0.5: Proof Templates
- Page ID
- 132680
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)From DR. THI DINH'S PROOF WRITING HANDBOOK THIS BIBLE:
In the beginning...
Let \(P\) and \(Q\) be statement variables. When needed, suppose that P = P(x) depends on a variable x. The symbol "\(\forall\)" means "for all" or "for any". The symbol \(\exists\) means "there exists".
| Type of statement | What must we do to prove that it is true |
|
(1) If \(P\), then \(Q\) (2) \(\forall P\), \(Q\) |
Suppose that \(P\) is true. Prove that \(Q\) is true. |
| (3) \(\exists x P(x) \) such that \(Q\) |
Choose \(x\) so that \(P(x)\) is true. Prove that \(Q\) is true. Note: You need not explain how you find \(x\). |
The first (and only) commandment
To prove that a statement is false, thou shalt write out the negation of the statement and prove that.
To prove injective relationship (ie prove one-to-one)
YES If \(f(x_1) = f(x_2)\) then you need to prove \(x_1=x_2\).
Let \(A\) be a set. Let \(x_1,x_2 \in A\) such that \(f(x_1)=f(x_2)\).
⠇
Show \(x_1=x_2\).
NO then you need to provide one counterexample where \(x_1 \ne x_2\) when \(f(x_1) = f(x_2)\).
Is there a surjective relationship (i.e. prove onto)?
YES Note we are starting from \(f: A \rightarrow B\). Thus \(f(A)\subseteq B\) by definition.
Let \(A\) and \(B\) be sets where \(A\) is onto \(B\) if and only if \(f(A) \subseteq B\) and \(B \subseteq f(A)\).
By definition, \(f(A) \subseteq B\). Next show \(B \subseteq f(A)\).
Let \(x \in B\).
⠇
Show \(x \in f(A)\).
Conclude with: Since \(f(A) \subseteq B\) and \(B \subseteq f(A)\) therefore \(f\) is onto.◻
NO then need to find a counterexample where \(b \in B\), but there is no \(a \in A\) whereby \(f(a)=b\).
-
Reflexive, Symmetric, Anti Symmetric Transitive Property Proofs
Need a non-empty set \(A\) and a relation \(R\).
-
Reflexive
Let \(a \in A\). ← Starting point
⠇
Show \(aRa\) ← Ending point
-
Symmetric
Let \(a, b \in A\) s.t. \(aRb\).
⠇
Show \(bRa\)
-
Antisymmetric
Let \(a,b \in A\) s.t. \(aRb\) and \(bRa\)
⠇
Show \(a=b\).
-
Transitive
Let \(a,b,c \in A\) s.t. \(aRb\) and \(bRa\).
⠇
Show \(aRc\).