How many cases do I need for a proof by induction












0














First of all, English is not my mother tongue so I'm sorry if my definitions or concepts aren't explained correctly.



I just started a discrete math course and we started with induction, the professor told us to start taking several specific numbers as cases to see if the hypothesis was right and if all those cases matched the hypothesis we could generalize that property for all the others cases that we didn't use.
My question is how do I know how many cases are necessary. For example, let's say I have some property "P" that only works from 0 to 45million, from 45million to 47million doesn't work and then from 47million to infinite it works.



So.. if the cases I used were from 0 to 200 and because all of them matched the hypothesis how can I be 100% sure that It would work for every other number? Is possible that such property could exist? And if so how can I know if certain property works like that?










share|cite|improve this question




















  • 1




    I expect you have misunderstood. No finite number of cases will ever be enough for a proof. You need one case to start the induction, and it might be a good idea to work several cases...A. To check that it works and B. to get a sense of how you might hope to prove it generally.
    – lulu
    Nov 27 '18 at 22:21










  • I presume that your professor suggested you to start by taking a few specific cases in order to determine if your hypothesis is trivially and obviously false. If it is not obviously false, it does not mean that it is true. But now some serious thinking and mathematical work is required.
    – M.G
    Nov 27 '18 at 22:22












  • I'm sure your professor was talking about doing a few small cases by hand, say for $nle 5.$ Sometimes, you might want to check for lots of cases, by computer, say, but that's usually for theorems where you don't think that induction will work. As for how many cases you need to be certain, no finite number is enough; that's why we need proofs.
    – saulspatz
    Nov 27 '18 at 22:23
















0














First of all, English is not my mother tongue so I'm sorry if my definitions or concepts aren't explained correctly.



I just started a discrete math course and we started with induction, the professor told us to start taking several specific numbers as cases to see if the hypothesis was right and if all those cases matched the hypothesis we could generalize that property for all the others cases that we didn't use.
My question is how do I know how many cases are necessary. For example, let's say I have some property "P" that only works from 0 to 45million, from 45million to 47million doesn't work and then from 47million to infinite it works.



So.. if the cases I used were from 0 to 200 and because all of them matched the hypothesis how can I be 100% sure that It would work for every other number? Is possible that such property could exist? And if so how can I know if certain property works like that?










share|cite|improve this question




















  • 1




    I expect you have misunderstood. No finite number of cases will ever be enough for a proof. You need one case to start the induction, and it might be a good idea to work several cases...A. To check that it works and B. to get a sense of how you might hope to prove it generally.
    – lulu
    Nov 27 '18 at 22:21










  • I presume that your professor suggested you to start by taking a few specific cases in order to determine if your hypothesis is trivially and obviously false. If it is not obviously false, it does not mean that it is true. But now some serious thinking and mathematical work is required.
    – M.G
    Nov 27 '18 at 22:22












  • I'm sure your professor was talking about doing a few small cases by hand, say for $nle 5.$ Sometimes, you might want to check for lots of cases, by computer, say, but that's usually for theorems where you don't think that induction will work. As for how many cases you need to be certain, no finite number is enough; that's why we need proofs.
    – saulspatz
    Nov 27 '18 at 22:23














0












0








0







First of all, English is not my mother tongue so I'm sorry if my definitions or concepts aren't explained correctly.



I just started a discrete math course and we started with induction, the professor told us to start taking several specific numbers as cases to see if the hypothesis was right and if all those cases matched the hypothesis we could generalize that property for all the others cases that we didn't use.
My question is how do I know how many cases are necessary. For example, let's say I have some property "P" that only works from 0 to 45million, from 45million to 47million doesn't work and then from 47million to infinite it works.



So.. if the cases I used were from 0 to 200 and because all of them matched the hypothesis how can I be 100% sure that It would work for every other number? Is possible that such property could exist? And if so how can I know if certain property works like that?










share|cite|improve this question















First of all, English is not my mother tongue so I'm sorry if my definitions or concepts aren't explained correctly.



