-0.000 282 005 914 557 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 557(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
-0.000 282 005 914 557(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:

|-0.000 282 005 914 557| = 0.000 282 005 914 557


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: 0.000 282 005 914 557.

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) 0.000 282 005 914 557 × 2 = 0 + 0.000 564 011 829 114;
  • 2) 0.000 564 011 829 114 × 2 = 0 + 0.001 128 023 658 228;
  • 3) 0.001 128 023 658 228 × 2 = 0 + 0.002 256 047 316 456;
  • 4) 0.002 256 047 316 456 × 2 = 0 + 0.004 512 094 632 912;
  • 5) 0.004 512 094 632 912 × 2 = 0 + 0.009 024 189 265 824;
  • 6) 0.009 024 189 265 824 × 2 = 0 + 0.018 048 378 531 648;
  • 7) 0.018 048 378 531 648 × 2 = 0 + 0.036 096 757 063 296;
  • 8) 0.036 096 757 063 296 × 2 = 0 + 0.072 193 514 126 592;
  • 9) 0.072 193 514 126 592 × 2 = 0 + 0.144 387 028 253 184;
  • 10) 0.144 387 028 253 184 × 2 = 0 + 0.288 774 056 506 368;
  • 11) 0.288 774 056 506 368 × 2 = 0 + 0.577 548 113 012 736;
  • 12) 0.577 548 113 012 736 × 2 = 1 + 0.155 096 226 025 472;
  • 13) 0.155 096 226 025 472 × 2 = 0 + 0.310 192 452 050 944;
  • 14) 0.310 192 452 050 944 × 2 = 0 + 0.620 384 904 101 888;
  • 15) 0.620 384 904 101 888 × 2 = 1 + 0.240 769 808 203 776;
  • 16) 0.240 769 808 203 776 × 2 = 0 + 0.481 539 616 407 552;
  • 17) 0.481 539 616 407 552 × 2 = 0 + 0.963 079 232 815 104;
  • 18) 0.963 079 232 815 104 × 2 = 1 + 0.926 158 465 630 208;
  • 19) 0.926 158 465 630 208 × 2 = 1 + 0.852 316 931 260 416;
  • 20) 0.852 316 931 260 416 × 2 = 1 + 0.704 633 862 520 832;
  • 21) 0.704 633 862 520 832 × 2 = 1 + 0.409 267 725 041 664;
  • 22) 0.409 267 725 041 664 × 2 = 0 + 0.818 535 450 083 328;
  • 23) 0.818 535 450 083 328 × 2 = 1 + 0.637 070 900 166 656;
  • 24) 0.637 070 900 166 656 × 2 = 1 + 0.274 141 800 333 312;
  • 25) 0.274 141 800 333 312 × 2 = 0 + 0.548 283 600 666 624;
  • 26) 0.548 283 600 666 624 × 2 = 1 + 0.096 567 201 333 248;
  • 27) 0.096 567 201 333 248 × 2 = 0 + 0.193 134 402 666 496;
  • 28) 0.193 134 402 666 496 × 2 = 0 + 0.386 268 805 332 992;
  • 29) 0.386 268 805 332 992 × 2 = 0 + 0.772 537 610 665 984;
  • 30) 0.772 537 610 665 984 × 2 = 1 + 0.545 075 221 331 968;
  • 31) 0.545 075 221 331 968 × 2 = 1 + 0.090 150 442 663 936;
  • 32) 0.090 150 442 663 936 × 2 = 0 + 0.180 300 885 327 872;
  • 33) 0.180 300 885 327 872 × 2 = 0 + 0.360 601 770 655 744;
  • 34) 0.360 601 770 655 744 × 2 = 0 + 0.721 203 541 311 488;
  • 35) 0.721 203 541 311 488 × 2 = 1 + 0.442 407 082 622 976;
  • 36) 0.442 407 082 622 976 × 2 = 0 + 0.884 814 165 245 952;
  • 37) 0.884 814 165 245 952 × 2 = 1 + 0.769 628 330 491 904;
  • 38) 0.769 628 330 491 904 × 2 = 1 + 0.539 256 660 983 808;
  • 39) 0.539 256 660 983 808 × 2 = 1 + 0.078 513 321 967 616;
  • 40) 0.078 513 321 967 616 × 2 = 0 + 0.157 026 643 935 232;
  • 41) 0.157 026 643 935 232 × 2 = 0 + 0.314 053 287 870 464;
  • 42) 0.314 053 287 870 464 × 2 = 0 + 0.628 106 575 740 928;
  • 43) 0.628 106 575 740 928 × 2 = 1 + 0.256 213 151 481 856;
  • 44) 0.256 213 151 481 856 × 2 = 0 + 0.512 426 302 963 712;
  • 45) 0.512 426 302 963 712 × 2 = 1 + 0.024 852 605 927 424;
  • 46) 0.024 852 605 927 424 × 2 = 0 + 0.049 705 211 854 848;
  • 47) 0.049 705 211 854 848 × 2 = 0 + 0.099 410 423 709 696;
  • 48) 0.099 410 423 709 696 × 2 = 0 + 0.198 820 847 419 392;
  • 49) 0.198 820 847 419 392 × 2 = 0 + 0.397 641 694 838 784;
  • 50) 0.397 641 694 838 784 × 2 = 0 + 0.795 283 389 677 568;
  • 51) 0.795 283 389 677 568 × 2 = 1 + 0.590 566 779 355 136;
  • 52) 0.590 566 779 355 136 × 2 = 1 + 0.181 133 558 710 272;
  • 53) 0.181 133 558 710 272 × 2 = 0 + 0.362 267 117 420 544;
  • 54) 0.362 267 117 420 544 × 2 = 0 + 0.724 534 234 841 088;
  • 55) 0.724 534 234 841 088 × 2 = 1 + 0.449 068 469 682 176;
  • 56) 0.449 068 469 682 176 × 2 = 0 + 0.898 136 939 364 352;
  • 57) 0.898 136 939 364 352 × 2 = 1 + 0.796 273 878 728 704;
  • 58) 0.796 273 878 728 704 × 2 = 1 + 0.592 547 757 457 408;
  • 59) 0.592 547 757 457 408 × 2 = 1 + 0.185 095 514 914 816;
  • 60) 0.185 095 514 914 816 × 2 = 0 + 0.370 191 029 829 632;
  • 61) 0.370 191 029 829 632 × 2 = 0 + 0.740 382 059 659 264;
  • 62) 0.740 382 059 659 264 × 2 = 1 + 0.480 764 119 318 528;
  • 63) 0.480 764 119 318 528 × 2 = 0 + 0.961 528 238 637 056;
  • 64) 0.961 528 238 637 056 × 2 = 1 + 0.923 056 477 274 112;

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:


0.000 282 005 914 557(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1000 0011 0010 1110 0101(2)

6. Positive number before normalization:

0.000 282 005 914 557(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1000 0011 0010 1110 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:


0.000 282 005 914 557(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1000 0011 0010 1110 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1000 0011 0010 1110 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0010 1000 0011 0010 1110 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 0110 0010 1110 0010 1000 0011 0010 1110 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 0110 0010 1110 0010 1000 0011 0010 1110 0101 =

0010 0111 1011 0100 0110 0010 1110 0010 1000 0011 0010 1110 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 0110 0010 1110 0010 1000 0011 0010 1110 0101


Decimal number -0.000 282 005 914 557 converted to 64 bit double precision IEEE 754 binary floating point representation:

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