Proof regarding the square root of a natural number $m>1$ and a prime number $p$, when $p nmid m$Proof regarding prime numbers:The contradiction method used to prove that the square root of a prime is irrationalLet $p$ be a prime. If $pmid(a_1a_2cdots a_n)$ then there exists an $i$ with $1 leq i leq n$ such that $p|a_i$.If $a$ and $b$ are relatively prime then so are $a$ and $b^n$ for every positive integer $n$The square root of a prime number is irrationalProof by induction with a prime numberHow is induction used to prove that $gcd(an, bn) = |n|gcd(a, b)$Proof: the square root of the product of two distinct primes is irrational$forall kin mathbbN^+, forall min mathbbN, (k = 2m + 1) implies 11|(10^k + 1)$ - ProofFor any prime numbers $p$, any integers $a$, any natural numbers $k$, if $p|a^k$ then $p|a$
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Proof regarding the square root of a natural number $m>1$ and a prime number $p$, when $p nmid m$
Proof regarding prime numbers:The contradiction method used to prove that the square root of a prime is irrationalLet $p$ be a prime. If $pmid(a_1a_2cdots a_n)$ then there exists an $i$ with $1 leq i leq n$ such that $p|a_i$.If $a$ and $b$ are relatively prime then so are $a$ and $b^n$ for every positive integer $n$The square root of a prime number is irrationalProof by induction with a prime numberHow is induction used to prove that $gcd(an, bn) = |n|gcd(a, b)$Proof: the square root of the product of two distinct primes is irrational$forall kin mathbbN^+, forall min mathbbN, (k = 2m + 1) implies 11|(10^k + 1)$ - ProofFor any prime numbers $p$, any integers $a$, any natural numbers $k$, if $p|a^k$ then $p|a$
$begingroup$
I am self studying from "A logical introduction to proof" by Daniel W.Cunningham,one of those textbooks that unfortunately do not contain answers to end-of-chapter exercises.Could someone help me with the following problem?
Let p be a prime number and m>1 be a natural number.Prove that if $pnmid m$ then $sqrtmp$ is irrational.
My attempt:
Let p be a prime number and m $in BbbN $ where m>1 (i.e.m>=2).The proof proceeds by regular mathematical induction.
The base step invovles proving that (p $nmid 2$) $Rightarrow$ ($sqrt2p $ is irrational).By contraposition,we need to prove that ($sqrt2p $ is rational) $Rightarrow$ ($pmid 2$).By defintion $sqrt2p=aover b $ for some integers a and b where b$neq 0 $ and gcd(a,b)=1.Therefore $2p=a^2over b^2 $ and hence $2b^2 p=a^2 $ by squaring both sides and subsequent multplication.
Furthermore, $pmid a^2$ and by implication $pmid a$ which by definition suggests $a=pk$ for some k $in BbbZ$. Consequently $2b^2 p=p^2k^2 $ then $2b^2 = pk^2$ hence $p mid 2b^2$ which by Euclid's Lemma suggests either (i) $p mid 2$ or (ii) $p mid b^2$ .
Now in the case where (i) $pmid 2$,the conclusion to the base step has been verified.
Consider the case that (ii) $pmid b^2$,then by the corrolary to Euclid's Lemma, $pmid b$ as well.With $pmid b$ and $pmid a$,we have a prime number p that divides both a and b when by assumption gcd(a,b)=1.This implies that $pmid b$ is false and hence that $pmid b^2$ is false by contraposition,leaving us with case(i) as verified above.
Does this prove the base step?
Regarding the inductive step i do not know how to link the induction conclusion to the induction hypothesis...it is as if the conclusion can be arrived at without even considering the hypothesis in the inductive step,which to my understanding means the 'dominos have not been aligned' in the typical induction-dominos imagery.
Any help would be great.
Thanks!
abstract-algebra elementary-number-theory proof-writing
$endgroup$
add a comment |
$begingroup$
I am self studying from "A logical introduction to proof" by Daniel W.Cunningham,one of those textbooks that unfortunately do not contain answers to end-of-chapter exercises.Could someone help me with the following problem?
