Avoiding direct proof while writing proof by inductionFactorial (Proof by Induction)Proof by contradiction and mathematical inductionWhat would be the induction hypothesis in my proof?Proof by Induction of an inequality with a sumProof By Induction Summations, Factorials and Inequalitiessummation and inequality induction proofIs my proof by induction on binary trees correct?Bogus Proof by Strong InductionProof By Induction: Summation of Polynomialtricky summation proof by induction

Solving a recurrence relation (poker chips)

Examples of smooth manifolds admitting inbetween one and a continuum of complex structures

How do I deal with an unproductive colleague in a small company?

Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?

What method can I use to design a dungeon difficult enough that the PCs can't make it through without killing them?

How seriously should I take size and weight limits of hand luggage?

Bullying boss launched a smear campaign and made me unemployable

How dangerous is XSS?

Avoiding the "not like other girls" trope?

Why is this clock signal connected to a capacitor to gnd?

Can we compute the area of a quadrilateral with one right angle when we only know the lengths of any three sides?

Is there an expression that means doing something right before you will need it rather than doing it in case you might need it?

GFCI outlets - can they be repaired? Are they really needed at the end of a circuit?

Forgetting the musical notes while performing in concert

Short story with a alien planet, government officials must wear exploding medallions

How to prevent "they're falling in love" trope

Little known, relatively unlikely, but scientifically plausible, apocalyptic (or near apocalyptic) events

Is it logically or scientifically possible to artificially send energy to the body?

Why are the 737's rear doors unusable in a water landing?

Do scales need to be in alphabetical order?

Why no variance term in Bayesian logistic regression?

Can compressed videos be decoded back to their uncompresed original format?

How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?

Plagiarism or not?



Avoiding direct proof while writing proof by induction


Factorial (Proof by Induction)Proof by contradiction and mathematical inductionWhat would be the induction hypothesis in my proof?Proof by Induction of an inequality with a sumProof By Induction Summations, Factorials and Inequalitiessummation and inequality induction proofIs my proof by induction on binary trees correct?Bogus Proof by Strong InductionProof By Induction: Summation of Polynomialtricky summation proof by induction













3












$begingroup$


$$S_n = sum_i=0^n(5i+3)$$



I received a homework problem that instructed me to use induction to prove that for all natural numbers n



$$S_n = fracn(5n+11)2+3$$



First I proved that my base case of $S_0$ holds, because substituting $0$ for $n$ in both the top formula and the following formula makes both equal to $3$. The next step is to form my inductive hypothesis. My hypothesis is that



$$sum_i=0^n(5i+3) = fracn(5n+11)2+3$$ for all natural numbers $n$. Then I'm assuming that $$sum_i=0^k(5i+3) = frack(5k+11)2+3$$ holds when $n$ = some arbitrary natural number $k$ (I've since been told not to do $n=k$ for some reason).



Next step is to prove that $S_k+1$ holds, because if it does, knowing that my base case holds will tell me that $S_1$ holds, telling me that $S_2$ holds, etc.



To prove this, I took the equation from my assumption and substituted $k+1$ for $k$. Evaluating the left hand side of $frac(k+1)(5(k+1)+11)2+3$ eventually yielded $frac5k^2+21k+222$, and solving the right hand side of $sum_i=0^k+1(5i+3)$ using Gauss's(?) sum and splitting the terms of the sum (I don't know what to call it) to come to the same result. Since both sides of the equation reduced to the same expression, I reasoned that this proves that my original assumption holds, therefore the statement at the top has been proven.



I've gone wrong somewhere above, since I was told that I proved the original assertion with a direct proof rather than by induction. Where did I go wrong? I thought that after making my assumption and learning the case that needs to hold to make such assumption true, all I need to do is see if both sides of the equation equal each other. Has doing a direct proof of the original statement caused me to make too many assumptions? Or have I done something else inappropriate?










share|cite|improve this question











$endgroup$











  • $begingroup$
    You were told... by whom? Your proof seems to line up with induction nicely.
    $endgroup$
    – abiessu
    2 hours ago










  • $begingroup$
    @abiessu I was told this by my TA
    $endgroup$
    – user2709168
    2 hours ago















3












$begingroup$


$$S_n = sum_i=0^n(5i+3)$$



I received a homework problem that instructed me to use induction to prove that for all natural numbers n



$$S_n = fracn(5n+11)2+3$$



First I proved that my base case of $S_0$ holds, because substituting $0$ for $n$ in both the top formula and the following formula makes both equal to $3$. The next step is to form my inductive hypothesis. My hypothesis is that



$$sum_i=0^n(5i+3) = fracn(5n+11)2+3$$ for all natural numbers $n$. Then I'm assuming that $$sum_i=0^k(5i+3) = frack(5k+11)2+3$$ holds when $n$ = some arbitrary natural number $k$ (I've since been told not to do $n=k$ for some reason).



