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I was learning proof, and I could not understand the part as to how $2d^2$ = $n^2$ implies that n is a multiple of $2$. screenshot of the problem

  1. Shouldn't it be " $n^2$ " instead of " n "? If not, can you explain to me why?

  2. Could you also include another short example to make it clear?

  3. If it supposes to be $n^2$ instead of $n$, then what would the sentence after it suppose to be(look at the screenshot)?

Thing I already tried: this question has been asked before, and I already read it, but the top answer seems to skip over the "$n$ vs $n^2$" part.

Link: $2d^2=n^2$ implies that $n$ is multiple of 2

Edited: I realized that the accepted answer in the older question actually addresses my issue. Initially, I didn't understand that "addressing $n^2$ implies n is even" has anything to do with "n being a multiple of 2". This is due to my inexperience with math. Thank you for all the comments and answers; it did help me understand.

Chhangsreng P
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    Both $n$ and $n^2$ work fine. Both sides must be even and the square of an odd number is odd. – lulu Nov 28 '22 at 23:35
  • Otherly said: if $n^2$ is even, then $n$ is even. – Anne Bauval Nov 28 '22 at 23:39
  • You need to prove $n^2$ is even if and only if $n$ is even. More generally, we have, if $p$ is prime, then $p$ divides $n^2$ if and only if $p$ divides $n.$ But these statements require proof. – Thomas Andrews Nov 28 '22 at 23:39
  • And even more generally, if $p$ is prime, then $p$ divides $ab$ if and only if $p$ divides $a$ or $b$. – Robert Israel Nov 28 '22 at 23:42
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    For prime $p,$ if $p|ab $ then $p|a$ or $p|b$ (where the word `or' means at least one of). So $p | n \cdot n$ means $p|n$ or $p|n,$ so..... – Will Jagy Nov 28 '22 at 23:42
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    If $n$ were odd, then $n^2$ would be odd too, but it is impossible because $n^2=2d^2$. Hence, $n$ has to be even. – Angelo Nov 29 '22 at 00:18
  • Research "Euclids Lemma" if $p$ is prime (and $2$ is prime), and $p|ab$ (and $n^2 = n\cdot n$ so $2|n\cdot n$), Then either $p|a$ (so either $2|n$) or $p|b$ (or $2|n$). So if $2|n^2$ then we have $2|n$ or $2|n$ and that means $2|n$. Now $2d^2 = n^2$ means by definition that $2|n^2$ and therefore $2|n$.... Alternatively if $n^2$ is an even square number then either $n$ is even or odd. If $n$ is odd then $n^2$ is ... odd. So if $n^2$ is even it is impossible for $n$ to be odd. – fleablood Nov 29 '22 at 18:45
  • " but the top answer seems to skip over the "n vs n2" part." No, the answer directly addresses that: "We now prove: n2 even implies n even. Proof: We tackle the contrapositive:, n odd implies n2 odd. Since n is odd, we can write n=2k+1 for some integer k.Then n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1, which is an odd number.(It's of the form 2m+1 for an integer m=2k2+2k.) This completes the proof, and the contrapositive is the statement you asked about. QED". I feel I have to mark this as a duplicate and the question you cite is exactly what you ask and the posted answer answers it exactly. – fleablood Nov 29 '22 at 18:50
  • Which is too bad, as none of the answer mention Euclid' Lemma which I find to be a better (albeit less intuitive... but more important to learn) argument. – fleablood Nov 29 '22 at 18:52
  • In your image, it refers to problem 1.10 and 1.11. What do those problems say? (Without even knowing what you text is, I'm going to bet 37 and a half cents that on of those problems is to prove that if $n^2$ is even then $n$ is even.) – fleablood Nov 30 '22 at 01:00
  • @fleablood You can bet, but the sources that I'm getting this document from do not provide 1.10 and 1.11. This problem is on a separate file from 1.10 and 1.11, and I could not find the file of 1.10 and 1.11. You can trust me that it takes less effort for me to read 1.10 and 1.11 than to type up my question on StackExchange. I would be more than happy to read 1.10 and 1.11 myself than to post a question. The reason that I type up this question is because I can't find it. – Chhangsreng P Nov 30 '22 at 09:59
  • "I didn't understand that "addressing n2 implies n is even" has anything to do with "n being a multiple of 2"" Every math term, no matter how intuitive, needs a meaning. "Even" has to mean something. As does "being a multiple of $2$". Most students are introduced to $n$ is "even" if there is an integer $k$ so that $n =2k$ and most students are introduced to $n$ is a multiple of $m$ if there is an integer $k$ so that $n =mk$. And if $m=2$ then "even" is by definition "a multiple of two". They are the exact same thing. – fleablood Nov 30 '22 at 16:59

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Since $n^{2}$ is divisible by $2$ (You can represent $n$ as $2x$

where $x$ is arbitrary and may be $d^2$) $n$ is also divisible by $2$.

The proof for $2|n^{2} \implies 2|n$ is simple by contraposition. You just need to think why $2\not|\,n\implies 2\not|\, n^{2}$.

If n would be odd then $n^{2}$ would also be odd, because the product of two odd numbers can't be even. Thus if $n^{2}$ is even, $n$ must be even.

TShiong
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dragontec12
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