# Convert the Number 20.654 343 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations

## Number 20.654 343(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

### 1. First, convert to binary (in base 2) the integer part: 20. Divide the number repeatedly by 2.

#### We stop when we get a quotient that is equal to zero.

• division = quotient + remainder;
• 20 ÷ 2 = 10 + 0;
• 10 ÷ 2 = 5 + 0;
• 5 ÷ 2 = 2 + 1;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

### 3. Convert to binary (base 2) the fractional part: 0.654 343.

#### Stop when we get a fractional part that is equal to zero.

• #) multiplying = integer + fractional part;
• 1) 0.654 343 × 2 = 1 + 0.308 686;
• 2) 0.308 686 × 2 = 0 + 0.617 372;
• 3) 0.617 372 × 2 = 1 + 0.234 744;
• 4) 0.234 744 × 2 = 0 + 0.469 488;
• 5) 0.469 488 × 2 = 0 + 0.938 976;
• 6) 0.938 976 × 2 = 1 + 0.877 952;
• 7) 0.877 952 × 2 = 1 + 0.755 904;
• 8) 0.755 904 × 2 = 1 + 0.511 808;
• 9) 0.511 808 × 2 = 1 + 0.023 616;
• 10) 0.023 616 × 2 = 0 + 0.047 232;
• 11) 0.047 232 × 2 = 0 + 0.094 464;
• 12) 0.094 464 × 2 = 0 + 0.188 928;
• 13) 0.188 928 × 2 = 0 + 0.377 856;
• 14) 0.377 856 × 2 = 0 + 0.755 712;
• 15) 0.755 712 × 2 = 1 + 0.511 424;
• 16) 0.511 424 × 2 = 1 + 0.022 848;
• 17) 0.022 848 × 2 = 0 + 0.045 696;
• 18) 0.045 696 × 2 = 0 + 0.091 392;
• 19) 0.091 392 × 2 = 0 + 0.182 784;
• 20) 0.182 784 × 2 = 0 + 0.365 568;
• 21) 0.365 568 × 2 = 0 + 0.731 136;
• 22) 0.731 136 × 2 = 1 + 0.462 272;
• 23) 0.462 272 × 2 = 0 + 0.924 544;
• 24) 0.924 544 × 2 = 1 + 0.849 088;
• 25) 0.849 088 × 2 = 1 + 0.698 176;
• 26) 0.698 176 × 2 = 1 + 0.396 352;
• 27) 0.396 352 × 2 = 0 + 0.792 704;
• 28) 0.792 704 × 2 = 1 + 0.585 408;
• 29) 0.585 408 × 2 = 1 + 0.170 816;
• 30) 0.170 816 × 2 = 0 + 0.341 632;
• 31) 0.341 632 × 2 = 0 + 0.683 264;
• 32) 0.683 264 × 2 = 1 + 0.366 528;
• 33) 0.366 528 × 2 = 0 + 0.733 056;
• 34) 0.733 056 × 2 = 1 + 0.466 112;
• 35) 0.466 112 × 2 = 0 + 0.932 224;
• 36) 0.932 224 × 2 = 1 + 0.864 448;
• 37) 0.864 448 × 2 = 1 + 0.728 896;
• 38) 0.728 896 × 2 = 1 + 0.457 792;
• 39) 0.457 792 × 2 = 0 + 0.915 584;
• 40) 0.915 584 × 2 = 1 + 0.831 168;
• 41) 0.831 168 × 2 = 1 + 0.662 336;
• 42) 0.662 336 × 2 = 1 + 0.324 672;
• 43) 0.324 672 × 2 = 0 + 0.649 344;
• 44) 0.649 344 × 2 = 1 + 0.298 688;
• 45) 0.298 688 × 2 = 0 + 0.597 376;
• 46) 0.597 376 × 2 = 1 + 0.194 752;
• 47) 0.194 752 × 2 = 0 + 0.389 504;
• 48) 0.389 504 × 2 = 0 + 0.779 008;
• 49) 0.779 008 × 2 = 1 + 0.558 016;
• 50) 0.558 016 × 2 = 1 + 0.116 032;
• 51) 0.116 032 × 2 = 0 + 0.232 064;
• 52) 0.232 064 × 2 = 0 + 0.464 128;
• 53) 0.464 128 × 2 = 0 + 0.928 256;

### 9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

#### Use the same technique of repeatedly dividing by 2:

• division = quotient + remainder;
• 1 027 ÷ 2 = 513 + 1;
• 513 ÷ 2 = 256 + 1;
• 256 ÷ 2 = 128 + 0;
• 128 ÷ 2 = 64 + 0;
• 64 ÷ 2 = 32 + 0;
• 32 ÷ 2 = 16 + 0;
• 16 ÷ 2 = 8 + 0;
• 8 ÷ 2 = 4 + 0;
• 4 ÷ 2 = 2 + 0;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

## The base ten decimal number 20.654 343 converted and written in 64 bit double precision IEEE 754 binary floating point representation: 0 - 100 0000 0011 - 0100 1010 0111 1000 0011 0000 0101 1101 1001 0101 1101 1101 0100

(64 bits IEEE 754)

