A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mathmath333

  • one year ago

greatest integer function

  • This Question is Closed
  1. mathmath333
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\large \color{black}{\begin{align} &\{x,y\}\in \mathbb{R} \hspace{.33em}\\~\\ & \lfloor{4x+5\rfloor}=5y+3 \hspace{.33em}\\~\\ & \lfloor{3y+7\rfloor}=x+4 \hspace{.33em}\\~\\ & \normalsize \text{where }\ \lfloor{\ \rfloor}\ \text{represents the greatest integer function } \hspace{.33em}\\~\\ &\normalsize \text{Find}\ x^2y^2 \hspace{.33em}\\~\\ & a.)\ 1 \hspace{.33em}\\~\\ & b.)\ 2 \hspace{.33em}\\~\\ & c.)\ 4 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{None of these} \hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I think it's an interesting system, so far I got one solution hopefully it's right. I stopped cause I believed there's no other solution. However there could be another soltion.. First start by assuming that we have either integer values orn non integers, each one will lead to a form of equation. The bonding ever however is hard to determine initially. Assume they are integers and solve the system by any method, x=-3 and y=-2 will he satisfied

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Since the question asks for unspecified x,y. Then it somehow gives a clue that there's only one x and one y that are satisfied. hence, the answer I THINK is none of the above

  4. mathmath333
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what if there exists another solution other than \(x=-3\) and \(y=-2\)

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Then you will in troubke. Because in order to get red of the greatest integer, then you will have to find c such that : [f(x)]=f(x)-c is an integer. This c will add a third and fourth parameter for the system. I told you I'm not saying it's impossible, but I need to think about it

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Another way to do it: You could try plotting both "lines". It might be a bit tedious, but it can be done. Doing so reveals two solutions, \(x=-3,y=-2\) and \(x=-2,y=-\dfrac{6}{5}\). There's probably a much better way to go about it.

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The lines plotted does not represent the functions we are dealing with, except if the expression inside it is integer value. Considering this case, then we ensure that we a have system of linear equations in two variables, which in turn satisfied at one point, zero points or infinite points, but never at two points ... So necessarily the pair (-3,-2) is UNIQUE Whenever there's another solution, then the functions we have are not linear anymore, they will be something else, in our case least integer function which are satisfied at non integer parameters

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    system:\[\lfloor{4x+5\rfloor}=5y+3 \\ \lfloor{3y+7\rfloor}=x+4\]second equation tells us that \(x+4\) is an integer, so \(4x+5\) is an integer too. It follows that\[\lfloor{4x+5\rfloor}=4x+5=5y+3 \\ y=\frac{4x+2}{5}\]put this in the second equation\[\lfloor{3\frac{4x+2}{5}+7\rfloor}=x+4 \\ 2x+8 +\lfloor{\frac{2x+1}{5}\rfloor}=x+4 \\ x+4+\lfloor{\frac{2x+1}{5}\rfloor}=0\]only integer that satisfies last equation is \(x=-3\)

  9. mathmath333
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how did u find that only \(x=-3\) satisfies this equation \(\Large x+4+\lfloor{\dfrac{2x+1}{5}\rfloor}=0\)

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for \(x>-3\) LHS becomes a positive number, for \(x<-3\) LHS is negative, only possibility is \(x=-3\)

  11. mathmath333
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnks

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yw

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Both \(x\) and \(y\) are said to be real numbers, not just integers. Besides, the case where \(y=-\dfrac{6}{5}\) still works because \(5y+3\) is an integer. Plot for the curious:

    1 Attachment
  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but \(\left(-2, -\frac{6}{5} \right)\) fails to satisfy second equation

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh you're right, the RHS and LHS are off by one.

  16. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.