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Mathematical induction method
The Next CEO of Stack OverflowStrong Mathematical Induction: Why More than One Base Case?Prove by induction help?Formal definition of Mathematical Induction & Strong InductionHow can be done by the method of mathematical induction?Mathematical Induction - InequalityMathematical induction and pigeon-hole principleMathematical Induction with Sigma NotationLesson in an induction problemBy mathematical induction prove that?Chess table Mathematical Induction Problem
$begingroup$
In the method of induction we take for granted the proposition P(k) and we want to prove P(k+1). Can we start from P(k+1) and prove from that that P(k) is true, or is it a wrong practise?
discrete-mathematics
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In the method of induction we take for granted the proposition P(k) and we want to prove P(k+1). Can we start from P(k+1) and prove from that that P(k) is true, or is it a wrong practise?
discrete-mathematics
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
No, you need to show $P(k)implies P(k+1)$ or, which is equivalent, $neg P(k+1)impliesneg P(k)$. This valid variant on your invalid idea is related to a technique called infinite descent, in which we show $neg P(n)impliesexists k<n (neg P(k))$.
$endgroup$
– J.G.
Mar 28 at 10:59
1
$begingroup$
It is not "wrong practice", it proves something different.
$endgroup$
– Yves Daoust
Mar 28 at 11:01
add a comment |
$begingroup$
In the method of induction we take for granted the proposition P(k) and we want to prove P(k+1). Can we start from P(k+1) and prove from that that P(k) is true, or is it a wrong practise?
discrete-mathematics
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In the method of induction we take for granted the proposition P(k) and we want to prove P(k+1). Can we start from P(k+1) and prove from that that P(k) is true, or is it a wrong practise?
discrete-mathematics
discrete-mathematics
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
asked Mar 28 at 10:51
Toni IvanovToni Ivanov
1
1
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Toni Ivanov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
No, you need to show $P(k)implies P(k+1)$ or, which is equivalent, $neg P(k+1)impliesneg P(k)$. This valid variant on your invalid idea is related to a technique called infinite descent, in which we show $neg P(n)impliesexists k<n (neg P(k))$.
$endgroup$
– J.G.
Mar 28 at 10:59
1
$begingroup$
It is not "wrong practice", it proves something different.
$endgroup$
– Yves Daoust
Mar 28 at 11:01
add a comment |
1
$begingroup$
No, you need to show $P(k)implies P(k+1)$ or, which is equivalent, $neg P(k+1)impliesneg P(k)$. This valid variant on your invalid idea is related to a technique called infinite descent, in which we show $neg P(n)impliesexists k<n (neg P(k))$.
$endgroup$
– J.G.
Mar 28 at 10:59
1
$begingroup$
It is not "wrong practice", it proves something different.
$endgroup$
– Yves Daoust
Mar 28 at 11:01
1
1
$begingroup$
No, you need to show $P(k)implies P(k+1)$ or, which is equivalent, $neg P(k+1)impliesneg P(k)$. This valid variant on your invalid idea is related to a technique called infinite descent, in which we show $neg P(n)impliesexists k<n (neg P(k))$.
$endgroup$
– J.G.
Mar 28 at 10:59
$begingroup$
No, you need to show $P(k)implies P(k+1)$ or, which is equivalent, $neg P(k+1)impliesneg P(k)$. This valid variant on your invalid idea is related to a technique called infinite descent, in which we show $neg P(n)impliesexists k<n (neg P(k))$.
$endgroup$
– J.G.
Mar 28 at 10:59
1
1
$begingroup$
It is not "wrong practice", it proves something different.
$endgroup$
– Yves Daoust
Mar 28 at 11:01
$begingroup$
It is not "wrong practice", it proves something different.
$endgroup$
– Yves Daoust
Mar 28 at 11:01
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
That is wrong. Suppose, for instance that $T(n)$ is $nleqslant1$. Then we have $T(1)$ (since it means that $1leqslant1$). Now, suppost that we have $T(k)$; in other words, $kleqslant 1$. But then $k-1<kleqslant1$ and so $k-1leqslant 1$. So, I proved that $T(k)implies T(k-1)$. But I hope that I don't think that$$(forall ninmathbb N):nleqslant1$$is true.
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
It is wrong/useless. In mathematical induction you want to move forward, not backward.
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
$begingroup$
The principle of induction is
prove for some $k_0$;
prove that if true for $k$, then true for $k+1$.
Combining these two pieces, the result is guaranteed true for
$$k_0, k_0+1, k_0+2, k_0+3, cdots$$
If you reverse the inductive argument, the result is guaranteed true for
$$k_0, k_0-1, k_0-2, k_0-3, cdots$$
Often, the statement to prove doesn't make sense for $k<0$ or $kle0$, and usually you want to prove for all $kge k_0$, not the other way.
