2

I am currently studying the book 'How to Think Like a Mathematician' and working on Part 5. I have some confusion regarding the process of checking my own proofs. Sometimes, I feel that a particular step is incorrect, but I cannot provide a solid reason why, nor can I explain why it is correct. For example, in the proof of 'If m²|n², then m|n,' I have proven it as follows: By assuming the existence of k² ∈ ℤ such that n² = m²k², we can conclude that n = mk, which means that m divides n. However, I have a feeling that taking k² ∈ ℤ might be incorrect, but I don't know why, nor do I have a solid reason for why it is correct.

Bill Dubuque
  • 272,048
Afzal
  • 588
  • Hmm... are $m,n\in\Bbb Z$? Also, $k=i$ means $k^2=-1\implies k^2\notin\Bbb Z$ ... right? I might be wrong – TheMather - or rather AMather May 31 '23 at 10:34
  • 3
    @TheMather-orratherAMather The context is likely working over $\Bbb Z$ only, so $k=i$ is not reasonable. But $-1\in\Bbb Z$ anyway – FShrike May 31 '23 at 10:50
  • 2
    First of all collect examples and "simple facts", and get used to recognizing patterns. Then understand the main structures that govern the theme, and the theorems for them. In the above case, the result needed is the "unique" decomposition in prime factors in the integers (up to units and permutations of the factors). Then when writing an argument you have to test it with those examples. In you case, what happens if $m=-5$, and $n=10$. Yes there is some $k\in\Bbb Z$, i am taking $k=1$. And your argument fails. Of course, it can be adjusted. You do so and next time take care... – dan_fulea May 31 '23 at 10:52
  • 3
    In general, mathematics is not assimilated "linearly". Take your time to understand the essence, get examples, see where there are counterexamples and test the theorems in each particular case. Exercises often use (more or less clear) algorithms. If you understand how these receipts work, stay there maybe to have some routine i little bit, but then move immediately further. 70% of the exercises in a book are useless to progress, although they are a good way to get routine. Always read books, ask questions, ask for feed back on sites like this one for each mathematical argument you have... – dan_fulea May 31 '23 at 10:57
  • @FShrike ah, that's right! Forgot about that. – TheMather - or rather AMather May 31 '23 at 11:06
  • @dan_fulea I collected some examples on this theorem. For example: If 25| 100, then there exist 4 ∈ ℤ s.t 100=25×4. I did few examples like that and I saw a pattern that for every m² and n² there exist k² ∈ ℤ s.t m²=k²n². Yes, for m=-5 and n= 10 my argument fails. I guess I was doing examples in wrong way, I was thinking about two squared number m² and n² s.t m²|n² then I was looking at the behaviour of k. – Afzal May 31 '23 at 12:21
  • @dan_fulea "70% of the exercise is useless to progress" how? I think you are taking about exercises like " prove 4|8 , 16|32 , etc " am i correct? If not please explain. – Afzal May 31 '23 at 12:30
  • Dupe of this. $\ \ $ – Bill Dubuque May 31 '23 at 13:24
  • 1
    There is a problem when you say "there exists $k^2$..." Logically, for the moment there is only a $K$ with $m^2=Kn^2$. Many comments and the answer have pointed to this aspect. From here you have to show somehow that this $K$ is (positive and) a square. Just apply the theorem of unique factorization in the integers! Do you know the theorem? When yes, take factorizations first on the R.H.S., namely for each of the factors, one for $K$ and one for $n$. Write this explicitly with many variables. We get then a factorization for the L.H.S. - and now insist to have a square. – dan_fulea May 31 '23 at 21:26
  • 1
    70% of the exercises is what the school wants. You may want it, too. But you may also want to progress quickly. Then sometimes, the same "experience" is covered by some 10% of the exercises, when the ambitions 10% are chosen by a "good hand". You will have this hand in time, for the time being ask for a guide. The 10% are not enough to have routine and clean skills, but you can complement this. Either having a focus on structure, or by using computer aided software, or by making giant jumps, then coming back and covering as much details as possible from the new perspective. Just keep going! – dan_fulea May 31 '23 at 21:30
  • @dan_fulea I got it now. The problem is that I am taking k²... But in actual I have to take K s.t m²=Kn² and have to show K is a square. Right? Yes I know unique factorization Theorem ( Fundamental theorem of arithmetic). So I have to assume n = Π p^a, K = Πq^b and m= Πh^c so i will have Π h^(2c) = Πq^b Πp^(2a). From here i have to show Π q^b is square? If yes, then Π q^b = ( Πh^c/ Π p^a)²? – Afzal Jun 01 '23 at 04:48
  • Yes, you are getting closer, very close to the end. The fact that $K$ is a square is determined by the fact that all powers appearing in the (unique) factor decomposition are even. We have to show this. Now we compare the two decompositions. Maybe it is better to stay in $\Bbb Z$, although this is just a matter of taste. (Else we need the translation of the unique prime factors decompositions in $\Bbb Q$, which is "the same".) So we compare - with your notations slightly refined:$$\prod_p p^{2ord(p,m)} = \prod_p p^{ord(p,K)} \cdot \prod_p p^{2ord(p,n)} \ .$$Here $ord$ is "order"... – dan_fulea Jun 01 '23 at 12:21
  • ... the "order of a prime in a given number". We may take it zero if a prime does not appear. And the products can be taken / are taken with only those primes that show up in either one of $m,K,n$. Now use use here, in the formula above, the unique prime factor decomposition (for positive numbers, this is the reason why the unit $\pm1$ is without loss taken to be $+1$). For each $p$ that is indeed relevant, we have:$$2ord(p,m)=ord(p,K)+2ord(p,n)\ .$$Move the $2ord$ terms on the one side. We can factor $2$ here. By unique factor decomposition, the other side $ord(p,K)$ is also even! – dan_fulea Jun 01 '23 at 12:26
  • So morally, the only thing which is in the essence of the structure is the unique factor decomposition in $\Bbb Z$. This "atomic" structure is so strong, that many, many properties in integer numbers are or should be first tested against it, i.e. we first sieve, filter, understand some properties of (more or less explicit or implicit) multilicative formulas by testing them w.r.t. divisibility by some "special prime(s)". Can you generalize now the property - by taking instead of the power two (this or) an other power? Which power? Should it be a prime power? – dan_fulea Jun 01 '23 at 12:31
  • @dan_fulea I don't know about "order of primes in a given number" i am only little bit familiar with fundamental theorem of arithmetic. But after reading this i guess the property can be generalized for any prime power – Afzal Jun 01 '23 at 14:33
  • @dan_fulea the reason is if i assume prime power P then after some calculation i will get P( ord(p,m) - ord(p,n) = ord(p,K). This shows ord(p,k) is multiple of P. Oh wait if i think true for all integers. Oh i am confused :( – Afzal Jun 01 '23 at 14:35
  • Let us continue this discussion in chat. – Afzal Jun 01 '23 at 14:40