Next step is to prove that $S_k+1$ holds, because if it does, knowing that my base case holds will tell me that $S_1$ holds, telling me that $S_2$ holds, etc.



To prove this, I took the equation from my assumption and substituted $k+1$ for $k$. Evaluating the left hand side of $frac(k+1)(5(k+1)+11)2+3$ eventually yielded $frac5k^2+21k+222$, and solving the right hand side of $sum_i=0^k+1(5i+3)$ using Gauss's(?) sum and splitting the terms of the sum (I don't know what to call it) to come to the same result. Since both sides of the equation reduced to the same expression, I reasoned that this proves that my original assumption holds, therefore the statement at the top has been proven.



I've gone wrong somewhere above, since I was told that I proved the original assertion with a direct proof rather than by induction. Where did I go wrong? I thought that after making my assumption and learning the case that needs to hold to make such assumption true, all I need to do is see if both sides of the equation equal each other. Has doing a direct proof of the original statement caused me to make too many assumptions? Or have I done something else inappropriate?










share|cite|improve this question











$endgroup$











  • $begingroup$
    You were told... by whom? Your proof seems to line up with induction nicely.
    $endgroup$
    – abiessu
    2 hours ago










  • $begingroup$
    @abiessu I was told this by my TA
    $endgroup$
    – user2709168
    2 hours ago













3












3








3





$begingroup$


$$S_n = sum_i=0^n(5i+3)$$



I received a homework problem that instructed me to use induction to prove that for all natural numbers n



$$S_n = fracn(5n+11)2+3$$



First I proved that my base case of $S_0$ holds, because substituting $0$ for $n$ in both the top formula and the following formula makes both equal to $3$. The next step is to form my inductive hypothesis. My hypothesis is that



$$sum_i=0^n(5i+3) = fracn(5n+11)2+3$$ for all natural numbers $n$. Then I'm assuming that $$sum_i=0^k(5i+3) = frack(5k+11)2+3$$ holds when $n$ = some arbitrary natural number $k$ (I've since been told not to do $n=k$ for some reason).



Next step is to prove that $S_k+1$ holds, because if it does, knowing that my base case holds will tell me that $S_1$ holds, telling me that $S_2$ holds, etc.



To prove this, I took the equation from my assumption and substituted $k+1$ for $k$. Evaluating the left hand side of $frac(k+1)(5(k+1)+11)2+3$ eventually yielded $frac5k^2+21k+222$, and solving the right hand side of $sum_i=0^k+1(5i+3)$ using Gauss's(?) sum and splitting the terms of the sum (I don't know what to call it) to come to the same result. Since both sides of the equation reduced to the same expression, I reasoned that this proves that my original assumption holds, therefore the statement at the top has been proven.



I've gone wrong somewhere above, since I was told that I proved the original assertion with a direct proof rather than by induction. Where did I go wrong? I thought that after making my assumption and learning the case that needs to hold to make such assumption true, all I need to do is see if both sides of the equation equal each other. Has doing a direct proof of the original statement caused me to make too many assumptions? Or have I done something else inappropriate?










share|cite|improve this question











$endgroup$




$$S_n = sum_i=0^n(5i+3)$$



I received a homework problem that instructed me to use induction to prove that for all natural numbers n



$$S_n = fracn(5n+11)2+3$$



First I proved that my base case of $S_0$ holds, because substituting $0$ for $n$ in both the top formula and the following formula makes both equal to $3$. The next step is to form my inductive hypothesis. My hypothesis is that



$$sum_i=0^n(5i+3) = fracn(5n+11)2+3$$ for all natural numbers $n$. Then I'm assuming that $$sum_i=0^k(5i+3) = frack(5k+11)2+3$$ holds when $n$ = some arbitrary natural number $k$ (I've since been told not to do $n=k$ for some reason).



