Proofs and Examples

Proofs

Simple Example: Odd Numbers

  • Claim: If nn is odd, then n2n^2 is odd.
  • Proof:
    • Assume nn is odd.
    • Then, we can write n=2k+1n = 2k + 1 for some integer kk (universal quantifier implied).
    • It follows that: n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1
    • Note that since kk is an integer, 2k2+2k2k^2 + 2k is also an integer.
    • Hence, n2n^2 is one more than twice an integer, so n2n^2 is odd.
  • The square symbol ($\blacksquare)orQEDarecommonlyusedasendofproofsymbols.</li></ul><h4id="morecomplicatedexampleperfectcubes">MoreComplicatedExample:PerfectCubes</h4><ul><li><strong>Claim:</strong>Everyperfectcubeisonemoreoronelessthanamultipleof9.<ul><li>) or QED are commonly used as end-of-proof symbols.</li> </ul> <h4 id="morecomplicatedexampleperfectcubes">More Complicated Example: Perfect Cubes</h4> <ul> <li><strong>Claim:</strong> Every perfect cube is one more or one less than a multiple of 9.<ul> <li>(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3</li></ul></li><li><strong>Proof:</strong><ul><li>Everyinteger</li></ul></li> <li><strong>Proof:</strong><ul> <li>Every integernisoneofthefollowing:<ul><li>Amultipleof3.</li><li>Onemorethanamultipleof3.</li><li>Twomorethanamultipleof3.</li></ul></li></ul></li></ul><h5id="case1ddnddisamultipleof3">Case1:is one of the following:<ul> <li>A multiple of 3.</li> <li>One more than a multiple of 3.</li> <li>Two more than a multiple of 3.</li></ul></li></ul></li> </ul> <h5 id="case1ddnddisamultipleof3">Case 1:nisamultipleof3</h5><ul><li>is a multiple of 3</h5> <ul> <li>n = 3kforsomeintegerfor some integerk(i.e.,(i.e.,n \mod 3 = 0).</li><li>).</li> <li>n^3 = (3k)^3 = 27k^3 = 9(3k^3)</li><li>Since</li> <li>Since3k^3isaninteger,is an integer,n^3isamultipleof9.</li></ul><h5id="case2ddnddisonemorethanamultipleof3">Case2:is a multiple of 9.</li> </ul> <h5 id="case2ddnddisonemorethanamultipleof3">Case 2:nisonemorethanamultipleof3</h5><ul><li>is one more than a multiple of 3</h5> <ul> <li>n = 3k + 1forsomeintegerfor some integerk(i.e.,(i.e.,n \mod 3 = 1).</li><li>).</li> <li>n^3 = (3k + 1)^3 = 27k^3 + 27k^2 + 9k + 1 = 9(3k^3 + 3k^2 + k) + 1</li><li>Since</li> <li>Since3k^3 + 3k^2 + kisaninteger,is an integer,n^3isonemorethanamultipleof9.</li></ul><h5id="case3ddnddistwomorethanamultipleof3">Case3:is one more than a multiple of 9.</li> </ul> <h5 id="case3ddnddistwomorethanamultipleof3">Case 3:nistwomorethanamultipleof3</h5><ul><li>is two more than a multiple of 3</h5> <ul> <li>n = 3k + 2forsomeintegerfor some integerk(i.e.,(i.e.,n \mod 3 = 2).</li><li>).</li> <li>n^3 = (3k + 2)^3 = 27k^3 + 54k^2 + 36k + 8 = 27k^3 + 54k^2 + 36k + 9 - 1 = 9(3k^3 + 6k^2 + 4k + 1) - 1</li><li>Since</li> <li>Since3k^3 + 6k^2 + 4k + 1isaninteger,is an integer,n^3isonelessthanamultipleof9.</li></ul><h5id="commentoncasestructures">CommentonCaseStructures</h5><ul><li>Theproofwasbrokenintocasesbasedontheremainderwhenis one less than a multiple of 9.</li> </ul> <h5 id="commentoncasestructures">Comment on Case Structures</h5> <ul> <li>The proof was broken into cases based on the remainder whenn$$ is divided by 3.
  • When faced with such problems, ask yourself, "How would I know to try this?"
  • The intuition about what case structure will work often develops this semester.
    • Start by trying odd/even (i.e., remainder when we divide by 2).
    • If that doesn't work, try others.