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I am not a mathematician, I'm a linguistics PhD student. As part of my research I need to put various convoluted sentences through various syntactic transformations and see then check whether people think they are true or not. Mathematical statements (well, some of them) suit my purposes very well, because they are less context dependent and can be straightforwardly assigned a truth value (i.e. be deemed true or false). The problem is that I'm not a mathematician. When these sentences get a bit convoluted, I have a bit of a problem knowing whether they are true or false myself (before they undergo various syntactic transformations).

I have a particular sentence which states that if a given number is:

  • an integer
  • divisible by 7 (meaning it will yield an integer if divided by 7)
  • a square number

then it is divisible by 49. I intuitively believe this to be correct (although I can't explain why). Is this actually true? I don't want to waste everybody's time by starting with an untrue untransformed sentence.

Araucaria - him
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    $49\cdot 2$ is not a square number. A square number is one whose prime factorization has each prime appearing to an even power. $2^1\cdot 7^2$ is not such a number, so your sentence is not both a necessary and sufficient condition. It is a sufficient condition however, just not necessary. – JMoravitz Aug 24 '17 at 16:45
  • \cdot $\cdot$ is the preferred way of representing multiplication rather than \times $\times$. That is to say $49\cdot 2=49\times 2 = 98$ – JMoravitz Aug 24 '17 at 16:46
  • @JMoravitz Sorry. Thanks. Got that now. All that's important for my purposes is that it's sufficient (i.e. a number fulfilling those conditions will always be divisible by 49). I take it you mean that that's correct (I think that is what your saying, but being a non-mathematician, I'm just checking that's definitely what you mean). – Araucaria - him Aug 24 '17 at 16:48
  • Just in case it wasn't perfectly clear what I was saying, A number satisfying your conditions will always be a multiple of $49$. A number which is a multiple of $49$ could possibly not satisfy your conditions (e.g. $98$). – JMoravitz Aug 24 '17 at 18:16
  • @JMoravitz Thanks. Was clear originally, but, not being a maths person, I thought I'd better check just in case any of those terms had a special mathematical meaning. – Araucaria - him Aug 24 '17 at 20:18
  • Note that all such numbers are divisible by $49$, but not all numbers divisible by $49$ adhere to your conditions. As another commenter noted, $49 \cdot 2$ is divisible by $49$, but it fails your third condition of being a square number. In other words, all cats are animals, but not all animals are cats. – SQB Aug 25 '17 at 11:21
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    From the answers, you can tell that it is important that $7$ is a prime number. Maybe it is instructive to give an example with a non-prime, so let us try to substitute $7$ and $49$ with $12$ and $144$, respectively (note that $12$ is not squarefree). Now, we can exhibit counterexamples, such as $36$ or $900$, which are perfect squares divisible by $12$, but not divisible by $144$. – Jeppe Stig Nielsen Aug 25 '17 at 11:40

8 Answers8

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You can derive this fairly directly from Euclid's Lemma, which says that if a product $a \cdot b$ is divisible by prime $p$, then either $a$ or $b$ (or both) is divisible by $p$.

So, $n^2$ divisble by 7 means $n \cdot n$ divisible by 7 which (by Euclid's Lemma) means $n$ divisible by 7 and thus $n^2$ divisible by 49.