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That is wrong. Suppose, for instance that $T(n)$ is $nleqslant1$. Then we have $T(1)$ (since it means that $1leqslant1$). Now, suppost that we have $T(k)$; in other words, $kleqslant 1$. But then $k-1<kleqslant1$ and so $k-1leqslant 1$. So, I proved that $T(k)implies T(k-1)$. But I hope that I don't think that$$(forall ninmathbb N):nleqslant1$$is true.
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
That is wrong. Suppose, for instance that $T(n)$ is $nleqslant1$. Then we have $T(1)$ (since it means that $1leqslant1$). Now, suppost that we have $T(k)$; in other words, $kleqslant 1$. But then $k-1<kleqslant1$ and so $k-1leqslant 1$. So, I proved that $T(k)implies T(k-1)$. But I hope that I don't think that$$(forall ninmathbb N):nleqslant1$$is true.
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
That is wrong. Suppose, for instance that $T(n)$ is $nleqslant1$. Then we have $T(1)$ (since it means that $1leqslant1$). Now, suppost that we have $T(k)$; in other words, $kleqslant 1$. But then $k-1<kleqslant1$ and so $k-1leqslant 1$. So, I proved that $T(k)implies T(k-1)$. But I hope that I don't think that$$(forall ninmathbb N):nleqslant1$$is true.
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
That is wrong. Suppose, for instance that $T(n)$ is $nleqslant1$. Then we have $T(1)$ (since it means that $1leqslant1$). Now, suppost that we have $T(k)$; in other words, $kleqslant 1$. But then $k-1<kleqslant1$ and so $k-1leqslant 1$. So, I proved that $T(k)implies T(k-1)$. But I hope that I don't think that$$(forall ninmathbb N):nleqslant1$$is true.
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 28 at 10:55
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
It is wrong/useless. In mathematical induction you want to move forward, not backward.
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
$begingroup$
It is wrong/useless. In mathematical induction you want to move forward, not backward.
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
$endgroup$
add a comment |
$begingroup$
It is wrong/useless. In mathematical induction you want to move forward, not backward.
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
$endgroup$
It is wrong/useless. In mathematical induction you want to move forward, not backward.
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
answered Mar 28 at 10:54
PierreCarrePierreCarre
1,665212
1,665212
add a comment |
add a comment |
$begingroup$
The principle of induction is
prove for some $k_0$;
prove that if true for $k$, then true for $k+1$.
Combining these two pieces, the result is guaranteed true for
$$k_0, k_0+1, k_0+2, k_0+3, cdots$$
If you reverse the inductive argument, the result is guaranteed true for
$$k_0, k_0-1, k_0-2, k_0-3, cdots$$
Often, the statement to prove doesn't make sense for $k<0$ or $kle0$, and usually you want to prove for all $kge k_0$, not the other way.
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
The principle of induction is
prove for some $k_0$;
prove that if true for $k$, then true for $k+1$.
Combining these two pieces, the result is guaranteed true for
$$k_0, k_0+1, k_0+2, k_0+3, cdots$$
If you reverse the inductive argument, the result is guaranteed true for
$$k_0, k_0-1, k_0-2, k_0-3, cdots$$
Often, the statement to prove doesn't make sense for $k<0$ or $kle0$, and usually you want to prove for all $kge k_0$, not the other way.
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
$endgroup$
add a comment |
$begingroup$
The principle of induction is
prove for some $k_0$;
prove that if true for $k$, then true for $k+1$.
Combining these two pieces, the result is guaranteed true for
$$k_0, k_0+1, k_0+2, k_0+3, cdots$$
If you reverse the inductive argument, the result is guaranteed true for
$$k_0, k_0-1, k_0-2, k_0-3, cdots$$
Often, the statement to prove doesn't make sense for $k<0$ or $kle0$, and usually you want to prove for all $kge k_0$, not the other way.
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
$endgroup$
The principle of induction is
prove for some $k_0$;
prove that if true for $k$, then true for $k+1$.
Combining these two pieces, the result is guaranteed true for
$$k_0, k_0+1, k_0+2, k_0+3, cdots$$
If you reverse the inductive argument, the result is guaranteed true for
$$k_0, k_0-1, k_0-2, k_0-3, cdots$$
Often, the statement to prove doesn't make sense for $k<0$ or $kle0$, and usually you want to prove for all $kge k_0$, not the other way.
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
answered Mar 28 at 10:59
Yves DaoustYves Daoust
131k676229
131k676229
add a comment |
add a comment |
Toni Ivanov is a new contributor. Be nice, and check out our Code of Conduct.
Toni Ivanov is a new contributor. Be nice, and check out our Code of Conduct.
Toni Ivanov is a new contributor. Be nice, and check out our Code of Conduct.
Toni Ivanov is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
No, you need to show $P(k)implies P(k+1)$ or, which is equivalent, $neg P(k+1)impliesneg P(k)$. This valid variant on your invalid idea is related to a technique called infinite descent, in which we show $neg P(n)impliesexists k<n (neg P(k))$.
$endgroup$
– J.G.
Mar 28 at 10:59
1
$begingroup$
It is not "wrong practice", it proves something different.
$endgroup$
– Yves Daoust
Mar 28 at 11:01