Let p be a prime number and m>1 be a natural number.Prove that if $pnmid m$ then $sqrtmp$ is irrational.
My attempt:
Let p be a prime number and m $in BbbN $ where m>1 (i.e.m>=2).The proof proceeds by regular mathematical induction.
The base step invovles proving that (p $nmid 2$) $Rightarrow$ ($sqrt2p $ is irrational).By contraposition,we need to prove that ($sqrt2p $ is rational) $Rightarrow$ ($pmid 2$).By defintion $sqrt2p=aover b $ for some integers a and b where b$neq 0 $ and gcd(a,b)=1.Therefore $2p=a^2over b^2 $ and hence $2b^2 p=a^2 $ by squaring both sides and subsequent multplication.
Furthermore, $pmid a^2$ and by implication $pmid a$ which by definition suggests $a=pk$ for some k $in BbbZ$. Consequently $2b^2 p=p^2k^2 $ then $2b^2 = pk^2$ hence $p mid 2b^2$ which by Euclid's Lemma suggests either (i) $p mid 2$ or (ii) $p mid b^2$ .
Now in the case where (i) $pmid 2$,the conclusion to the base step has been verified.
Consider the case that (ii) $pmid b^2$,then by the corrolary to Euclid's Lemma, $pmid b$ as well.With $pmid b$ and $pmid a$,we have a prime number p that divides both a and b when by assumption gcd(a,b)=1.This implies that $pmid b$ is false and hence that $pmid b^2$ is false by contraposition,leaving us with case(i) as verified above.
Does this prove the base step?
Regarding the inductive step i do not know how to link the induction conclusion to the induction hypothesis...it is as if the conclusion can be arrived at without even considering the hypothesis in the inductive step,which to my understanding means the 'dominos have not been aligned' in the typical induction-dominos imagery.
Any help would be great.
Thanks!
abstract-algebra elementary-number-theory proof-writing
$endgroup$
2
$begingroup$
As there is no clear route from $sqrtmpnotin Bbb Q$ to $sqrt(m+1)pnotin Bbb Q$ it seems that this approach is not ideal.
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
yikes...since the question appears in the induction chapters...i just thought this approach would be fine...Also,how can you tell that there is no clear route??? Thank you for the reply.
$endgroup$
– HalfAFoot
2 days ago
add a comment |
$begingroup$
I am self studying from "A logical introduction to proof" by Daniel W.Cunningham,one of those textbooks that unfortunately do not contain answers to end-of-chapter exercises.Could someone help me with the following problem?
Let p be a prime number and m>1 be a natural number.Prove that if $pnmid m$ then $sqrtmp$ is irrational.
My attempt:
Let p be a prime number and m $in BbbN $ where m>1 (i.e.m>=2).The proof proceeds by regular mathematical induction.
The base step invovles proving that (p $nmid 2$) $Rightarrow$ ($sqrt2p $ is irrational).By contraposition,we need to prove that ($sqrt2p $ is rational) $Rightarrow$ ($pmid 2$).By defintion $sqrt2p=aover b $ for some integers a and b where b$neq 0 $ and gcd(a,b)=1.Therefore $2p=a^2over b^2 $ and hence $2b^2 p=a^2 $ by squaring both sides and subsequent multplication.
Furthermore, $pmid a^2$ and by implication $pmid a$ which by definition suggests $a=pk$ for some k $in BbbZ$. Consequently $2b^2 p=p^2k^2 $ then $2b^2 = pk^2$ hence $p mid 2b^2$ which by Euclid's Lemma suggests either (i) $p mid 2$ or (ii) $p mid b^2$ .
Now in the case where (i) $pmid 2$,the conclusion to the base step has been verified.