Of course it took even Euclid a little work to prove the Lemma

Nemo
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    +1 Nice to use Euclid's Lemma and avoid the full strength of the fundamental theorem of arithmetic. – Ethan Bolker Aug 24 '17 at 18:25
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    @EthanBolker But it does use the full strength since it is trivial to prove that the Prime Divisor Property: prime $,p\mid ab,\Rightarrow, p\mid a,$ or $,p\mid b,,$ is equivalent to uniqueness of prime factorizations. – Bill Dubuque Aug 24 '17 at 21:20
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    @BillDubuque Yes it's equivalent, so my calling it "not full strength" isn't quite right, but in the most common proofs of the fundamental theorem of arithmetic the prime divisor property, following from the extended Euclidean algorithm, comes first, with the FTA as a consequence of some further induction arguments. – Ethan Bolker Aug 24 '17 at 21:32
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    @Ethan Not "quite right:" is an understatement. The remark is incorrect and misleading. Further, that one is proved before the other in some presentations has nothing at all to do with (logical) strength. – Bill Dubuque Aug 24 '17 at 21:39
  • @BillDubuque: You are characterizing strength in absolute terms, but it can be described in relative terms as well: How long is the (shortest) proof from A to B? From B to A? If the latter is shorter than the former, then B is stronger than A in a relative sense, even though they are logically equivalent and therefore of the same absolute strength. – Kevin Aug 25 '17 at 05:30
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    @BillDubuque All theorems are equivalent, so saying that Euclid's lemma and the FTA are equivalent doesn't really mean anything other than "they are both true". On a more subjective note, I believe Euclid's lemma to be more elementary than the FTA, in the sense Kevin explained well above. If you're starting from axioms and definitions (like Peano), then proving this by first proving the FTA will probably be longer than proving this by first proving Euclid's lemma. – Arthur Aug 25 '17 at 07:35
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    @Arthur You misunderstand. I am comparing them as properties of an integral domain (the usual base theory for (algebraic) number theory). In this sense the Prime Divisor Property (or Euclid's Lemma) is trivially equivalent to uniqueness of factorizations into irreducibles (in any domain where nonzero nonunits have factorizations into irreducibles), i.e. in such a domain, irreducibles satisfy the prime divisor property $\iff$ factorizations into irreducibles are unique (up to order and unit multiples). The proof is short and straightforward (essentially the same as for the integers). – Bill Dubuque Aug 25 '17 at 16:39
  • @Kevin See my prior comment. Said in ring theory language, a domain is a UFD $\iff $ it is atomic (enjoys factorizations into atoms = irreducibles) and atoms are prime (i.e.satisfy said Prime DIvisor Property). This is one of the 10 characterizations of UFDs that I mention in this answer. – Bill Dubuque Aug 25 '17 at 16:50
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    @Bill: Who said anything about arbitrary rings? This is a question about $\mathbb{Z}$. – Kevin Aug 25 '17 at 19:07
  • @Kevin I did (as would any number theorist). We obtain much better insight on number theory in $\Bbb Z$ by studying number theory in more general rings, e.g studying number theory of quadratic integers greatly simplifies the theory of representations of integers by quadratic forms. – Bill Dubuque Aug 25 '17 at 19:11
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    @BillDubuque: You are welcome to bring high-level esoterica into a low-level discussion, of course, but you should not complain that the folks working at that low level are using terminology differently from how you expect. – Kevin Aug 25 '17 at 21:46
  • @Kevin Apparently you misunderstand. There are no complaints about "terminology" nor any "high-level esoterica". Best of luck in your mathematical studies. – Bill Dubuque Aug 25 '17 at 22:05
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    Please, everybody, whenever Bill Dubuque shows up in the comments sphere, someone please post this quote of Goethe’s: “The mathematician is like a Frenchman: you tell him something, he translates it into his own language, and at once it become something altogether different.” – Mike Jones Aug 30 '17 at 13:58
  • In defense of low-level discussion: “Many branches of human learning have suffered from taking norms too seriously.” -- Everett Hughes – Mike Jones Aug 31 '17 at 22:27
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Yes.

To see this, call the number $x$ and suppose that $x$ is the square of the number $y$. Suppose that $y$ has a prime factorization

$$y = p_1^{n_1}p_2^{n_2}\cdots p_K^{n_K}$$

where each $n_k>0$ and $p_1<p_2<\cdots < p_K$. (In other words, we're displaying the prime factorization as compactly as possible, and "in order".)

Since $x = y^2$, we have

$$x = (p_1^{n_1}p_2^{n_2}\cdots p_K^{n_K})^2 = p_1^{2n_1}p_2^{2n_2}\cdots p_K^{2n_K}$$ Note all of the powers $2n_1,2n_2,\ldots,2n_K$ are even; and since $7$ divides $x$, one of the prime factors $p_i$ must be $7$.

But then $2n_i$ is an even number greater than zero, so is at least $2$. Thus there are at least two factors of $7$ in $x$, so $49$ divides $x$.

