-0.000 282 005 876 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 876 2(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
-0.000 282 005 876 2(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:

|-0.000 282 005 876 2| = 0.000 282 005 876 2


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: 0.000 282 005 876 2.

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) 0.000 282 005 876 2 × 2 = 0 + 0.000 564 011 752 4;
  • 2) 0.000 564 011 752 4 × 2 = 0 + 0.001 128 023 504 8;
  • 3) 0.001 128 023 504 8 × 2 = 0 + 0.002 256 047 009 6;
  • 4) 0.002 256 047 009 6 × 2 = 0 + 0.004 512 094 019 2;
  • 5) 0.004 512 094 019 2 × 2 = 0 + 0.009 024 188 038 4;
  • 6) 0.009 024 188 038 4 × 2 = 0 + 0.018 048 376 076 8;
  • 7) 0.018 048 376 076 8 × 2 = 0 + 0.036 096 752 153 6;
  • 8) 0.036 096 752 153 6 × 2 = 0 + 0.072 193 504 307 2;
  • 9) 0.072 193 504 307 2 × 2 = 0 + 0.144 387 008 614 4;
  • 10) 0.144 387 008 614 4 × 2 = 0 + 0.288 774 017 228 8;
  • 11) 0.288 774 017 228 8 × 2 = 0 + 0.577 548 034 457 6;
  • 12) 0.577 548 034 457 6 × 2 = 1 + 0.155 096 068 915 2;
  • 13) 0.155 096 068 915 2 × 2 = 0 + 0.310 192 137 830 4;
  • 14) 0.310 192 137 830 4 × 2 = 0 + 0.620 384 275 660 8;
  • 15) 0.620 384 275 660 8 × 2 = 1 + 0.240 768 551 321 6;
  • 16) 0.240 768 551 321 6 × 2 = 0 + 0.481 537 102 643 2;
  • 17) 0.481 537 102 643 2 × 2 = 0 + 0.963 074 205 286 4;
  • 18) 0.963 074 205 286 4 × 2 = 1 + 0.926 148 410 572 8;
  • 19) 0.926 148 410 572 8 × 2 = 1 + 0.852 296 821 145 6;
  • 20) 0.852 296 821 145 6 × 2 = 1 + 0.704 593 642 291 2;
  • 21) 0.704 593 642 291 2 × 2 = 1 + 0.409 187 284 582 4;
  • 22) 0.409 187 284 582 4 × 2 = 0 + 0.818 374 569 164 8;
  • 23) 0.818 374 569 164 8 × 2 = 1 + 0.636 749 138 329 6;
  • 24) 0.636 749 138 329 6 × 2 = 1 + 0.273 498 276 659 2;
  • 25) 0.273 498 276 659 2 × 2 = 0 + 0.546 996 553 318 4;
  • 26) 0.546 996 553 318 4 × 2 = 1 + 0.093 993 106 636 8;
  • 27) 0.093 993 106 636 8 × 2 = 0 + 0.187 986 213 273 6;
  • 28) 0.187 986 213 273 6 × 2 = 0 + 0.375 972 426 547 2;
  • 29) 0.375 972 426 547 2 × 2 = 0 + 0.751 944 853 094 4;
  • 30) 0.751 944 853 094 4 × 2 = 1 + 0.503 889 706 188 8;
  • 31) 0.503 889 706 188 8 × 2 = 1 + 0.007 779 412 377 6;
  • 32) 0.007 779 412 377 6 × 2 = 0 + 0.015 558 824 755 2;
  • 33) 0.015 558 824 755 2 × 2 = 0 + 0.031 117 649 510 4;
  • 34) 0.031 117 649 510 4 × 2 = 0 + 0.062 235 299 020 8;
  • 35) 0.062 235 299 020 8 × 2 = 0 + 0.124 470 598 041 6;
  • 36) 0.124 470 598 041 6 × 2 = 0 + 0.248 941 196 083 2;
  • 37) 0.248 941 196 083 2 × 2 = 0 + 0.497 882 392 166 4;
  • 38) 0.497 882 392 166 4 × 2 = 0 + 0.995 764 784 332 8;
  • 39) 0.995 764 784 332 8 × 2 = 1 + 0.991 529 568 665 6;
  • 40) 0.991 529 568 665 6 × 2 = 1 + 0.983 059 137 331 2;
  • 41) 0.983 059 137 331 2 × 2 = 1 + 0.966 118 274 662 4;
  • 42) 0.966 118 274 662 4 × 2 = 1 + 0.932 236 549 324 8;
  • 43) 0.932 236 549 324 8 × 2 = 1 + 0.864 473 098 649 6;
  • 44) 0.864 473 098 649 6 × 2 = 1 + 0.728 946 197 299 2;
  • 45) 0.728 946 197 299 2 × 2 = 1 + 0.457 892 394 598 4;
  • 46) 0.457 892 394 598 4 × 2 = 0 + 0.915 784 789 196 8;
  • 47) 0.915 784 789 196 8 × 2 = 1 + 0.831 569 578 393 6;
  • 48) 0.831 569 578 393 6 × 2 = 1 + 0.663 139 156 787 2;
  • 49) 0.663 139 156 787 2 × 2 = 1 + 0.326 278 313 574 4;
  • 50) 0.326 278 313 574 4 × 2 = 0 + 0.652 556 627 148 8;
  • 51) 0.652 556 627 148 8 × 2 = 1 + 0.305 113 254 297 6;
  • 52) 0.305 113 254 297 6 × 2 = 0 + 0.610 226 508 595 2;
  • 53) 0.610 226 508 595 2 × 2 = 1 + 0.220 453 017 190 4;
  • 54) 0.220 453 017 190 4 × 2 = 0 + 0.440 906 034 380 8;
  • 55) 0.440 906 034 380 8 × 2 = 0 + 0.881 812 068 761 6;
  • 56) 0.881 812 068 761 6 × 2 = 1 + 0.763 624 137 523 2;
  • 57) 0.763 624 137 523 2 × 2 = 1 + 0.527 248 275 046 4;
  • 58) 0.527 248 275 046 4 × 2 = 1 + 0.054 496 550 092 8;
  • 59) 0.054 496 550 092 8 × 2 = 0 + 0.108 993 100 185 6;
  • 60) 0.108 993 100 185 6 × 2 = 0 + 0.217 986 200 371 2;
  • 61) 0.217 986 200 371 2 × 2 = 0 + 0.435 972 400 742 4;
  • 62) 0.435 972 400 742 4 × 2 = 0 + 0.871 944 801 484 8;
  • 63) 0.871 944 801 484 8 × 2 = 1 + 0.743 889 602 969 6;
  • 64) 0.743 889 602 969 6 × 2 = 1 + 0.487 779 205 939 2;

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:


0.000 282 005 876 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0011 1111 1011 1010 1001 1100 0011(2)

6. Positive number before normalization:

0.000 282 005 876 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0011 1111 1011 1010 1001 1100 0011(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:


0.000 282 005 876 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0011 1111 1011 1010 1001 1100 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0000 0011 1111 1011 1010 1001 1100 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0000 0011 1111 1011 1010 1001 1100 0011(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 0011 1111 1011 1010 1001 1100 0011


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 0011 1111 1011 1010 1001 1100 0011 =

0010 0111 1011 0100 0110 0000 0011 1111 1011 1010 1001 1100 0011


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 0011 1111 1011 1010 1001 1100 0011


Decimal number -0.000 282 005 876 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0000 0011 1111 1011 1010 1001 1100 0011

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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