Write in normalized form, using 4 decimal digit IEEE rounding.

Question

The numbers are:

• a = 124.68
• a = -1.2345
• a = 0.55555
• a = -0.0054321

I'm not sure what the format is for rounding numbers in this way. Explanations would be great!

Answer 1

A number N which can be written as $d_nd_{n-1}d_{n-2}...d_2d_1.d_{-1}d_{-2}...d_{-t}$ with n digits before decimal place and t digits after decimal place, can be written in the normalised form as $N = 0.d_nd_{n-1}d_{n-2}...d_1d_{-1}d_{-2}...d_{-t} \times \beta^{k}$ where $\beta$ is the base for the number system and $d_n \neq 0$.

Now, there are some rounding rules which we shall follow:-

1. If we want to round the number upto mth digit and (m+1)th digit is less than 5, then simply truncate whatever is after the mth digit.
2. If the (m+1)th digit is equal to five then we have two cases - if all other digits are zero after (m+1)th digit, then round mth digit to the nearest even digit, and if not so, round mth digit to the next digit.

So, we will first convert all the numbers in the normalised form:-

• $a = 0.12468 \times 10^3$
• $a = -0.12345 \times 10$
• $a = 0.55555 \times 10^0$
• $a = -0.54321 \times 10^{-2}$

Now, we will round them using 4 - digit rounding arithmetic and the rules stated above. Therefore, we get:-

• $a = 0.1247 \times 10^3 = 124.7$
• $a = -0.1234 \times 10 = 1.234$
• $a = 0.5556 \times 10^0 = 0.5556$
• $a = -0.5432 \times 10^{-2} = -0.005432$

I hope it is helpful.