Consider the case that (ii) $pmid b^2$,then by the corrolary to Euclid's Lemma, $pmid b$ as well.With $pmid b$ and $pmid a$,we have a prime number p that divides both a and b when by assumption gcd(a,b)=1.This implies that $pmid b$ is false and hence that $pmid b^2$ is false by contraposition,leaving us with case(i) as verified above.
Does this prove the base step?
Regarding the inductive step i do not know how to link the induction conclusion to the induction hypothesis...it is as if the conclusion can be arrived at without even considering the hypothesis in the inductive step,which to my understanding means the 'dominos have not been aligned' in the typical induction-dominos imagery.
Any help would be great.
Thanks!
abstract-algebra elementary-number-theory proof-writing
$endgroup$
I am self studying from "A logical introduction to proof" by Daniel W.Cunningham,one of those textbooks that unfortunately do not contain answers to end-of-chapter exercises.Could someone help me with the following problem?
Let p be a prime number and m>1 be a natural number.Prove that if $pnmid m$ then $sqrtmp$ is irrational.
My attempt:
Let p be a prime number and m $in BbbN $ where m>1 (i.e.m>=2).The proof proceeds by regular mathematical induction.
The base step invovles proving that (p $nmid 2$) $Rightarrow$ ($sqrt2p $ is irrational).By contraposition,we need to prove that ($sqrt2p $ is rational) $Rightarrow$ ($pmid 2$).By defintion $sqrt2p=aover b $ for some integers a and b where b$neq 0 $ and gcd(a,b)=1.Therefore $2p=a^2over b^2 $ and hence $2b^2 p=a^2 $ by squaring both sides and subsequent multplication.
Furthermore, $pmid a^2$ and by implication $pmid a$ which by definition suggests $a=pk$ for some k $in BbbZ$. Consequently $2b^2 p=p^2k^2 $ then $2b^2 = pk^2$ hence $p mid 2b^2$ which by Euclid's Lemma suggests either (i) $p mid 2$ or (ii) $p mid b^2$ .
Now in the case where (i) $pmid 2$,the conclusion to the base step has been verified.
Consider the case that (ii) $pmid b^2$,then by the corrolary to Euclid's Lemma, $pmid b$ as well.With $pmid b$ and $pmid a$,we have a prime number p that divides both a and b when by assumption gcd(a,b)=1.This implies that $pmid b$ is false and hence that $pmid b^2$ is false by contraposition,leaving us with case(i) as verified above.
Does this prove the base step?
Regarding the inductive step i do not know how to link the induction conclusion to the induction hypothesis...it is as if the conclusion can be arrived at without even considering the hypothesis in the inductive step,which to my understanding means the 'dominos have not been aligned' in the typical induction-dominos imagery.
Any help would be great.
Thanks!
abstract-algebra elementary-number-theory proof-writing
abstract-algebra elementary-number-theory proof-writing
edited 2 days ago
rtybase
11.4k31533
11.4k31533
asked 2 days ago
HalfAFootHalfAFoot
134
134
2
$begingroup$
As there is no clear route from $sqrtmpnotin Bbb Q$ to $sqrt(m+1)pnotin Bbb Q$ it seems that this approach is not ideal.
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
yikes...since the question appears in the induction chapters...i just thought this approach would be fine...Also,how can you tell that there is no clear route??? Thank you for the reply.
$endgroup$
– HalfAFoot
2 days ago
add a comment |
2
$begingroup$
As there is no clear route from $sqrtmpnotin Bbb Q$ to $sqrt(m+1)pnotin Bbb Q$ it seems that this approach is not ideal.
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
yikes...since the question appears in the induction chapters...i just thought this approach would be fine...Also,how can you tell that there is no clear route??? Thank you for the reply.
$endgroup$
– HalfAFoot
2 days ago
2
2
$begingroup$
As there is no clear route from $sqrtmpnotin Bbb Q$ to $sqrt(m+1)pnotin Bbb Q$ it seems that this approach is not ideal.
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
As there is no clear route from $sqrtmpnotin Bbb Q$ to $sqrt(m+1)pnotin Bbb Q$ it seems that this approach is not ideal.