MPW
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  • The argument is implicitly using the existence and uniqueness of prime factorizations, i.e. the Fundamental Theorem of Arithmetic. These properties should be explicitly mentioned when invoked in proofs at this level. – Bill Dubuque Aug 25 '17 at 16:55
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    No, they shouldn't be. At this level, which is proving theories to a layman, simply saying how things work is better than citing theorems. Citing a theorem to a non-mathematician is just citing magic words. Prime factorization being unique is something we learn in grade school, and does not necessarily need to be invoked explicitly. +1 for not just using magic words. – trlkly Aug 25 '17 at 19:17
  • @BillDubuque : Disagree. I think that a nonmathematician would almost certainly take the existence and uniqueness of prime factorization as a given (even if you had to explain what that meant). In my opinion, OP wants a reasonable heuristic explanation, not a mathematically watertight proof. trlkly's comment is right on the money, I think (+1 for trlkly). – MPW Aug 25 '17 at 19:20
  • @MPW We'll have to disagree on that. I've been teaching number theory for many decades, so my views on such matters are based on extensive experience. But, alas, it is impossible to do justice to this intricate topic in SE comments. – Bill Dubuque Aug 25 '17 at 19:25
  • @BillDubuque : In the classroom, for a mathematics course, I would agree with you. – MPW Aug 25 '17 at 19:27
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Yes, it is true. It hinges on the fact that $7$ is a prime number.

In general, if $n$ is an integer that is divisible by a prime number $p$ and $n$ is a square, then $n$ is divisible by $p^2$.

This follows from the Fundamental theorem of arithmetic.

lhf
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  • +1 Thanks for the swift and helpful reply. That's what I'd have thought. Is there any kind of reference that I could use with regard to this? (or a proof?) – Araucaria - him Aug 24 '17 at 16:49
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    @Araucaria any text relating to prime numbers and/or elementary number theory. Even texts as old as Euclid's Elements from 300 bc would have the appropriate definition of prime numbers which leads to this conclusion. For something more modern, I'd recommend any book on introduction to proof writing as elementary number theory is almost always used as a source of examples. – JMoravitz Aug 24 '17 at 16:52
  • @JMoravitz Thank you. – Araucaria - him Aug 24 '17 at 16:57
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    The divisor need not be prime; it just needs to have no duplicate prime factors. So a square number divisible by 6 would also be divisible by 36, but a square number divisible by 4 would not necessarily be divisible by 16. – Kyle Strand Aug 24 '17 at 20:18
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    @Kyle And conversely, i.e. $,(q\mid n^2\Rightarrow q\mid n)\iff q,$ is squarefree. See this answer for a handful of characterizations of squarefree integers. – Bill Dubuque Aug 24 '17 at 21:49
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Yes. The reason is that the only way a square number can be divisible by $7$ is if its square root is divisible by $7$. So your number is the result of squaring a multiple of $7$, and when you do that you get a multiple of $49$.

Especially Lime
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    You've simply restated the result in an equivalent form, but have not given any justification for the equivalent claim. That's a circular argument. – Bill Dubuque Aug 24 '17 at 17:56
  • @BillDubuque Is it? The question was why it is divisble by 49. Because 7 is a factor of the square root (that requires another small proof, I guess), the square must contain 7*7 in its factorization, hence 49. Is there anything else to prove? This sounds like the simplest explanation to understand for a non-mathematician. – IS4 Aug 24 '17 at 20:44
  • @IllidanS4 The argument is $,7\mid n^2,\Rightarrow, 7\mid n,\Rightarrow, 7^2\mid n^2.\ $ The 2nd arrow is trivial. The 1st arrow is highly nontrivial.Its proof requires using the FTA = Fundamental Theorem of Arithmetic or some closely related property. Beginner's often think this is "obvious" based on some unspoken intuition. But in mathematics we insist on rigorous proof - not handwaving. In fact this property may fail in slightly more general number systems, e.g. quadratic integers $,a+b\sqrt d,,$ i.e. generally they do not satisfy the analogue of FTA (i.e. are not UFDs). – Bill Dubuque Aug 24 '17 at 21:06
  • @BillDubuque Sure, this isn't a complete rigorous proof, but I don't think that's what OP was asking for. It is certainly not the answer I would give to anyone studying mathematics, but the question does say "I am not a mathematician" twice, and I think some of the other answers failed to take this into account. – Especially Lime Aug 25 '17 at 08:16
  • @EspeciallyLime It's not a proof (so not mathematics) if you claim without any justifcation that $,7\mid n^2,\Rightarrow,7\mid n.\ $ That's no different than claiming that your favorite (unproven) conjecture is true. This is a site about mathematics, not magic, numerology, pseudoscience, etc. – Bill Dubuque Aug 25 '17 at 16:59
  • +1. Citing theorems is not how you prove something to a non mathematician. He just stated Euclid's lemma in common everyday terms, rather than using the name of it. This is perfectly valid math. – trlkly Aug 25 '17 at 19:14
  • @trlkly There is no statement of any Lemma in the answer. There are merely (unfounded) claims (trivially equivalent to the original claim). – Bill Dubuque Aug 25 '17 at 19:31
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This is true. Let us call your number $n$. The number $n$ is supposed to be the square of some number, say $a$, so $n=a^2$. Now since $a^2$ is divisible by $7$ and $7$ is prime, $a$ has to be divisible by $7$. (It is always true that if a prime number divides a product, it has to divide one of the factors.) But if $a$ is divisible by $7$, then $a^2$ is divisible by $7^2=49$.

