# Base ten decimal number 1.043 188 691 converted to 64 bit double precision IEEE 754 binary floating point standard

## How to convert the decimal number 1.043 188 691(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

### 1. First, convert to binary (base 2) the integer part: 1. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

• division = quotient + remainder;
• 1 ÷ 2 = 0 + 1;

### 3. Convert to binary (base 2) the fractional part: 0.043 188 691. Multiply it 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) 0.043 188 691 × 2 = 0 + 0.086 377 382;
• 2) 0.086 377 382 × 2 = 0 + 0.172 754 764;
• 3) 0.172 754 764 × 2 = 0 + 0.345 509 528;
• 4) 0.345 509 528 × 2 = 0 + 0.691 019 056;
• 5) 0.691 019 056 × 2 = 1 + 0.382 038 112;
• 6) 0.382 038 112 × 2 = 0 + 0.764 076 224;
• 7) 0.764 076 224 × 2 = 1 + 0.528 152 448;
• 8) 0.528 152 448 × 2 = 1 + 0.056 304 896;
• 9) 0.056 304 896 × 2 = 0 + 0.112 609 792;
• 10) 0.112 609 792 × 2 = 0 + 0.225 219 584;
• 11) 0.225 219 584 × 2 = 0 + 0.450 439 168;
• 12) 0.450 439 168 × 2 = 0 + 0.900 878 336;
• 13) 0.900 878 336 × 2 = 1 + 0.801 756 672;
• 14) 0.801 756 672 × 2 = 1 + 0.603 513 344;
• 15) 0.603 513 344 × 2 = 1 + 0.207 026 688;
• 16) 0.207 026 688 × 2 = 0 + 0.414 053 376;
• 17) 0.414 053 376 × 2 = 0 + 0.828 106 752;
• 18) 0.828 106 752 × 2 = 1 + 0.656 213 504;
• 19) 0.656 213 504 × 2 = 1 + 0.312 427 008;
• 20) 0.312 427 008 × 2 = 0 + 0.624 854 016;
• 21) 0.624 854 016 × 2 = 1 + 0.249 708 032;
• 22) 0.249 708 032 × 2 = 0 + 0.499 416 064;
• 23) 0.499 416 064 × 2 = 0 + 0.998 832 128;
• 24) 0.998 832 128 × 2 = 1 + 0.997 664 256;
• 25) 0.997 664 256 × 2 = 1 + 0.995 328 512;
• 26) 0.995 328 512 × 2 = 1 + 0.990 657 024;
• 27) 0.990 657 024 × 2 = 1 + 0.981 314 048;
• 28) 0.981 314 048 × 2 = 1 + 0.962 628 096;
• 29) 0.962 628 096 × 2 = 1 + 0.925 256 192;
• 30) 0.925 256 192 × 2 = 1 + 0.850 512 384;
• 31) 0.850 512 384 × 2 = 1 + 0.701 024 768;
• 32) 0.701 024 768 × 2 = 1 + 0.402 049 536;
• 33) 0.402 049 536 × 2 = 0 + 0.804 099 072;
• 34) 0.804 099 072 × 2 = 1 + 0.608 198 144;
• 35) 0.608 198 144 × 2 = 1 + 0.216 396 288;
• 36) 0.216 396 288 × 2 = 0 + 0.432 792 576;
• 37) 0.432 792 576 × 2 = 0 + 0.865 585 152;
• 38) 0.865 585 152 × 2 = 1 + 0.731 170 304;
• 39) 0.731 170 304 × 2 = 1 + 0.462 340 608;
• 40) 0.462 340 608 × 2 = 0 + 0.924 681 216;
• 41) 0.924 681 216 × 2 = 1 + 0.849 362 432;
• 42) 0.849 362 432 × 2 = 1 + 0.698 724 864;
• 43) 0.698 724 864 × 2 = 1 + 0.397 449 728;
• 44) 0.397 449 728 × 2 = 0 + 0.794 899 456;
• 45) 0.794 899 456 × 2 = 1 + 0.589 798 912;
• 46) 0.589 798 912 × 2 = 1 + 0.179 597 824;
• 47) 0.179 597 824 × 2 = 0 + 0.359 195 648;
• 48) 0.359 195 648 × 2 = 0 + 0.718 391 296;
• 49) 0.718 391 296 × 2 = 1 + 0.436 782 592;
• 50) 0.436 782 592 × 2 = 0 + 0.873 565 184;
• 51) 0.873 565 184 × 2 = 1 + 0.747 130 368;
• 52) 0.747 130 368 × 2 = 1 + 0.494 260 736;
• 53) 0.494 260 736 × 2 = 0 + 0.988 521 472;

### 6. 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:

#### 1 023(10)

• division = quotient + remainder;
• 1 023 ÷ 2 = 511 + 1;
• 511 ÷ 2 = 255 + 1;
• 255 ÷ 2 = 127 + 1;
• 127 ÷ 2 = 63 + 1;
• 63 ÷ 2 = 31 + 1;
• 31 ÷ 2 = 15 + 1;
• 15 ÷ 2 = 7 + 1;
• 7 ÷ 2 = 3 + 1;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

## 0 - 011 1111 1111 - 0000 1011 0000 1110 0110 1001 1111 1111 0110 0110 1110 1100 1011

(64 bits IEEE 754)

• 0

63

• 0

62
• 1

61
• 1

60
• 1

59
• 1

58
• 1

57
• 1

56
• 1

55
• 1

54
• 1

53
• 1

52

• 0

51
• 0

50
• 0

49
• 0

48
• 1

47
• 0

46
• 1

45
• 1

44
• 0

43
• 0

42
• 0

41
• 0

40
• 1

39
• 1

38
• 1

37
• 0

36
• 0

35
• 1

34
• 1

33
• 0

32
• 1

31
• 0

30
• 0

29
• 1

28
• 1

27
• 1

26
• 1

25
• 1

24
• 1

23
• 1

22
• 1

21
• 1

20
• 0

19
• 1

18
• 1

17
• 0

16
• 0

15
• 1

14
• 1

13
• 0

12
• 1

11
• 1

10
• 1

9
• 0

8
• 1

7
• 1

6
• 0

5
• 0

4
• 1

3
• 0

2
• 1

1
• 1

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