Next step is to prove that $S_k+1$ holds, because if it does, knowing that my base case holds will tell me that $S_1$ holds, telling me that $S_2$ holds, etc.



To prove this, I took the equation from my assumption and substituted $k+1$ for $k$. Evaluating the left hand side of $frac(k+1)(5(k+1)+11)2+3$ eventually yielded $frac5k^2+21k+222$, and solving the right hand side of $sum_i=0^k+1(5i+3)$ using Gauss's(?) sum and splitting the terms of the sum (I don't know what to call it) to come to the same result. Since both sides of the equation reduced to the same expression, I reasoned that this proves that my original assumption holds, therefore the statement at the top has been proven.



I've gone wrong somewhere above, since I was told that I proved the original assertion with a direct proof rather than by induction. Where did I go wrong? I thought that after making my assumption and learning the case that needs to hold to make such assumption true, all I need to do is see if both sides of the equation equal each other. Has doing a direct proof of the original statement caused me to make too many assumptions? Or have I done something else inappropriate?







proof-verification summation induction alternative-proof gauss-sums






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Eevee Trainer

9,51331740




9,51331740










asked 2 hours ago









user2709168user2709168

374




374











  • $begingroup$
    You were told... by whom? Your proof seems to line up with induction nicely.
    $endgroup$
    – abiessu
    2 hours ago










  • $begingroup$
    @abiessu I was told this by my TA
    $endgroup$
    – user2709168
    2 hours ago
















  • $begingroup$
    You were told... by whom? Your proof seems to line up with induction nicely.
    $endgroup$
    – abiessu
    2 hours ago










  • $begingroup$
    @abiessu I was told this by my TA
    $endgroup$
    – user2709168
    2 hours ago















$begingroup$
You were told... by whom? Your proof seems to line up with induction nicely.
$endgroup$
– abiessu
2 hours ago




$begingroup$
You were told... by whom? Your proof seems to line up with induction nicely.
$endgroup$
– abiessu
2 hours ago












$begingroup$
@abiessu I was told this by my TA
$endgroup$
– user2709168
2 hours ago




$begingroup$
@abiessu I was told this by my TA
$endgroup$
– user2709168
2 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Typically, you want to remember that, for proof by induction, you have to make use of the induction assumption. You assume some case greater than your base case holds, and then show it implies the succeeding step - that gives you the whole "$S_1 implies S_2 implies S_3 implies ...$" chain.



So our assumption is



$$sum_i=0^k(5i+3) = frack(5k+11)2+3$$



We seek to show



$$sum_i=0^k+1(5i+3) = frac(k+1)(5(k+1)+11)2+3 = frac(k+1)(5k+16)2+3$$



Starting with the sum at the left, we can pull out the $(k+1)^th$ term:



$$sum_i=0^k+1(5i+3) = 5(k+1) + 3 + sum_i=0^k(5i+3) = 5k+8 + sum_i=0^k(5i+3)$$



As it happens, this new summation is precisely what we assume holds. So we substitute the corresponding expression and do some algebra:



$$beginalign
5k+9 + sum_i=0^k(5i+3) &= 5k+8 + frack(5k+11)2+3\
&=frac10k+16 + 5k^2 + 11k2 + 3\
&=frac5k^2+21k+162 + 3\
&= frac(k+1)(5k+16)2+3
endalign$$



Thus, the case for $(k+1)$ holds, completing the induction step.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds?
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is.
    $endgroup$
    – Eevee Trainer
    2 hours ago










  • $begingroup$
    Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2."
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_kimplies S_k+1$.
    $endgroup$
    – Eevee Trainer
    2 hours ago











  • $begingroup$
    What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n
    $endgroup$
    – user2709168
    1 hour ago


















3












$begingroup$

I think the place where you say you used the "Gauss sum" is where your instructor says you just gave a direct proof. It's hard to tell, because you didn't show us your proof, you just said "and then I did and then ...".



What's expected is that you write the result for a particular value of $k$ - the inductive hypothesis, then add the next term and do some algebra to show that you get the result for $k+1$.



