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I have issues in understanding the proof of the F.T.A. by Wallace in the book titled: Groups, Rings and Fields, on page #66.pg. #66
The issues are :
(i) It is stated that : "Then $p_1$ divides $q_1q_2...q_n$".
I feel $q_n$is a typo, and should be $q_s$.

(ii) After finding that with suitable reordering, $p_1 = q_1$, the prime $p_r$ is replaced with $q_r$, and $q_1$ with $p_1$. So, it means the replacement of $p_r$ is done with $q_r$, for the same reason as shown for replacing $q_1$ with $p_1$.

(iii) It is stated that : "By our induction assumption we have $r=s$". But, it was only 'supposed' by the statement: "Suppose now that $n=p_1p_2...p_r = q_1q_2...q_s$", that the two prime factorization are equal to $n$, nothing more. Also, to vindicate my conjecture, in the issue (ii) it is shown in the book that $q_r$ replaces $p_r$; rather than $q_s$ replacing $p_r$.

jiten
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1 Answers1

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The issues are : (i) It is stated that : "Then p1 divides q1q2...qn". feel qn is a typo, and should be qr.

Note: You are correct. This is a Typo.

(ii) After finding that with suitable reordering, p1=q1, the prime pr is replaced with qr, and q1 with p1. So, it means the replacement of pr is done with qr, for the same reason as shown for replacing q1 with p1.

Note: At this step we just one to eliminate one of the primes from each side.

(iii) It is stated that : "By our induction hypothesis we have r=s". But, it was only 'supposed' by the statement: "Suppose now that n=p1p2...pr=q1q2...qs", that the two prime factorization are equal to n, nothing more.

Note: The induction hypotheses clearly indicates that if n=p1p2...pr=q1q2...qs then r=s, therefore there is no flaws in the proof.

Mohammad Riazi-Kermani
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  • Regarding the issue (iii), it is only stated before the start of the proof (in the theorem) that $r=s$, not later. I have read the proof multiple times, and am unable to find where the assumption $r=s$ is stated. Also, to vindicate my statement, in the proof, the replacement is of $p_r$ by $q_r$, rather than of $p_r$ by $q_s$. – jiten Jan 26 '18 at 03:27
  • Please help me find the inductive hypothesis where it is stated that $r=s$. – jiten Jan 26 '18 at 04:23
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    @jiten When you prove the theorem by induction, you assume the unique factorization for n and prove it for n+1. Note that the unique factorization for n implies r=s and p_i= q_i simultaneously. – Mohammad Riazi-Kermani Jan 26 '18 at 05:28
  • Thanks for that. I request that at least one more such inductive proof be given that has different (or, additional) set of assumptions than the usual one of just assuming some relation to be true for $n=k$. But, two things are confusing still :(i) it is still not found by me where $r=s$ is stated there along with the equality of two different forms of prime factorization of $n$, so is it an implicit assumption?; (ii) why it is that in-spite of this assumption, in the proof $p_r$ is replaced not by $q_s$. – jiten Jan 26 '18 at 08:03
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    @jiten The "uniqueness " assumption means two things at the same time. It means r=s and p_ i = q_i for i=1,2,3,...,r. With this in mind the proof is OK. – Mohammad Riazi-Kermani Jan 26 '18 at 10:29
  • Sir, I mean that it is all okay that if the inductive hypothesis assumes (not shown that step here explicitly, just read elsewhere) for prime $n, p_ip_2...p_r = q_1q_2...q_s$, with addtl. assumption $r=s$. But, why $p_r$ is not replaced by $q_s$, for $n+1$. Is it as not easy to go to the next step of strong induction, as if $n=24$ and $n+1=25$, then the factorization of the two differs totally. So will induction serve, if going from one number($n$) to next ($n+1$) does not keep the pattern of earlier prime factors. Also, not clear why weak induction not enough here. – jiten Jan 26 '18 at 11:11
  • Sir, please take my earlier comment as final, and am sorry for the changes; but the understanding reached slowly to final level. – jiten Jan 26 '18 at 11:15