

<hint>
  <p>
    For openers, a basis for a subspace must be a
    <em>subset</em> of the subspace.
  </p>
</hint>

<hint>
  <p>
    Python boolean variables begin with capital latters.
  </p>
</hint>

<hint>
  <p>
    <m>P_n</m>, the vector space of polynomials with degree at most <m>n</m>,
    has dimension <m>n+1</m> by <xref ref="theorem-exponent-laws" />. [Cross-reference is just a demo,
    content is not relevant.] What happens if we relax the defintion and remove the parameter <m>n</m>?
  </p>
</hint>

<hint>
  <p>
    What did you see last time you went driving?
  </p>
</hint>

<hint>
  <p>
    Maybe go out for a drive?
  </p>
</hint>