I just started a discrete math course and we started with induction, the professor told us to start taking several specific numbers as cases to see if the hypothesis was right and if all those cases matched the hypothesis we could generalize that property for all the others cases that we didn't use.
My question is how do I know how many cases are necessary. For example, let's say I have some property "P" that only works from 0 to 45million, from 45million to 47million doesn't work and then from 47million to infinite it works.



So.. if the cases I used were from 0 to 200 and because all of them matched the hypothesis how can I be 100% sure that It would work for every other number? Is possible that such property could exist? And if so how can I know if certain property works like that?







discrete-mathematics induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 22:19









M.G

2,2131134




2,2131134










asked Nov 27 '18 at 22:17









Shai Lèger

106




106








  • 1




    I expect you have misunderstood. No finite number of cases will ever be enough for a proof. You need one case to start the induction, and it might be a good idea to work several cases...A. To check that it works and B. to get a sense of how you might hope to prove it generally.
    – lulu
    Nov 27 '18 at 22:21










  • I presume that your professor suggested you to start by taking a few specific cases in order to determine if your hypothesis is trivially and obviously false. If it is not obviously false, it does not mean that it is true. But now some serious thinking and mathematical work is required.
    – M.G
    Nov 27 '18 at 22:22












  • I'm sure your professor was talking about doing a few small cases by hand, say for $nle 5.$ Sometimes, you might want to check for lots of cases, by computer, say, but that's usually for theorems where you don't think that induction will work. As for how many cases you need to be certain, no finite number is enough; that's why we need proofs.
    – saulspatz
    Nov 27 '18 at 22:23














  • 1




    I expect you have misunderstood. No finite number of cases will ever be enough for a proof. You need one case to start the induction, and it might be a good idea to work several cases...A. To check that it works and B. to get a sense of how you might hope to prove it generally.
    – lulu
    Nov 27 '18 at 22:21










  • I presume that your professor suggested you to start by taking a few specific cases in order to determine if your hypothesis is trivially and obviously false. If it is not obviously false, it does not mean that it is true. But now some serious thinking and mathematical work is required.
    – M.G
    Nov 27 '18 at 22:22












  • I'm sure your professor was talking about doing a few small cases by hand, say for $nle 5.$ Sometimes, you might want to check for lots of cases, by computer, say, but that's usually for theorems where you don't think that induction will work. As for how many cases you need to be certain, no finite number is enough; that's why we need proofs.
    – saulspatz
    Nov 27 '18 at 22:23








1




1




I expect you have misunderstood. No finite number of cases will ever be enough for a proof. You need one case to start the induction, and it might be a good idea to work several cases...A. To check that it works and B. to get a sense of how you might hope to prove it generally.
– lulu
Nov 27 '18 at 22:21




I expect you have misunderstood. No finite number of cases will ever be enough for a proof. You need one case to start the induction, and it might be a good idea to work several cases...A. To check that it works and B. to get a sense of how you might hope to prove it generally.
– lulu
Nov 27 '18 at 22:21












I presume that your professor suggested you to start by taking a few specific cases in order to determine if your hypothesis is trivially and obviously false. If it is not obviously false, it does not mean that it is true. But now some serious thinking and mathematical work is required.
– M.G
Nov 27 '18 at 22:22






I presume that your professor suggested you to start by taking a few specific cases in order to determine if your hypothesis is trivially and obviously false. If it is not obviously false, it does not mean that it is true. But now some serious thinking and mathematical work is required.
– M.G
Nov 27 '18 at 22:22














I'm sure your professor was talking about doing a few small cases by hand, say for $nle 5.$ Sometimes, you might want to check for lots of cases, by computer, say, but that's usually for theorems where you don't think that induction will work. As for how many cases you need to be certain, no finite number is enough; that's why we need proofs.
– saulspatz
Nov 27 '18 at 22:23




I'm sure your professor was talking about doing a few small cases by hand, say for $nle 5.$ Sometimes, you might want to check for lots of cases, by computer, say, but that's usually for theorems where you don't think that induction will work. As for how many cases you need to be certain, no finite number is enough; that's why we need proofs.
– saulspatz
Nov 27 '18 at 22:23










