-0.000 282 005 914 393 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 393(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
-0.000 282 005 914 393(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:

|-0.000 282 005 914 393| = 0.000 282 005 914 393


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: 0.000 282 005 914 393.

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) 0.000 282 005 914 393 × 2 = 0 + 0.000 564 011 828 786;
  • 2) 0.000 564 011 828 786 × 2 = 0 + 0.001 128 023 657 572;
  • 3) 0.001 128 023 657 572 × 2 = 0 + 0.002 256 047 315 144;
  • 4) 0.002 256 047 315 144 × 2 = 0 + 0.004 512 094 630 288;
  • 5) 0.004 512 094 630 288 × 2 = 0 + 0.009 024 189 260 576;
  • 6) 0.009 024 189 260 576 × 2 = 0 + 0.018 048 378 521 152;
  • 7) 0.018 048 378 521 152 × 2 = 0 + 0.036 096 757 042 304;
  • 8) 0.036 096 757 042 304 × 2 = 0 + 0.072 193 514 084 608;
  • 9) 0.072 193 514 084 608 × 2 = 0 + 0.144 387 028 169 216;
  • 10) 0.144 387 028 169 216 × 2 = 0 + 0.288 774 056 338 432;
  • 11) 0.288 774 056 338 432 × 2 = 0 + 0.577 548 112 676 864;
  • 12) 0.577 548 112 676 864 × 2 = 1 + 0.155 096 225 353 728;
  • 13) 0.155 096 225 353 728 × 2 = 0 + 0.310 192 450 707 456;
  • 14) 0.310 192 450 707 456 × 2 = 0 + 0.620 384 901 414 912;
  • 15) 0.620 384 901 414 912 × 2 = 1 + 0.240 769 802 829 824;
  • 16) 0.240 769 802 829 824 × 2 = 0 + 0.481 539 605 659 648;
  • 17) 0.481 539 605 659 648 × 2 = 0 + 0.963 079 211 319 296;
  • 18) 0.963 079 211 319 296 × 2 = 1 + 0.926 158 422 638 592;
  • 19) 0.926 158 422 638 592 × 2 = 1 + 0.852 316 845 277 184;
  • 20) 0.852 316 845 277 184 × 2 = 1 + 0.704 633 690 554 368;
  • 21) 0.704 633 690 554 368 × 2 = 1 + 0.409 267 381 108 736;
  • 22) 0.409 267 381 108 736 × 2 = 0 + 0.818 534 762 217 472;
  • 23) 0.818 534 762 217 472 × 2 = 1 + 0.637 069 524 434 944;
  • 24) 0.637 069 524 434 944 × 2 = 1 + 0.274 139 048 869 888;
  • 25) 0.274 139 048 869 888 × 2 = 0 + 0.548 278 097 739 776;
  • 26) 0.548 278 097 739 776 × 2 = 1 + 0.096 556 195 479 552;
  • 27) 0.096 556 195 479 552 × 2 = 0 + 0.193 112 390 959 104;
  • 28) 0.193 112 390 959 104 × 2 = 0 + 0.386 224 781 918 208;
  • 29) 0.386 224 781 918 208 × 2 = 0 + 0.772 449 563 836 416;
  • 30) 0.772 449 563 836 416 × 2 = 1 + 0.544 899 127 672 832;
  • 31) 0.544 899 127 672 832 × 2 = 1 + 0.089 798 255 345 664;
  • 32) 0.089 798 255 345 664 × 2 = 0 + 0.179 596 510 691 328;
  • 33) 0.179 596 510 691 328 × 2 = 0 + 0.359 193 021 382 656;
  • 34) 0.359 193 021 382 656 × 2 = 0 + 0.718 386 042 765 312;
  • 35) 0.718 386 042 765 312 × 2 = 1 + 0.436 772 085 530 624;
  • 36) 0.436 772 085 530 624 × 2 = 0 + 0.873 544 171 061 248;
  • 37) 0.873 544 171 061 248 × 2 = 1 + 0.747 088 342 122 496;
  • 38) 0.747 088 342 122 496 × 2 = 1 + 0.494 176 684 244 992;
  • 39) 0.494 176 684 244 992 × 2 = 0 + 0.988 353 368 489 984;
  • 40) 0.988 353 368 489 984 × 2 = 1 + 0.976 706 736 979 968;
  • 41) 0.976 706 736 979 968 × 2 = 1 + 0.953 413 473 959 936;
  • 42) 0.953 413 473 959 936 × 2 = 1 + 0.906 826 947 919 872;
  • 43) 0.906 826 947 919 872 × 2 = 1 + 0.813 653 895 839 744;
  • 44) 0.813 653 895 839 744 × 2 = 1 + 0.627 307 791 679 488;
  • 45) 0.627 307 791 679 488 × 2 = 1 + 0.254 615 583 358 976;
  • 46) 0.254 615 583 358 976 × 2 = 0 + 0.509 231 166 717 952;
  • 47) 0.509 231 166 717 952 × 2 = 1 + 0.018 462 333 435 904;
  • 48) 0.018 462 333 435 904 × 2 = 0 + 0.036 924 666 871 808;
  • 49) 0.036 924 666 871 808 × 2 = 0 + 0.073 849 333 743 616;
  • 50) 0.073 849 333 743 616 × 2 = 0 + 0.147 698 667 487 232;
  • 51) 0.147 698 667 487 232 × 2 = 0 + 0.295 397 334 974 464;
  • 52) 0.295 397 334 974 464 × 2 = 0 + 0.590 794 669 948 928;
  • 53) 0.590 794 669 948 928 × 2 = 1 + 0.181 589 339 897 856;
  • 54) 0.181 589 339 897 856 × 2 = 0 + 0.363 178 679 795 712;
  • 55) 0.363 178 679 795 712 × 2 = 0 + 0.726 357 359 591 424;
  • 56) 0.726 357 359 591 424 × 2 = 1 + 0.452 714 719 182 848;
  • 57) 0.452 714 719 182 848 × 2 = 0 + 0.905 429 438 365 696;
  • 58) 0.905 429 438 365 696 × 2 = 1 + 0.810 858 876 731 392;
  • 59) 0.810 858 876 731 392 × 2 = 1 + 0.621 717 753 462 784;
  • 60) 0.621 717 753 462 784 × 2 = 1 + 0.243 435 506 925 568;
  • 61) 0.243 435 506 925 568 × 2 = 0 + 0.486 871 013 851 136;
  • 62) 0.486 871 013 851 136 × 2 = 0 + 0.973 742 027 702 272;
  • 63) 0.973 742 027 702 272 × 2 = 1 + 0.947 484 055 404 544;
  • 64) 0.947 484 055 404 544 × 2 = 1 + 0.894 968 110 809 088;

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:


0.000 282 005 914 393(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1010 0000 1001 0111 0011(2)

6. Positive number before normalization:

0.000 282 005 914 393(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1010 0000 1001 0111 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:


0.000 282 005 914 393(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1010 0000 1001 0111 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1010 0000 1001 0111 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1010 0000 1001 0111 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 1101 1111 1010 0000 1001 0111 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 1101 1111 1010 0000 1001 0111 0011 =

0010 0111 1011 0100 0110 0010 1101 1111 1010 0000 1001 0111 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 1101 1111 1010 0000 1001 0111 0011


Decimal number -0.000 282 005 914 393 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1010 0000 1001 0111 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