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**For all integers a,b,c, prove or disprove each of the following:
i.  If a | c and b | c, then ab | c^2.
ii. If a | (b+c), then a | b and a | c.**

Note : "If true prove without using numbers as an example otherwise you can use them if false" That's What My Prof. says.

Unfortunately, The first one I couldn't solve it but the second one i solve it but i want to ensure that my answers are right Thank you in advance

Jim Henry
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  • Hint: Remember that the definition of $a\mid c$ is that $c$ is an integer multiple of $a$, in other words there exists some integer $n$ such that $c=a\cdot n$ (remember, $n$ may be different than another factor that appears from another definition such as for $b\mid c$). Show that $c^2$ is an integer multiple of $ab$. – JMoravitz Apr 06 '20 at 21:42
  • "But the second one i solve it" And your answer there was? I can't confirm your answer as correct or incorrect since you did not share. – JMoravitz Apr 06 '20 at 21:43
  • ii. 25/(3+2) = 5 : a = 3 , b = 2 but 25∤2, 25∤3 ∴ If a | (b+c),then a | b and a | c statement is False @JMoravitz – Jim Henry Apr 06 '20 at 21:49
  • $5$ is not a multiple of $25$... it is the other way around. $25$ is a multiple of $5$. If $a\mid b$ then $b$ is an integer multiple of $a$. $5\mid 25$ is true. $25\mid 5$ is false. – JMoravitz Apr 06 '20 at 21:52
  • ii.∵ 5/(13+17) = 6 : a = 13 , b = 17 but 5∤13, 5∤17 ∴ If a | (b+c),then a | b and a | c statement is False , Is it correct now ? @JMoravitz – Jim Henry Apr 06 '20 at 22:04
  • Yes, that is fine. A much smaller example would have sufficed... $2\mid (1+1)$ but $2\nmid 1$ as well as $2\nmid 1$. – JMoravitz Apr 06 '20 at 22:10

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