2 Answers
2






active

oldest

votes


















0














When you think of induction, one of the best analogies to think about is a ladder. When you climb up the
ladder, you have to step on the lower step and need to go up based on it. After we climb up the several
steps, we can go up further by assuming that the step you are stepping on exists. Proof by induction works on a similar principle. First, we prove a given statement for a base case that is any natural number $n_0$, this is the lowest step of the ladder. Now we prove that whenever the statement is true for a natural number $k$ then it will be also true for $k+1$. This is same as assuming that if the next step exists then we can climb the ladder. Now from these two parts our statement becomes true $forall ninmathbb{N}, ngeq n_0$.






share|cite|improve this answer





























    0














    The few cases that you initially try don't prove the theorem—they show you what the pattern might be. Then, when you think you know the pattern, you express it as a theorem and use proof by induction to show that it's true.



    The idea is:



    First you show "If this is true for $n=k$, then it is also true for $n=k+1$", without specifying a value for $k$.



    Then you apply this to a case you already know to be true (typically the one for $n=1$ or 0) and effectively say "Since this case is true, so is the next one, and the next, and the next, and the next, all the way up to infinity". This is usually worded something like "But this is true for $n=1$. Therefore, by induction, it is true for all $n ge 1$."



    Note that having $k$ as a separate symbol from $n$ makes things a lot less confusing than trying to prove "if it's true for $n$, it's true for $n+1$", since you can be consistent about what $n$ is rather than having to remember what you're currently doing to it, at least in my experience.






    share|cite|improve this answer























      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016381%2fhow-many-cases-do-i-need-for-a-proof-by-induction%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      When you think of induction, one of the best analogies to think about is a ladder. When you climb up the
      ladder, you have to step on the lower step and need to go up based on it. After we climb up the several
      steps, we can go up further by assuming that the step you are stepping on exists. Proof by induction works on a similar principle. First, we prove a given statement for a base case that is any natural number $n_0$, this is the lowest step of the ladder. Now we prove that whenever the statement is true for a natural number $k$ then it will be also true for $k+1$. This is same as assuming that if the next step exists then we can climb the ladder. Now from these two parts our statement becomes true $forall ninmathbb{N}, ngeq n_0$.






      share|cite|improve this answer


























        0














        When you think of induction, one of the best analogies to think about is a ladder. When you climb up the
        ladder, you have to step on the lower step and need to go up based on it. After we climb up the several
        steps, we can go up further by assuming that the step you are stepping on exists. Proof by induction works on a similar principle. First, we prove a given statement for a base case that is any natural number $n_0$, this is the lowest step of the ladder. Now we prove that whenever the statement is true for a natural number $k$ then it will be also true for $k+1$. This is same as assuming that if the next step exists then we can climb the ladder. Now from these two parts our statement becomes true $forall ninmathbb{N}, ngeq n_0$.






        share|cite|improve this answer
























          0












          0








          0






          When you think of induction, one of the best analogies to think about is a ladder. When you climb up the
          ladder, you have to step on the lower step and need to go up based on it. After we climb up the several
          steps, we can go up further by assuming that the step you are stepping on exists. Proof by induction works on a similar principle. First, we prove a given statement for a base case that is any natural number $n_0$, this is the lowest step of the ladder. Now we prove that whenever the statement is true for a natural number $k$ then it will be also true for $k+1$. This is same as assuming that if the next step exists then we can climb the ladder. Now from these two parts our statement becomes true $forall ninmathbb{N}, ngeq n_0$.






          share|cite|improve this answer












          When you think of induction, one of the best analogies to think about is a ladder. When you climb up the
          ladder, you have to step on the lower step and need to go up based on it. After we climb up the several
          steps, we can go up further by assuming that the step you are stepping on exists. Proof by induction works on a similar principle. First, we prove a given statement for a base case that is any natural number $n_0$, this is the lowest step of the ladder. Now we prove that whenever the statement is true for a natural number $k$ then it will be also true for $k+1$. This is same as assuming that if the next step exists then we can climb the ladder. Now from these two parts our statement becomes true $forall ninmathbb{N}, ngeq n_0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 22:27









          Jimmy

          17212




          17212























              0














              The few cases that you initially try don't prove the theorem—they show you what the pattern might be. Then, when you think you know the pattern, you express it as a theorem and use proof by induction to show that it's true.



