Science:Math Exam Resources/Courses/MATH312/December 2008/Question 01 (f)

MATH312 December 2008

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q2 (a) i  •  Q2 (a) ii  •  Q2 (a) iii  •  Q2 (a) iv  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q6 (d)  •  

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Question 01 (f)
For each of the following statements, indicate if it holds for every positive integers $\displaystyle a,b,c$ (if so, a simple true without a proof suffices, if not, a false together with a counterexample is expected). If $\displaystyle \gcd(a,b,c)=1$, then $\displaystyle \phi (abc)=\phi (a)\phi (b)\phi (c)$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Look for a counter example with $\displaystyle a=b$ but $\displaystyle \gcd(b,c)=1$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The answer is false. Let $\displaystyle a=b=2$ and $\displaystyle c=1$. Then $\displaystyle \gcd(a,b,c)=\gcd(2,2,1)=1$ and further $\displaystyle \phi (abc)=\phi ((2)(2)(1))=\phi (4)=2$. However $\displaystyle \phi (a)\phi (b)\phi (c)=\phi (2)\phi (2)\phi (1)=(1)(1)(1)=1\neq 2$ and thus this is a counter example.

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