• 0

63

• 1

62
• 0

61
• 0

60
• 0

59
• 0

58
• 0

57
• 0

56
• 0

55
• 0

54
• 1

53
• 1

52

• 0

51
• 1

50
• 0

49
• 0

48
• 1

47
• 0

46
• 1

45
• 0

44
• 0

43
• 1

42
• 1

41
• 1

40
• 1

39
• 0

38
• 0

37
• 0

36
• 0

35
• 0

34
• 1

33
• 1

32
• 0

31
• 0

30
• 0

29
• 0

28
• 0

27
• 1

26
• 0

25
• 1

24
• 1

23
• 1

22
• 0

21
• 1

20
• 1

19
• 0

18
• 0

17
• 1

16
• 0

15
• 1

14
• 0

13
• 1

12
• 1

11
• 1

10
• 0

9
• 1

8
• 1

7
• 1

6
• 0

5
• 1

4
• 0

3
• 1

2
• 0

1
• 0

0

## How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

### Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

• 1. If the number to be converted is negative, start with its the positive version.
• 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
• 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
• 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
• 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
• 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
• 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
• 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
• 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

### Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

|-31.640 215| = 31.640 215

• 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
• division = quotient + remainder;
• 31 ÷ 2 = 15 + 1;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;
• We have encountered a quotient that is ZERO => FULL STOP
• 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

31(10) = 1 1111(2)

• 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
• #) multiplying = integer + fractional part;
• 1) 0.640 215 × 2 = 1 + 0.280 43;
• 2) 0.280 43 × 2 = 0 + 0.560 86;
• 3) 0.560 86 × 2 = 1 + 0.121 72;
• 4) 0.121 72 × 2 = 0 + 0.243 44;
• 5) 0.243 44 × 2 = 0 + 0.486 88;
• 6) 0.486 88 × 2 = 0 + 0.973 76;
• 7) 0.973 76 × 2 = 1 + 0.947 52;
• 8) 0.947 52 × 2 = 1 + 0.895 04;
• 9) 0.895 04 × 2 = 1 + 0.790 08;
• 10) 0.790 08 × 2 = 1 + 0.580 16;
• 11) 0.580 16 × 2 = 1 + 0.160 32;
• 12) 0.160 32 × 2 = 0 + 0.320 64;
• 13) 0.320 64 × 2 = 0 + 0.641 28;
• 14) 0.641 28 × 2 = 1 + 0.282 56;
• 15) 0.282 56 × 2 = 0 + 0.565 12;
• 16) 0.565 12 × 2 = 1 + 0.130 24;
• 17) 0.130 24 × 2 = 0 + 0.260 48;
• 18) 0.260 48 × 2 = 0 + 0.520 96;
• 19) 0.520 96 × 2 = 1 + 0.041 92;
• 20) 0.041 92 × 2 = 0 + 0.083 84;
• 21) 0.083 84 × 2 = 0 + 0.167 68;
• 22) 0.167 68 × 2 = 0 + 0.335 36;
• 23) 0.335 36 × 2 = 0 + 0.670 72;
• 24) 0.670 72 × 2 = 1 + 0.341 44;
• 25) 0.341 44 × 2 = 0 + 0.682 88;
• 26) 0.682 88 × 2 = 1 + 0.365 76;
• 27) 0.365 76 × 2 = 0 + 0.731 52;
• 28) 0.731 52 × 2 = 1 + 0.463 04;
• 29) 0.463 04 × 2 = 0 + 0.926 08;
• 30) 0.926 08 × 2 = 1 + 0.852 16;
• 31) 0.852 16 × 2 = 1 + 0.704 32;
• 32) 0.704 32 × 2 = 1 + 0.408 64;
• 33) 0.408 64 × 2 = 0 + 0.817 28;
• 34) 0.817 28 × 2 = 1 + 0.634 56;
• 35) 0.634 56 × 2 = 1 + 0.269 12;
• 36) 0.269 12 × 2 = 0 + 0.538 24;
• 37) 0.538 24 × 2 = 1 + 0.076 48;
• 38) 0.076 48 × 2 = 0 + 0.152 96;
• 39) 0.152 96 × 2 = 0 + 0.305 92;
• 40) 0.305 92 × 2 = 0 + 0.611 84;
• 41) 0.611 84 × 2 = 1 + 0.223 68;
• 42) 0.223 68 × 2 = 0 + 0.447 36;
• 43) 0.447 36 × 2 = 0 + 0.894 72;
• 44) 0.894 72 × 2 = 1 + 0.789 44;
• 45) 0.789 44 × 2 = 1 + 0.578 88;
• 46) 0.578 88 × 2 = 1 + 0.157 76;
• 47) 0.157 76 × 2 = 0 + 0.315 52;
• 48) 0.315 52 × 2 = 0 + 0.631 04;
• 49) 0.631 04 × 2 = 1 + 0.262 08;
• 50) 0.262 08 × 2 = 0 + 0.524 16;
• 51) 0.524 16 × 2 = 1 + 0.048 32;
• 52) 0.048 32 × 2 = 0 + 0.096 64;
• 53) 0.096 64 × 2 = 0 + 0.193 28;
• We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
• 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

• 6. Summarizing - the positive number before normalization:

31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

• 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

31.640 215(10) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

• 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign: 1 (a negative number)

Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

• 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
100 0000 0011(2)

• 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

• Conclusion:

Sign (1 bit) = 1 (a negative number)

Exponent (8 bits) = 100 0000 0011

Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100