As an aside, I really don't like a question that asks you to prove something by induction when there is an easier straightforward way - in this case, Gauss's method.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1.
    $endgroup$
    – user2709168
    2 hours ago











  • $begingroup$
    I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments.
    $endgroup$
    – Ethan Bolker
    1 hour ago











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%2f3174030%2favoiding-direct-proof-while-writing-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









3












$begingroup$

Typically, you want to remember that, for proof by induction, you have to make use of the induction assumption. You assume some case greater than your base case holds, and then show it implies the succeeding step - that gives you the whole "$S_1 implies S_2 implies S_3 implies ...$" chain.



So our assumption is



$$sum_i=0^k(5i+3) = frack(5k+11)2+3$$



We seek to show



$$sum_i=0^k+1(5i+3) = frac(k+1)(5(k+1)+11)2+3 = frac(k+1)(5k+16)2+3$$



Starting with the sum at the left, we can pull out the $(k+1)^th$ term:



$$sum_i=0^k+1(5i+3) = 5(k+1) + 3 + sum_i=0^k(5i+3) = 5k+8 + sum_i=0^k(5i+3)$$



As it happens, this new summation is precisely what we assume holds. So we substitute the corresponding expression and do some algebra:



$$beginalign
5k+9 + sum_i=0^k(5i+3) &= 5k+8 + frack(5k+11)2+3\
&=frac10k+16 + 5k^2 + 11k2 + 3\
&=frac5k^2+21k+162 + 3\
&= frac(k+1)(5k+16)2+3
endalign$$



Thus, the case for $(k+1)$ holds, completing the induction step.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds?
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is.
    $endgroup$
    – Eevee Trainer
    2 hours ago










  • $begingroup$
    Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2."
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_kimplies S_k+1$.
    $endgroup$
    – Eevee Trainer
    2 hours ago











  • $begingroup$
    What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n
    $endgroup$
    – user2709168
    1 hour ago















3












$begingroup$

Typically, you want to remember that, for proof by induction, you have to make use of the induction assumption. You assume some case greater than your base case holds, and then show it implies the succeeding step - that gives you the whole "$S_1 implies S_2 implies S_3 implies ...$" chain.



So our assumption is



$$sum_i=0^k(5i+3) = frack(5k+11)2+3$$



We seek to show



$$sum_i=0^k+1(5i+3) = frac(k+1)(5(k+1)+11)2+3 = frac(k+1)(5k+16)2+3$$



Starting with the sum at the left, we can pull out the $(k+1)^th$ term:



$$sum_i=0^k+1(5i+3) = 5(k+1) + 3 + sum_i=0^k(5i+3) = 5k+8 + sum_i=0^k(5i+3)$$



As it happens, this new summation is precisely what we assume holds. So we substitute the corresponding expression and do some algebra:



$$beginalign
5k+9 + sum_i=0^k(5i+3) &= 5k+8 + frack(5k+11)2+3\
&=frac10k+16 + 5k^2 + 11k2 + 3\
&=frac5k^2+21k+162 + 3\
&= frac(k+1)(5k+16)2+3
endalign$$



Thus, the case for $(k+1)$ holds, completing the induction step.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds?
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is.
    $endgroup$
    – Eevee Trainer
    2 hours ago










  • $begingroup$
    Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2."
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_kimplies S_k+1$.
    $endgroup$
    – Eevee Trainer
    2 hours ago











  • $begingroup$
    What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n
    $endgroup$
    – user2709168
    1 hour ago













3












3








3





$begingroup$

Typically, you want to remember that, for proof by induction, you have to make use of the induction assumption. You assume some case greater than your base case holds, and then show it implies the succeeding step - that gives you the whole "$S_1 implies S_2 implies S_3 implies ...$" chain.



