-0.000 282 006 186 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 186(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
-0.000 282 006 186(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:

|-0.000 282 006 186| = 0.000 282 006 186


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: 0.000 282 006 186.

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) 0.000 282 006 186 × 2 = 0 + 0.000 564 012 372;
  • 2) 0.000 564 012 372 × 2 = 0 + 0.001 128 024 744;
  • 3) 0.001 128 024 744 × 2 = 0 + 0.002 256 049 488;
  • 4) 0.002 256 049 488 × 2 = 0 + 0.004 512 098 976;
  • 5) 0.004 512 098 976 × 2 = 0 + 0.009 024 197 952;
  • 6) 0.009 024 197 952 × 2 = 0 + 0.018 048 395 904;
  • 7) 0.018 048 395 904 × 2 = 0 + 0.036 096 791 808;
  • 8) 0.036 096 791 808 × 2 = 0 + 0.072 193 583 616;
  • 9) 0.072 193 583 616 × 2 = 0 + 0.144 387 167 232;
  • 10) 0.144 387 167 232 × 2 = 0 + 0.288 774 334 464;
  • 11) 0.288 774 334 464 × 2 = 0 + 0.577 548 668 928;
  • 12) 0.577 548 668 928 × 2 = 1 + 0.155 097 337 856;
  • 13) 0.155 097 337 856 × 2 = 0 + 0.310 194 675 712;
  • 14) 0.310 194 675 712 × 2 = 0 + 0.620 389 351 424;
  • 15) 0.620 389 351 424 × 2 = 1 + 0.240 778 702 848;
  • 16) 0.240 778 702 848 × 2 = 0 + 0.481 557 405 696;
  • 17) 0.481 557 405 696 × 2 = 0 + 0.963 114 811 392;
  • 18) 0.963 114 811 392 × 2 = 1 + 0.926 229 622 784;
  • 19) 0.926 229 622 784 × 2 = 1 + 0.852 459 245 568;
  • 20) 0.852 459 245 568 × 2 = 1 + 0.704 918 491 136;
  • 21) 0.704 918 491 136 × 2 = 1 + 0.409 836 982 272;
  • 22) 0.409 836 982 272 × 2 = 0 + 0.819 673 964 544;
  • 23) 0.819 673 964 544 × 2 = 1 + 0.639 347 929 088;
  • 24) 0.639 347 929 088 × 2 = 1 + 0.278 695 858 176;
  • 25) 0.278 695 858 176 × 2 = 0 + 0.557 391 716 352;
  • 26) 0.557 391 716 352 × 2 = 1 + 0.114 783 432 704;
  • 27) 0.114 783 432 704 × 2 = 0 + 0.229 566 865 408;
  • 28) 0.229 566 865 408 × 2 = 0 + 0.459 133 730 816;
  • 29) 0.459 133 730 816 × 2 = 0 + 0.918 267 461 632;
  • 30) 0.918 267 461 632 × 2 = 1 + 0.836 534 923 264;
  • 31) 0.836 534 923 264 × 2 = 1 + 0.673 069 846 528;
  • 32) 0.673 069 846 528 × 2 = 1 + 0.346 139 693 056;
  • 33) 0.346 139 693 056 × 2 = 0 + 0.692 279 386 112;
  • 34) 0.692 279 386 112 × 2 = 1 + 0.384 558 772 224;
  • 35) 0.384 558 772 224 × 2 = 0 + 0.769 117 544 448;
  • 36) 0.769 117 544 448 × 2 = 1 + 0.538 235 088 896;
  • 37) 0.538 235 088 896 × 2 = 1 + 0.076 470 177 792;
  • 38) 0.076 470 177 792 × 2 = 0 + 0.152 940 355 584;
  • 39) 0.152 940 355 584 × 2 = 0 + 0.305 880 711 168;
  • 40) 0.305 880 711 168 × 2 = 0 + 0.611 761 422 336;
  • 41) 0.611 761 422 336 × 2 = 1 + 0.223 522 844 672;
  • 42) 0.223 522 844 672 × 2 = 0 + 0.447 045 689 344;
  • 43) 0.447 045 689 344 × 2 = 0 + 0.894 091 378 688;
  • 44) 0.894 091 378 688 × 2 = 1 + 0.788 182 757 376;
  • 45) 0.788 182 757 376 × 2 = 1 + 0.576 365 514 752;
  • 46) 0.576 365 514 752 × 2 = 1 + 0.152 731 029 504;
  • 47) 0.152 731 029 504 × 2 = 0 + 0.305 462 059 008;
  • 48) 0.305 462 059 008 × 2 = 0 + 0.610 924 118 016;
  • 49) 0.610 924 118 016 × 2 = 1 + 0.221 848 236 032;
  • 50) 0.221 848 236 032 × 2 = 0 + 0.443 696 472 064;
  • 51) 0.443 696 472 064 × 2 = 0 + 0.887 392 944 128;
  • 52) 0.887 392 944 128 × 2 = 1 + 0.774 785 888 256;
  • 53) 0.774 785 888 256 × 2 = 1 + 0.549 571 776 512;
  • 54) 0.549 571 776 512 × 2 = 1 + 0.099 143 553 024;
  • 55) 0.099 143 553 024 × 2 = 0 + 0.198 287 106 048;
  • 56) 0.198 287 106 048 × 2 = 0 + 0.396 574 212 096;
  • 57) 0.396 574 212 096 × 2 = 0 + 0.793 148 424 192;
  • 58) 0.793 148 424 192 × 2 = 1 + 0.586 296 848 384;
  • 59) 0.586 296 848 384 × 2 = 1 + 0.172 593 696 768;
  • 60) 0.172 593 696 768 × 2 = 0 + 0.345 187 393 536;
  • 61) 0.345 187 393 536 × 2 = 0 + 0.690 374 787 072;
  • 62) 0.690 374 787 072 × 2 = 1 + 0.380 749 574 144;
  • 63) 0.380 749 574 144 × 2 = 0 + 0.761 499 148 288;
  • 64) 0.761 499 148 288 × 2 = 1 + 0.522 998 296 576;

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:


0.000 282 006 186(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 1000 1001 1100 1001 1100 0110 0101(2)

6. Positive number before normalization:

0.000 282 006 186(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 1000 1001 1100 1001 1100 0110 0101(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:


0.000 282 006 186(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 1000 1001 1100 1001 1100 0110 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 1000 1001 1100 1001 1100 0110 0101(2) × 20 =

1.0010 0111 1011 0100 0111 0101 1000 1001 1100 1001 1100 0110 0101(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 0111 0101 1000 1001 1100 1001 1100 0110 0101


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 0111 0101 1000 1001 1100 1001 1100 0110 0101 =

0010 0111 1011 0100 0111 0101 1000 1001 1100 1001 1100 0110 0101


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 0111 0101 1000 1001 1100 1001 1100 0110 0101


Decimal number -0.000 282 006 186 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 0101 1000 1001 1100 1001 1100 0110 0101

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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