# Convert 1.003 143 86 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

## 1.003 143 86(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 the binary (base 2) the integer part: 1. Divide the number repeatedly by 2.

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

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

### 3. Convert to the binary (base 2) the fractional part: 0.003 143 86.

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

• #) multiplying = integer + fractional part;
• 1) 0.003 143 86 × 2 = 0 + 0.006 287 72;
• 2) 0.006 287 72 × 2 = 0 + 0.012 575 44;
• 3) 0.012 575 44 × 2 = 0 + 0.025 150 88;
• 4) 0.025 150 88 × 2 = 0 + 0.050 301 76;
• 5) 0.050 301 76 × 2 = 0 + 0.100 603 52;
• 6) 0.100 603 52 × 2 = 0 + 0.201 207 04;
• 7) 0.201 207 04 × 2 = 0 + 0.402 414 08;
• 8) 0.402 414 08 × 2 = 0 + 0.804 828 16;
• 9) 0.804 828 16 × 2 = 1 + 0.609 656 32;
• 10) 0.609 656 32 × 2 = 1 + 0.219 312 64;
• 11) 0.219 312 64 × 2 = 0 + 0.438 625 28;
• 12) 0.438 625 28 × 2 = 0 + 0.877 250 56;
• 13) 0.877 250 56 × 2 = 1 + 0.754 501 12;
• 14) 0.754 501 12 × 2 = 1 + 0.509 002 24;
• 15) 0.509 002 24 × 2 = 1 + 0.018 004 48;
• 16) 0.018 004 48 × 2 = 0 + 0.036 008 96;
• 17) 0.036 008 96 × 2 = 0 + 0.072 017 92;
• 18) 0.072 017 92 × 2 = 0 + 0.144 035 84;
• 19) 0.144 035 84 × 2 = 0 + 0.288 071 68;
• 20) 0.288 071 68 × 2 = 0 + 0.576 143 36;
• 21) 0.576 143 36 × 2 = 1 + 0.152 286 72;
• 22) 0.152 286 72 × 2 = 0 + 0.304 573 44;
• 23) 0.304 573 44 × 2 = 0 + 0.609 146 88;
• 24) 0.609 146 88 × 2 = 1 + 0.218 293 76;
• 25) 0.218 293 76 × 2 = 0 + 0.436 587 52;
• 26) 0.436 587 52 × 2 = 0 + 0.873 175 04;
• 27) 0.873 175 04 × 2 = 1 + 0.746 350 08;
• 28) 0.746 350 08 × 2 = 1 + 0.492 700 16;
• 29) 0.492 700 16 × 2 = 0 + 0.985 400 32;
• 30) 0.985 400 32 × 2 = 1 + 0.970 800 64;
• 31) 0.970 800 64 × 2 = 1 + 0.941 601 28;
• 32) 0.941 601 28 × 2 = 1 + 0.883 202 56;
• 33) 0.883 202 56 × 2 = 1 + 0.766 405 12;
• 34) 0.766 405 12 × 2 = 1 + 0.532 810 24;
• 35) 0.532 810 24 × 2 = 1 + 0.065 620 48;
• 36) 0.065 620 48 × 2 = 0 + 0.131 240 96;
• 37) 0.131 240 96 × 2 = 0 + 0.262 481 92;
• 38) 0.262 481 92 × 2 = 0 + 0.524 963 84;
• 39) 0.524 963 84 × 2 = 1 + 0.049 927 68;
• 40) 0.049 927 68 × 2 = 0 + 0.099 855 36;
• 41) 0.099 855 36 × 2 = 0 + 0.199 710 72;
• 42) 0.199 710 72 × 2 = 0 + 0.399 421 44;
• 43) 0.399 421 44 × 2 = 0 + 0.798 842 88;
• 44) 0.798 842 88 × 2 = 1 + 0.597 685 76;
• 45) 0.597 685 76 × 2 = 1 + 0.195 371 52;
• 46) 0.195 371 52 × 2 = 0 + 0.390 743 04;
• 47) 0.390 743 04 × 2 = 0 + 0.781 486 08;
• 48) 0.781 486 08 × 2 = 1 + 0.562 972 16;
• 49) 0.562 972 16 × 2 = 1 + 0.125 944 32;
• 50) 0.125 944 32 × 2 = 0 + 0.251 888 64;
• 51) 0.251 888 64 × 2 = 0 + 0.503 777 28;
• 52) 0.503 777 28 × 2 = 1 + 0.007 554 56;
• 53) 0.007 554 56 × 2 = 0 + 0.015 109 12;

### 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;
• 1 023 ÷ 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;

## Number 1.003 143 86 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point: 0 - 011 1111 1111 - 0000 0000 1100 1110 0000 1001 0011 0111 1110 0010 0001 1001 1001

(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
• 0

47
• 0

46
• 0

45
• 0

44
• 1

43
• 1

42
• 0

41
• 0

40
• 1

39
• 1

38
• 1

37
• 0

36
• 0

35
• 0

34
• 0

33
• 0

32
• 1

31
• 0

30
• 0

29
• 1

28
• 0

27
• 0

26
• 1

25
• 1

24
• 0

23
• 1

22
• 1

21
• 1

20
• 1

19
• 1

18
• 1

17
• 0

16
• 0

15
• 0

14
• 1

13
• 0

12
• 0

11
• 0

10
• 0

9
• 1

8
• 1

7
• 0

6
• 0

5
• 1

4
• 1

3
• 0

2
• 0

1
• 1

0

## Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

 1.003 143 86 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:56 UTC (GMT) 10 653 532 161 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:56 UTC (GMT) 4 916 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:56 UTC (GMT) -51.21 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:55 UTC (GMT) -44 036.780 11 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:55 UTC (GMT) 38.28 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:55 UTC (GMT) 11 111 111 091 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:55 UTC (GMT) 8 987 551 792 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT) 2.000 000 007 3 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT) 999 999 999 999 999 999 999 999 999 999 999 999 991 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT) -86 749 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:54 UTC (GMT) 0.063 232 421 2 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:53 UTC (GMT) -284.011 100 000 001 110 001 110 000 000 010 100 011 110 101 2 to 64 bit double precision IEEE 754 binary floating point = ? Nov 30 09:53 UTC (GMT) All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

## 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