-0.000 282 005 874 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 874 4(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 282 005 874 4(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 282 005 874 4| = 0.000 282 005 874 4


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

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


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

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 282 005 874 4.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

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


  • #) multiplying = integer + fractional part;
  • 1) 0.000 282 005 874 4 × 2 = 0 + 0.000 564 011 748 8;
  • 2) 0.000 564 011 748 8 × 2 = 0 + 0.001 128 023 497 6;
  • 3) 0.001 128 023 497 6 × 2 = 0 + 0.002 256 046 995 2;
  • 4) 0.002 256 046 995 2 × 2 = 0 + 0.004 512 093 990 4;
  • 5) 0.004 512 093 990 4 × 2 = 0 + 0.009 024 187 980 8;
  • 6) 0.009 024 187 980 8 × 2 = 0 + 0.018 048 375 961 6;
  • 7) 0.018 048 375 961 6 × 2 = 0 + 0.036 096 751 923 2;
  • 8) 0.036 096 751 923 2 × 2 = 0 + 0.072 193 503 846 4;
  • 9) 0.072 193 503 846 4 × 2 = 0 + 0.144 387 007 692 8;
  • 10) 0.144 387 007 692 8 × 2 = 0 + 0.288 774 015 385 6;
  • 11) 0.288 774 015 385 6 × 2 = 0 + 0.577 548 030 771 2;
  • 12) 0.577 548 030 771 2 × 2 = 1 + 0.155 096 061 542 4;
  • 13) 0.155 096 061 542 4 × 2 = 0 + 0.310 192 123 084 8;
  • 14) 0.310 192 123 084 8 × 2 = 0 + 0.620 384 246 169 6;
  • 15) 0.620 384 246 169 6 × 2 = 1 + 0.240 768 492 339 2;
  • 16) 0.240 768 492 339 2 × 2 = 0 + 0.481 536 984 678 4;
  • 17) 0.481 536 984 678 4 × 2 = 0 + 0.963 073 969 356 8;
  • 18) 0.963 073 969 356 8 × 2 = 1 + 0.926 147 938 713 6;
  • 19) 0.926 147 938 713 6 × 2 = 1 + 0.852 295 877 427 2;
  • 20) 0.852 295 877 427 2 × 2 = 1 + 0.704 591 754 854 4;
  • 21) 0.704 591 754 854 4 × 2 = 1 + 0.409 183 509 708 8;
  • 22) 0.409 183 509 708 8 × 2 = 0 + 0.818 367 019 417 6;
  • 23) 0.818 367 019 417 6 × 2 = 1 + 0.636 734 038 835 2;
  • 24) 0.636 734 038 835 2 × 2 = 1 + 0.273 468 077 670 4;
  • 25) 0.273 468 077 670 4 × 2 = 0 + 0.546 936 155 340 8;
  • 26) 0.546 936 155 340 8 × 2 = 1 + 0.093 872 310 681 6;
  • 27) 0.093 872 310 681 6 × 2 = 0 + 0.187 744 621 363 2;
  • 28) 0.187 744 621 363 2 × 2 = 0 + 0.375 489 242 726 4;
  • 29) 0.375 489 242 726 4 × 2 = 0 + 0.750 978 485 452 8;
  • 30) 0.750 978 485 452 8 × 2 = 1 + 0.501 956 970 905 6;
  • 31) 0.501 956 970 905 6 × 2 = 1 + 0.003 913 941 811 2;
  • 32) 0.003 913 941 811 2 × 2 = 0 + 0.007 827 883 622 4;
  • 33) 0.007 827 883 622 4 × 2 = 0 + 0.015 655 767 244 8;
  • 34) 0.015 655 767 244 8 × 2 = 0 + 0.031 311 534 489 6;
  • 35) 0.031 311 534 489 6 × 2 = 0 + 0.062 623 068 979 2;
  • 36) 0.062 623 068 979 2 × 2 = 0 + 0.125 246 137 958 4;
  • 37) 0.125 246 137 958 4 × 2 = 0 + 0.250 492 275 916 8;
  • 38) 0.250 492 275 916 8 × 2 = 0 + 0.500 984 551 833 6;
  • 39) 0.500 984 551 833 6 × 2 = 1 + 0.001 969 103 667 2;
  • 40) 0.001 969 103 667 2 × 2 = 0 + 0.003 938 207 334 4;
  • 41) 0.003 938 207 334 4 × 2 = 0 + 0.007 876 414 668 8;
  • 42) 0.007 876 414 668 8 × 2 = 0 + 0.015 752 829 337 6;
  • 43) 0.015 752 829 337 6 × 2 = 0 + 0.031 505 658 675 2;
  • 44) 0.031 505 658 675 2 × 2 = 0 + 0.063 011 317 350 4;
  • 45) 0.063 011 317 350 4 × 2 = 0 + 0.126 022 634 700 8;
  • 46) 0.126 022 634 700 8 × 2 = 0 + 0.252 045 269 401 6;
  • 47) 0.252 045 269 401 6 × 2 = 0 + 0.504 090 538 803 2;
  • 48) 0.504 090 538 803 2 × 2 = 1 + 0.008 181 077 606 4;
  • 49) 0.008 181 077 606 4 × 2 = 0 + 0.016 362 155 212 8;
  • 50) 0.016 362 155 212 8 × 2 = 0 + 0.032 724 310 425 6;
  • 51) 0.032 724 310 425 6 × 2 = 0 + 0.065 448 620 851 2;
  • 52) 0.065 448 620 851 2 × 2 = 0 + 0.130 897 241 702 4;
  • 53) 0.130 897 241 702 4 × 2 = 0 + 0.261 794 483 404 8;
  • 54) 0.261 794 483 404 8 × 2 = 0 + 0.523 588 966 809 6;
  • 55) 0.523 588 966 809 6 × 2 = 1 + 0.047 177 933 619 2;
  • 56) 0.047 177 933 619 2 × 2 = 0 + 0.094 355 867 238 4;
  • 57) 0.094 355 867 238 4 × 2 = 0 + 0.188 711 734 476 8;
  • 58) 0.188 711 734 476 8 × 2 = 0 + 0.377 423 468 953 6;
  • 59) 0.377 423 468 953 6 × 2 = 0 + 0.754 846 937 907 2;
  • 60) 0.754 846 937 907 2 × 2 = 1 + 0.509 693 875 814 4;
  • 61) 0.509 693 875 814 4 × 2 = 1 + 0.019 387 751 628 8;
  • 62) 0.019 387 751 628 8 × 2 = 0 + 0.038 775 503 257 6;
  • 63) 0.038 775 503 257 6 × 2 = 0 + 0.077 551 006 515 2;
  • 64) 0.077 551 006 515 2 × 2 = 0 + 0.155 102 013 030 4;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 282 005 874 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000(2)

6. Positive number before normalization:

0.000 282 005 874 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 874 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000(2) × 2-12


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)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. 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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1011(10) =

011 1111 0011(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000 =

0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

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

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000


Decimal number -0.000 282 005 874 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 0010 0000 0001 0000 0010 0001 1000

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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:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 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:

  • 1. Start with the positive version of the number:

    |-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)

    Exponent (unadjusted): 4

    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

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100