# Thread: relative error for A

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?  Reply With Quote

2. Originally Posted by evinda 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..  Reply With Quote

4. Originally Posted by evinda 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.    Reply With Quote Originally Posted by I like Serena 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!!!!!   Reply With Quote

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