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
yikes...since the question appears in the induction chapters...i just thought this approach would be fine...Also,how can you tell that there is no clear route??? Thank you for the reply.
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
yikes...since the question appears in the induction chapters...i just thought this approach would be fine...Also,how can you tell that there is no clear route??? Thank you for the reply.
$endgroup$
– HalfAFoot
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's not a good idea to try induction to prove this because induction will prove it for ALL $m$, but this is false if $p mid m$.
Your proof works more directly for $m$:
Suppose $sqrtpm=fracab$ for some coprime positive integers $a$ and $b$. Then $b^2 p m = a^2$. Then $p mid a^2$. As $p$ is prime we have $pmid a$, that is $a=p k$ for some integer $k$. Then $b^2 p m = p^2 k^2$, so $b^2 m = pk^2$. This means $pmid b^2m$. Again, as $p$ is prime this implies $pmid b$ or $pmid m$, but the hypothesis tell us $p notmid m$, so we must have $pmid b$. So we now have $pmid a$ and $pmid b$, a contradiction to the fact that $a$ and $b$ are coprime.
$endgroup$
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
add a comment |
$begingroup$
You actually don’t need to recur, the proof is standard, and almost the same as the proof that $sqrt2$ is irrationnal.
Let’s assume there is a $x = fracab$ so that $x^2 = mp$ and $pgcd(a,b) =1$
Then $fraca^2b^2 = mp$
$a^2 = mp b^2$
then
either p is a prime factor of b, which means it is not one of a (since $pgcd(a,b) = 1$) and therefore $a^2 = m p b^2$ is impossible
or p is not a prime factor of b. since
$a^2 = mp b^2$, we have $p | a^2$ so $p | a$ ($p|xy$ implies $p|x$ or $p|y$ when $p$ is prime, by euclide’s theorem. We just set $x = y = a$)
then $p^2 | a^2$
then $p^2 | m p b^2$, that is $p | mb^2$ which contradicts with our hypothesis
$endgroup$
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Generally, induction is appropriate when it seems easier to prove the property knowing that it is true for a smaller m. The way of thinking you can adopt is : « if needed, I can assume the property is true for any smaller value. ». Then, if you actually use that hypothesis, you are using induction, if you don’t, then you didn’t need it
$endgroup$
– Nephanth
2 days ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
It's not a good idea to try induction to prove this because induction will prove it for ALL $m$, but this is false if $p mid m$.
Your proof works more directly for $m$:
Suppose $sqrtpm=fracab$ for some coprime positive integers $a$ and $b$. Then $b^2 p m = a^2$. Then $p mid a^2$. As $p$ is prime we have $pmid a$, that is $a=p k$ for some integer $k$. Then $b^2 p m = p^2 k^2$, so $b^2 m = pk^2$. This means $pmid b^2m$. Again, as $p$ is prime this implies $pmid b$ or $pmid m$, but the hypothesis tell us $p notmid m$, so we must have $pmid b$. So we now have $pmid a$ and $pmid b$, a contradiction to the fact that $a$ and $b$ are coprime.
$endgroup$
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
add a comment |
$begingroup$
It's not a good idea to try induction to prove this because induction will prove it for ALL $m$, but this is false if $p mid m$.
Your proof works more directly for $m$:
Suppose $sqrtpm=fracab$ for some coprime positive integers $a$ and $b$. Then $b^2 p m = a^2$. Then $p mid a^2$. As $p$ is prime we have $pmid a$, that is $a=p k$ for some integer $k$. Then $b^2 p m = p^2 k^2$, so $b^2 m = pk^2$. This means $pmid b^2m$. Again, as $p$ is prime this implies $pmid b$ or $pmid m$, but the hypothesis tell us $p notmid m$, so we must have $pmid b$. So we now have $pmid a$ and $pmid b$, a contradiction to the fact that $a$ and $b$ are coprime.