So our assumption is



$$sum_i=0^k(5i+3) = frack(5k+11)2+3$$



We seek to show



$$sum_i=0^k+1(5i+3) = frac(k+1)(5(k+1)+11)2+3 = frac(k+1)(5k+16)2+3$$



Starting with the sum at the left, we can pull out the $(k+1)^th$ term:



$$sum_i=0^k+1(5i+3) = 5(k+1) + 3 + sum_i=0^k(5i+3) = 5k+8 + sum_i=0^k(5i+3)$$



As it happens, this new summation is precisely what we assume holds. So we substitute the corresponding expression and do some algebra:



$$beginalign
5k+9 + sum_i=0^k(5i+3) &= 5k+8 + frack(5k+11)2+3\
&=frac10k+16 + 5k^2 + 11k2 + 3\
&=frac5k^2+21k+162 + 3\
&= frac(k+1)(5k+16)2+3
endalign$$



Thus, the case for $(k+1)$ holds, completing the induction step.






share|cite|improve this answer









$endgroup$



Typically, you want to remember that, for proof by induction, you have to make use of the induction assumption. You assume some case greater than your base case holds, and then show it implies the succeeding step - that gives you the whole "$S_1 implies S_2 implies S_3 implies ...$" chain.



So our assumption is



$$sum_i=0^k(5i+3) = frack(5k+11)2+3$$



We seek to show



$$sum_i=0^k+1(5i+3) = frac(k+1)(5(k+1)+11)2+3 = frac(k+1)(5k+16)2+3$$



Starting with the sum at the left, we can pull out the $(k+1)^th$ term:



$$sum_i=0^k+1(5i+3) = 5(k+1) + 3 + sum_i=0^k(5i+3) = 5k+8 + sum_i=0^k(5i+3)$$



As it happens, this new summation is precisely what we assume holds. So we substitute the corresponding expression and do some algebra:



$$beginalign
5k+9 + sum_i=0^k(5i+3) &= 5k+8 + frack(5k+11)2+3\
&=frac10k+16 + 5k^2 + 11k2 + 3\
&=frac5k^2+21k+162 + 3\
&= frac(k+1)(5k+16)2+3
endalign$$



Thus, the case for $(k+1)$ holds, completing the induction step.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Eevee TrainerEevee Trainer

9,51331740




9,51331740











  • $begingroup$
    I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds?
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is.
    $endgroup$
    – Eevee Trainer
    2 hours ago










  • $begingroup$
    Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2."
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_kimplies S_k+1$.
    $endgroup$
    – Eevee Trainer
    2 hours ago











  • $begingroup$
    What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n
    $endgroup$
    – user2709168
    1 hour ago
















  • $begingroup$
    I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds?
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is.
    $endgroup$
    – Eevee Trainer
    2 hours ago










  • $begingroup$
    Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2."
    $endgroup$
    – user2709168
    2 hours ago










  • $begingroup$
    I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_kimplies S_k+1$.
    $endgroup$
    – Eevee Trainer
    2 hours ago











  • $begingroup$
    What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n
    $endgroup$
    – user2709168
    1 hour ago















$begingroup$
I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds?
$endgroup$
– user2709168
2 hours ago




$begingroup$
I think I mixed up my expressions in the post, but my intention was to have what you had as your assumption(?) as my inductive hypothesis. Do I not use that hypothesis when proving that the k+1 substitution holds?
$endgroup$
– user2709168
2 hours ago












$begingroup$
You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is.
$endgroup$
– Eevee Trainer
2 hours ago




$begingroup$
You used something you refer to as "Gauss's (?) sum" in that, so, no, you did not make use of your induction hypothesis. At least in any obvious way because I have no idea what this sum you refer to is.
$endgroup$
– Eevee Trainer
2 hours ago












$begingroup$
Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2."
$endgroup$
– user2709168
2 hours ago




$begingroup$
Are you saying my inductive hypothesis was Gauss's sum? Because that's not what I thought I was asserting Copy pasting a different comment of mine explaining what I meant: "I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2."
$endgroup$
– user2709168
2 hours ago












$begingroup$
I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_kimplies S_k+1$.
$endgroup$
– Eevee Trainer
2 hours ago





$begingroup$
I'm saying that you're not making use of the inductive hypothesis. You verified the inductive step by another method, which makes no use of the inductive hypothesis. You have to assume the inductive hypothesis holds when you verify the inductive step: that's the whole point of the "this implies that implies that" domino effect. Alongside the base case and the fact that one implies the next - and you have to have a step implying the next, and have to show that implication holds - that gives us the domino effect. Verifying the induction step independently does not show $S_kimplies S_k+1$.
$endgroup$
– Eevee Trainer
2 hours ago













$begingroup$
What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n
$endgroup$
– user2709168
1 hour ago




$begingroup$
What is my inductive hypothesis in this situation? I thought the hypothesis was that the statement holds when n=k, therefore by proving it holds for k+1 then it holds for all n
$endgroup$
– user2709168
1 hour ago











3












$begingroup$

I think the place where you say you used the "Gauss sum" is where your instructor says you just gave a direct proof. It's hard to tell, because you didn't show us your proof, you just said "and then I did and then ...".