              The idea is:



              First you show "If this is true for $n=k$, then it is also true for $n=k+1$", without specifying a value for $k$.



              Then you apply this to a case you already know to be true (typically the one for $n=1$ or 0) and effectively say "Since this case is true, so is the next one, and the next, and the next, and the next, all the way up to infinity". This is usually worded something like "But this is true for $n=1$. Therefore, by induction, it is true for all $n ge 1$."



              Note that having $k$ as a separate symbol from $n$ makes things a lot less confusing than trying to prove "if it's true for $n$, it's true for $n+1$", since you can be consistent about what $n$ is rather than having to remember what you're currently doing to it, at least in my experience.






              share|cite|improve this answer




























                0














                The few cases that you initially try don't prove the theorem—they show you what the pattern might be. Then, when you think you know the pattern, you express it as a theorem and use proof by induction to show that it's true.



                The idea is:



                First you show "If this is true for $n=k$, then it is also true for $n=k+1$", without specifying a value for $k$.



                Then you apply this to a case you already know to be true (typically the one for $n=1$ or 0) and effectively say "Since this case is true, so is the next one, and the next, and the next, and the next, all the way up to infinity". This is usually worded something like "But this is true for $n=1$. Therefore, by induction, it is true for all $n ge 1$."



                Note that having $k$ as a separate symbol from $n$ makes things a lot less confusing than trying to prove "if it's true for $n$, it's true for $n+1$", since you can be consistent about what $n$ is rather than having to remember what you're currently doing to it, at least in my experience.






                share|cite|improve this answer


























                  0












                  0








                  0






                  The few cases that you initially try don't prove the theorem—they show you what the pattern might be. Then, when you think you know the pattern, you express it as a theorem and use proof by induction to show that it's true.



                  The idea is:



                  First you show "If this is true for $n=k$, then it is also true for $n=k+1$", without specifying a value for $k$.



                  Then you apply this to a case you already know to be true (typically the one for $n=1$ or 0) and effectively say "Since this case is true, so is the next one, and the next, and the next, and the next, all the way up to infinity". This is usually worded something like "But this is true for $n=1$. Therefore, by induction, it is true for all $n ge 1$."



                  Note that having $k$ as a separate symbol from $n$ makes things a lot less confusing than trying to prove "if it's true for $n$, it's true for $n+1$", since you can be consistent about what $n$ is rather than having to remember what you're currently doing to it, at least in my experience.






                  share|cite|improve this answer














                  The few cases that you initially try don't prove the theorem—they show you what the pattern might be. Then, when you think you know the pattern, you express it as a theorem and use proof by induction to show that it's true.



                  The idea is:



                  First you show "If this is true for $n=k$, then it is also true for $n=k+1$", without specifying a value for $k$.



                  Then you apply this to a case you already know to be true (typically the one for $n=1$ or 0) and effectively say "Since this case is true, so is the next one, and the next, and the next, and the next, all the way up to infinity". This is usually worded something like "But this is true for $n=1$. Therefore, by induction, it is true for all $n ge 1$."



                  Note that having $k$ as a separate symbol from $n$ makes things a lot less confusing than trying to prove "if it's true for $n$, it's true for $n+1$", since you can be consistent about what $n$ is rather than having to remember what you're currently doing to it, at least in my experience.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 28 '18 at 1:42

























                  answered Nov 27 '18 at 23:04









                  timtfj

                  1,016318




                  1,016318






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016381%2fhow-many-cases-do-i-need-for-a-proof-by-induction%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Aardman Animations

                      Are they similar matrix

                      “minimization” problem in Euclidean space related to orthonormal basis