$endgroup$
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
add a comment |
$begingroup$
It's not a good idea to try induction to prove this because induction will prove it for ALL $m$, but this is false if $p mid m$.
Your proof works more directly for $m$:
Suppose $sqrtpm=fracab$ for some coprime positive integers $a$ and $b$. Then $b^2 p m = a^2$. Then $p mid a^2$. As $p$ is prime we have $pmid a$, that is $a=p k$ for some integer $k$. Then $b^2 p m = p^2 k^2$, so $b^2 m = pk^2$. This means $pmid b^2m$. Again, as $p$ is prime this implies $pmid b$ or $pmid m$, but the hypothesis tell us $p notmid m$, so we must have $pmid b$. So we now have $pmid a$ and $pmid b$, a contradiction to the fact that $a$ and $b$ are coprime.
$endgroup$
It's not a good idea to try induction to prove this because induction will prove it for ALL $m$, but this is false if $p mid m$.
Your proof works more directly for $m$:
Suppose $sqrtpm=fracab$ for some coprime positive integers $a$ and $b$. Then $b^2 p m = a^2$. Then $p mid a^2$. As $p$ is prime we have $pmid a$, that is $a=p k$ for some integer $k$. Then $b^2 p m = p^2 k^2$, so $b^2 m = pk^2$. This means $pmid b^2m$. Again, as $p$ is prime this implies $pmid b$ or $pmid m$, but the hypothesis tell us $p notmid m$, so we must have $pmid b$. So we now have $pmid a$ and $pmid b$, a contradiction to the fact that $a$ and $b$ are coprime.
edited 2 days ago
John Omielan
3,9151215
3,9151215
answered 2 days ago
jjagmathjjagmath
3837
3837
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
add a comment |
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
add a comment |
$begingroup$
You actually don’t need to recur, the proof is standard, and almost the same as the proof that $sqrt2$ is irrationnal.
Let’s assume there is a $x = fracab$ so that $x^2 = mp$ and $pgcd(a,b) =1$
Then $fraca^2b^2 = mp$
$a^2 = mp b^2$
then
either p is a prime factor of b, which means it is not one of a (since $pgcd(a,b) = 1$) and therefore $a^2 = m p b^2$ is impossible
or p is not a prime factor of b. since
$a^2 = mp b^2$, we have $p | a^2$ so $p | a$ ($p|xy$ implies $p|x$ or $p|y$ when $p$ is prime, by euclide’s theorem. We just set $x = y = a$)
then $p^2 | a^2$
then $p^2 | m p b^2$, that is $p | mb^2$ which contradicts with our hypothesis
$endgroup$
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Generally, induction is appropriate when it seems easier to prove the property knowing that it is true for a smaller m. The way of thinking you can adopt is : « if needed, I can assume the property is true for any smaller value. ». Then, if you actually use that hypothesis, you are using induction, if you don’t, then you didn’t need it
$endgroup$
– Nephanth
2 days ago
add a comment |
$begingroup$
You actually don’t need to recur, the proof is standard, and almost the same as the proof that $sqrt2$ is irrationnal.
Let’s assume there is a $x = fracab$ so that $x^2 = mp$ and $pgcd(a,b) =1$
Then $fraca^2b^2 = mp$
$a^2 = mp b^2$
then
either p is a prime factor of b, which means it is not one of a (since $pgcd(a,b) = 1$) and therefore $a^2 = m p b^2$ is impossible
or p is not a prime factor of b. since
$a^2 = mp b^2$, we have $p | a^2$ so $p | a$ ($p|xy$ implies $p|x$ or $p|y$ when $p$ is prime, by euclide’s theorem. We just set $x = y = a$)
then $p^2 | a^2$
then $p^2 | m p b^2$, that is $p | mb^2$ which contradicts with our hypothesis
$endgroup$
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Generally, induction is appropriate when it seems easier to prove the property knowing that it is true for a smaller m. The way of thinking you can adopt is : « if needed, I can assume the property is true for any smaller value. ». Then, if you actually use that hypothesis, you are using induction, if you don’t, then you didn’t need it
$endgroup$
– Nephanth
2 days ago
add a comment |
$begingroup$
You actually don’t need to recur, the proof is standard, and almost the same as the proof that $sqrt2$ is irrationnal.