1 Answers1

4

I can't answer the general question about what is a suitable learning practice; that will depend on the person and on the goals.

But I can answer the specific question about your proof:

If $m,n\in\Bbb Z$ have $m^2|n^2$, then say $k\in\Bbb Z$ with $n^2=m^2k^2$ and thus $n=mk$, so $m|n$.

There are two mistakes here. One is important, the other is not. The unimportant mistake is going from: $n^2=m^2k^2$ and deducing: $n=mk$. It may be that $n=-mk$, for all you know. But in that case it would still be true that $m|n$, so it's not a big deal (you should still watch out for it though: it might have been a big deal).

The slightly more serious mistake is saying that there should be $k\in\Bbb Z$ with $n^2=m^2k^2$. You were right to feel uneasy about this step. In general, I would advise seriously asking yourself why is that true before you write it down: it's easy for carelessness to creep in, so self-checking can be helpful.

$m^2|n^2$ tells you only that there is a (nonnegative) integer $\lambda$ with $n^2=m^2\lambda$. But, a priori, it's not necessarily the case that $\sqrt{\lambda}\in\Bbb Z$: suppose, for instance, that $n^2=3m^2$. As $\sqrt{3}$ is not an integer (or even a rational) your proof does not apply.

Hint: is it possible for $n^2=m^2\lambda$ to be true, with all quantities being integers, if $\sqrt{\lambda}\notin\Bbb Z$? Prove or disprove it!

FShrike
  • 40,125
  • 1
    It is not possible. Sorry for late reply it took me some time to figure it out. We have n² = m² λ implies m/n = ± √ λ . If √λ ∉ ℤ, then λ is not a square. And we have know square root of non square is irrational so it is impossible to write n²=m² λ if √ λ ∈ ℤ. – Afzal Jun 01 '23 at 14:25
  • 1
    @MuhammadAfzalSoomro Exactly right. Here's how to format your maths: $n^2=m^2\lambda$ implies $m/n=\pm\sqrt{\lambda}$. If $\sqrt{\lambda}\not\in\Bbb Z$ then $\lambda$ is not a square. This code produces: "$n^2=m^2\lambda$ implies $m/n=\pm\sqrt{\lambda}$. If $\sqrt{\lambda}\not\in\Bbb Z$ then $\lambda$ is not a square." – FShrike Jun 01 '23 at 14:31
  • 1
    thanks, I am new here so still learning about codes. I was studying and suddenly saw a corollary that square root of non square number is irrational and there was proof too. So when i here and read again your question suddenly i remember and saw a connection between ideas. – Afzal Jun 01 '23 at 14:39