What's expected is that you write the result for a particular value of $k$ - the inductive hypothesis, then add the next term and do some algebra to show that you get the result for $k+1$.



As an aside, I really don't like a question that asks you to prove something by induction when there is an easier straightforward way - in this case, Gauss's method.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1.
    $endgroup$
    – user2709168
    2 hours ago











  • $begingroup$
    I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments.
    $endgroup$
    – Ethan Bolker
    1 hour ago















3












$begingroup$

I think the place where you say you used the "Gauss sum" is where your instructor says you just gave a direct proof. It's hard to tell, because you didn't show us your proof, you just said "and then I did and then ...".



What's expected is that you write the result for a particular value of $k$ - the inductive hypothesis, then add the next term and do some algebra to show that you get the result for $k+1$.



As an aside, I really don't like a question that asks you to prove something by induction when there is an easier straightforward way - in this case, Gauss's method.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1.
    $endgroup$
    – user2709168
    2 hours ago











  • $begingroup$
    I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments.
    $endgroup$
    – Ethan Bolker
    1 hour ago













3












3








3





$begingroup$

I think the place where you say you used the "Gauss sum" is where your instructor says you just gave a direct proof. It's hard to tell, because you didn't show us your proof, you just said "and then I did and then ...".



What's expected is that you write the result for a particular value of $k$ - the inductive hypothesis, then add the next term and do some algebra to show that you get the result for $k+1$.



As an aside, I really don't like a question that asks you to prove something by induction when there is an easier straightforward way - in this case, Gauss's method.






share|cite|improve this answer









$endgroup$



I think the place where you say you used the "Gauss sum" is where your instructor says you just gave a direct proof. It's hard to tell, because you didn't show us your proof, you just said "and then I did and then ...".



What's expected is that you write the result for a particular value of $k$ - the inductive hypothesis, then add the next term and do some algebra to show that you get the result for $k+1$.



As an aside, I really don't like a question that asks you to prove something by induction when there is an easier straightforward way - in this case, Gauss's method.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Ethan BolkerEthan Bolker

45.5k553120




45.5k553120











  • $begingroup$
    What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1.
    $endgroup$
    – user2709168
    2 hours ago











  • $begingroup$
    I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments.
    $endgroup$
    – Ethan Bolker
    1 hour ago
















  • $begingroup$
    What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1.
    $endgroup$
    – user2709168
    2 hours ago











  • $begingroup$
    I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments.
    $endgroup$
    – Ethan Bolker
    1 hour ago















$begingroup$
What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1.
$endgroup$
– user2709168
2 hours ago





$begingroup$
What do you mean by a particular value of k? I mentioned Gauss's sum because that's one of the things I used to evaluate the right side of my equation- through turning the sum of 5i into 5((k+1)(k+2))/2. I thought the particular value was writing that the statement holds when the value of k is n (or the other way around? trying to figure out what supposedly went wrong is confusing me), and then I can prove I get the same result for k+1.
$endgroup$
– user2709168
2 hours ago













$begingroup$
I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments.
$endgroup$
– Ethan Bolker
1 hour ago




$begingroup$
I can't explain any better what I mean than what is in @EeveeTrainer 's answer and comments.
$endgroup$
– Ethan Bolker
1 hour ago

















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.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3174030%2favoiding-direct-proof-while-writing-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

Oświęcim Innehåll Historia | Källor | Externa länkar | Navigeringsmeny50°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.2213950°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.221393089658Nordisk familjebok, AuschwitzInsidan tro och existensJewish Community i OświęcimAuschwitz Jewish Center: MuseumAuschwitz Jewish Center

Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

Typsetting diagram chases (with TikZ?) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to define the default vertical distance between nodes?Draw edge on arcNumerical conditional within tikz keys?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to place nodes in an absolute coordinate system in tikzCommutative diagram with curve connecting between nodesTikz with standalone: pinning tikz coordinates to page cmDrawing a Decision Diagram with Tikz and layout manager