Let’s assume there is a $x = fracab$ so that $x^2 = mp$ and $pgcd(a,b) =1$
Then $fraca^2b^2 = mp$
$a^2 = mp b^2$
then
either p is a prime factor of b, which means it is not one of a (since $pgcd(a,b) = 1$) and therefore $a^2 = m p b^2$ is impossible
or p is not a prime factor of b. since
$a^2 = mp b^2$, we have $p | a^2$ so $p | a$ ($p|xy$ implies $p|x$ or $p|y$ when $p$ is prime, by euclide’s theorem. We just set $x = y = a$)
then $p^2 | a^2$
then $p^2 | m p b^2$, that is $p | mb^2$ which contradicts with our hypothesis
$endgroup$
You actually don’t need to recur, the proof is standard, and almost the same as the proof that $sqrt2$ is irrationnal.
Let’s assume there is a $x = fracab$ so that $x^2 = mp$ and $pgcd(a,b) =1$
Then $fraca^2b^2 = mp$
$a^2 = mp b^2$
then
either p is a prime factor of b, which means it is not one of a (since $pgcd(a,b) = 1$) and therefore $a^2 = m p b^2$ is impossible
or p is not a prime factor of b. since
$a^2 = mp b^2$, we have $p | a^2$ so $p | a$ ($p|xy$ implies $p|x$ or $p|y$ when $p$ is prime, by euclide’s theorem. We just set $x = y = a$)
then $p^2 | a^2$
then $p^2 | m p b^2$, that is $p | mb^2$ which contradicts with our hypothesis
answered 2 days ago
NephanthNephanth
1563
1563
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
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– HalfAFoot
2 days ago
$begingroup$
Generally, induction is appropriate when it seems easier to prove the property knowing that it is true for a smaller m. The way of thinking you can adopt is : « if needed, I can assume the property is true for any smaller value. ». Then, if you actually use that hypothesis, you are using induction, if you don’t, then you didn’t need it
$endgroup$
– Nephanth
2 days ago
add a comment |
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Generally, induction is appropriate when it seems easier to prove the property knowing that it is true for a smaller m. The way of thinking you can adopt is : « if needed, I can assume the property is true for any smaller value. ». Then, if you actually use that hypothesis, you are using induction, if you don’t, then you didn’t need it
$endgroup$
– Nephanth
2 days ago
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Thank you,i think the logic is very similar to my base step...I just instantly jumped to using induction to prove it...how does one tell whether induction is appropriate or not at the outset?
$endgroup$
– HalfAFoot
2 days ago
$begingroup$
Generally, induction is appropriate when it seems easier to prove the property knowing that it is true for a smaller m. The way of thinking you can adopt is : « if needed, I can assume the property is true for any smaller value. ». Then, if you actually use that hypothesis, you are using induction, if you don’t, then you didn’t need it
$endgroup$
– Nephanth
2 days ago
$begingroup$
Generally, induction is appropriate when it seems easier to prove the property knowing that it is true for a smaller m. The way of thinking you can adopt is : « if needed, I can assume the property is true for any smaller value. ». Then, if you actually use that hypothesis, you are using induction, if you don’t, then you didn’t need it
$endgroup$
– Nephanth
2 days ago
add a comment |
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$begingroup$
As there is no clear route from $sqrtmpnotin Bbb Q$ to $sqrt(m+1)pnotin Bbb Q$ it seems that this approach is not ideal.
$endgroup$
– Lord Shark the Unknown
2 days ago
$begingroup$
yikes...since the question appears in the induction chapters...i just thought this approach would be fine...Also,how can you tell that there is no clear route??? Thank you for the reply.
$endgroup$
– HalfAFoot
2 days ago