-0.000 282 005 914 444 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 444(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
-0.000 282 005 914 444(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:

|-0.000 282 005 914 444| = 0.000 282 005 914 444


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: 0.000 282 005 914 444.

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) 0.000 282 005 914 444 × 2 = 0 + 0.000 564 011 828 888;
  • 2) 0.000 564 011 828 888 × 2 = 0 + 0.001 128 023 657 776;
  • 3) 0.001 128 023 657 776 × 2 = 0 + 0.002 256 047 315 552;
  • 4) 0.002 256 047 315 552 × 2 = 0 + 0.004 512 094 631 104;
  • 5) 0.004 512 094 631 104 × 2 = 0 + 0.009 024 189 262 208;
  • 6) 0.009 024 189 262 208 × 2 = 0 + 0.018 048 378 524 416;
  • 7) 0.018 048 378 524 416 × 2 = 0 + 0.036 096 757 048 832;
  • 8) 0.036 096 757 048 832 × 2 = 0 + 0.072 193 514 097 664;
  • 9) 0.072 193 514 097 664 × 2 = 0 + 0.144 387 028 195 328;
  • 10) 0.144 387 028 195 328 × 2 = 0 + 0.288 774 056 390 656;
  • 11) 0.288 774 056 390 656 × 2 = 0 + 0.577 548 112 781 312;
  • 12) 0.577 548 112 781 312 × 2 = 1 + 0.155 096 225 562 624;
  • 13) 0.155 096 225 562 624 × 2 = 0 + 0.310 192 451 125 248;
  • 14) 0.310 192 451 125 248 × 2 = 0 + 0.620 384 902 250 496;
  • 15) 0.620 384 902 250 496 × 2 = 1 + 0.240 769 804 500 992;
  • 16) 0.240 769 804 500 992 × 2 = 0 + 0.481 539 609 001 984;
  • 17) 0.481 539 609 001 984 × 2 = 0 + 0.963 079 218 003 968;
  • 18) 0.963 079 218 003 968 × 2 = 1 + 0.926 158 436 007 936;
  • 19) 0.926 158 436 007 936 × 2 = 1 + 0.852 316 872 015 872;
  • 20) 0.852 316 872 015 872 × 2 = 1 + 0.704 633 744 031 744;
  • 21) 0.704 633 744 031 744 × 2 = 1 + 0.409 267 488 063 488;
  • 22) 0.409 267 488 063 488 × 2 = 0 + 0.818 534 976 126 976;
  • 23) 0.818 534 976 126 976 × 2 = 1 + 0.637 069 952 253 952;
  • 24) 0.637 069 952 253 952 × 2 = 1 + 0.274 139 904 507 904;
  • 25) 0.274 139 904 507 904 × 2 = 0 + 0.548 279 809 015 808;
  • 26) 0.548 279 809 015 808 × 2 = 1 + 0.096 559 618 031 616;
  • 27) 0.096 559 618 031 616 × 2 = 0 + 0.193 119 236 063 232;
  • 28) 0.193 119 236 063 232 × 2 = 0 + 0.386 238 472 126 464;
  • 29) 0.386 238 472 126 464 × 2 = 0 + 0.772 476 944 252 928;
  • 30) 0.772 476 944 252 928 × 2 = 1 + 0.544 953 888 505 856;
  • 31) 0.544 953 888 505 856 × 2 = 1 + 0.089 907 777 011 712;
  • 32) 0.089 907 777 011 712 × 2 = 0 + 0.179 815 554 023 424;
  • 33) 0.179 815 554 023 424 × 2 = 0 + 0.359 631 108 046 848;
  • 34) 0.359 631 108 046 848 × 2 = 0 + 0.719 262 216 093 696;
  • 35) 0.719 262 216 093 696 × 2 = 1 + 0.438 524 432 187 392;
  • 36) 0.438 524 432 187 392 × 2 = 0 + 0.877 048 864 374 784;
  • 37) 0.877 048 864 374 784 × 2 = 1 + 0.754 097 728 749 568;
  • 38) 0.754 097 728 749 568 × 2 = 1 + 0.508 195 457 499 136;
  • 39) 0.508 195 457 499 136 × 2 = 1 + 0.016 390 914 998 272;
  • 40) 0.016 390 914 998 272 × 2 = 0 + 0.032 781 829 996 544;
  • 41) 0.032 781 829 996 544 × 2 = 0 + 0.065 563 659 993 088;
  • 42) 0.065 563 659 993 088 × 2 = 0 + 0.131 127 319 986 176;
  • 43) 0.131 127 319 986 176 × 2 = 0 + 0.262 254 639 972 352;
  • 44) 0.262 254 639 972 352 × 2 = 0 + 0.524 509 279 944 704;
  • 45) 0.524 509 279 944 704 × 2 = 1 + 0.049 018 559 889 408;
  • 46) 0.049 018 559 889 408 × 2 = 0 + 0.098 037 119 778 816;
  • 47) 0.098 037 119 778 816 × 2 = 0 + 0.196 074 239 557 632;
  • 48) 0.196 074 239 557 632 × 2 = 0 + 0.392 148 479 115 264;
  • 49) 0.392 148 479 115 264 × 2 = 0 + 0.784 296 958 230 528;
  • 50) 0.784 296 958 230 528 × 2 = 1 + 0.568 593 916 461 056;
  • 51) 0.568 593 916 461 056 × 2 = 1 + 0.137 187 832 922 112;
  • 52) 0.137 187 832 922 112 × 2 = 0 + 0.274 375 665 844 224;
  • 53) 0.274 375 665 844 224 × 2 = 0 + 0.548 751 331 688 448;
  • 54) 0.548 751 331 688 448 × 2 = 1 + 0.097 502 663 376 896;
  • 55) 0.097 502 663 376 896 × 2 = 0 + 0.195 005 326 753 792;
  • 56) 0.195 005 326 753 792 × 2 = 0 + 0.390 010 653 507 584;
  • 57) 0.390 010 653 507 584 × 2 = 0 + 0.780 021 307 015 168;
  • 58) 0.780 021 307 015 168 × 2 = 1 + 0.560 042 614 030 336;
  • 59) 0.560 042 614 030 336 × 2 = 1 + 0.120 085 228 060 672;
  • 60) 0.120 085 228 060 672 × 2 = 0 + 0.240 170 456 121 344;
  • 61) 0.240 170 456 121 344 × 2 = 0 + 0.480 340 912 242 688;
  • 62) 0.480 340 912 242 688 × 2 = 0 + 0.960 681 824 485 376;
  • 63) 0.960 681 824 485 376 × 2 = 1 + 0.921 363 648 970 752;
  • 64) 0.921 363 648 970 752 × 2 = 1 + 0.842 727 297 941 504;

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:


0.000 282 005 914 444(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 0110 0100 0110 0011(2)

6. Positive number before normalization:

0.000 282 005 914 444(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 0110 0100 0110 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:


0.000 282 005 914 444(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 0110 0100 0110 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1000 0110 0100 0110 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 1000 0110 0100 0110 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 0010 1110 0000 1000 0110 0100 0110 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 0010 1110 0000 1000 0110 0100 0110 0011 =

0010 0111 1011 0100 0110 0010 1110 0000 1000 0110 0100 0110 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 0010 1110 0000 1000 0110 0100 0110 0011


Decimal number -0.000 282 005 914 444 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 1000 0110 0100 0110 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