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    #1
    Hey!!! I have also an other question
    Suppose the base $\beta =10$ ,the precision $t=3$, $-L=U=10$ and $$A=(317+0.3)-(171.499+145.501)$$
    I have to find the relative error for $A$.
    We don't make rounding.For example,if we have the value $345.924$ it is equal to $0.345924*10^3$ and the corresponding floating number is $0.345*10^3$.
    I found that it is equal to
    $$\frac{||A-fl(A)||}{||A||}=\frac{7}{5}=2.33$$
    Could you tell me if it is right?

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    #2 Thread Author
    Quote Originally Posted by evinda View Post
    Hey!!! I have also an other question
    Suppose the base $\beta =10$ ,the precision $t=3$, $-L=U=10$ and $$A=(317+0.3)-(171.499+145.501)$$
    I have to find the relative error for $A$.
    We don't make rounding.For example,if we have the value $345.924$ it is equal to $0.345924*10^3$ and the corresponding floating number is $0.345*10^3$.
    I found that it is equal to
    $$\frac{||A-fl(A)||}{||A||}=\frac{7}{5}=2.33$$
    Could you tell me if it is right?
    I found that A=0.3 and fl(A)=1..

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    #3
    Quote Originally Posted by evinda View Post
    I found that A=0.3 and fl(A)=1..
    Looks good!!


    ... but doesn't that mean that:
    $$\frac{||A-fl(A)||}{||A||} = \frac{|0.3 - 1|}{|0.3|} = \frac{0.7}{0.3} = 2.33$$

    Oh wait! You also got $2.33$... while you shouldn't have.

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    #4 Thread Author
    Quote Originally Posted by I like Serena View Post
    Looks good!!


    ... but doesn't that mean that:
    $$\frac{||A-fl(A)||}{||A||} = \frac{|0.3 - 1|}{|0.3|} = \frac{0.7}{0.3} = 2.33$$

    Oh wait! You also got $2.33$... while you shouldn't have.
    I accidentally wrote $\frac{7}{5}$ I meant that it is equal to $\frac{7}{3}$..

    Thank you very much!!!!!

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