Carsten S
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Definition: A square number is a number whose square root is a whole number. Definition: The square root of a number, call it N, is a number which can be created by squaring some other number, call it m, i.e., if m^2 = N then sqrt(N) = m

Fundamental Theorem of Arithmetic (FTA): Any whole number greater than 1 (i.e., from 2 up) can be written as a unique product of prime numbers (the unique part means no matter how you split it up you get the same list in the end).

So, if you start with a (perfect) square whole number, call it N, being divisible by 7: Since N is a square, it must be factorable as a product of some number times itself, i.e., N = m*m. Now, each m is either prime or factors again into a product of primes (by FTA). Once you have this list of primes written out, 7 must be there in that list as least once, since 7 divides N. But, since both m's are the same m, the 7 must appear twice (once for each m), i.e., since there's a pair of m's there must be a pair of all the factors of each m.

This is just a wordy version of what user MPW was saying using formulas. Also, this can be generalized for any prime number, not just the number 7. Now that I think about it, it could be generalized to any whole number, not just primes.

Also, a caveat, your original post says the number should be an integer. If the original number is negative, it cannot be a perfect square (unless you want to talk about complex numbers). I suspect you meant "whole number", i.e., just positives.

user474566
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  • Does a "product" of two numbers mean a multiple of two numbers? – Araucaria - him Aug 24 '17 at 20:49
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    If you wish to give a rigorous proof using FTA then you should do it correctly, and explicitly mention every use of existence and uniqueness of prime factorizations (which occurs several times). – Bill Dubuque Aug 24 '17 at 21:16
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One simple way of expressing the fundamental theorem of arithmetic is this.

Let $p_i$ be the ith prime number. Then every positive integer, $n$, can be expressed uniquely as an infinite product

$$n = \prod_{i=1}^\infty p_i^{\pi_i}$$

where all of the $\alpha_i$ are non negative integers and all but a finite number of them are equal to $0$. (A fancy way of saying this is that the sequence $\{\pi_i\}_{i=1}^\infty$ has finite support.

Suppose $n$ is:

  • an integer
  • divisible by 7, which equals $p_4$.
  • a square number

Then there must be an integer $m$ such that $n = m^2$

We know that, for some $\alpha_i$,

$$m = \prod_{i=1}^\infty p_i^{\alpha_i}$$

in which case,

$$n = \prod_{i=1}^\infty p_i^{2\alpha_i}$$

Since $n$ is a multiple of $7$, then $2\alpha_4 > 0$. Hence $\alpha_4 > 0$

Since there is the only way to express $m$, then $m$ is a multiple of $7$.

Steven Alexis Gregory
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suppose this number is x.

Now we know:

  1. x is an integer (Assumption: positive integer)
  2. x is divisible by 7
  3. x is a square number
  • From Point 1 we know its a positive integer. From that we can say x is 1 or greater than 1. (x=1 or x>1)
  • From Point 2 we know x is divisible by 7. That brings us to conclude that x is at least 7. Or multiple of 7 since its divisible by 7. We can say (x=7 or x=7*y) Here y can be any positive integer (y=1 or greater)
  • Now from Point 3 we know that its a square. that means it has some form of (x= 7^2) or (x=7^(2) .y) or (x=7^2 . z^2). where y = z^2.

What I am trying to say here is

  • 7 is a prime number.
  • x is a square and divisible by 7.

This forces the number x to be a product of 7^2. Which makes it divisible by 49. because 49 is 7^2.

Deepak
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