-0.000 282 005 914 542 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 542(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
-0.000 282 005 914 542(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:

|-0.000 282 005 914 542| = 0.000 282 005 914 542


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: 0.000 282 005 914 542.

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) 0.000 282 005 914 542 × 2 = 0 + 0.000 564 011 829 084;
  • 2) 0.000 564 011 829 084 × 2 = 0 + 0.001 128 023 658 168;
  • 3) 0.001 128 023 658 168 × 2 = 0 + 0.002 256 047 316 336;
  • 4) 0.002 256 047 316 336 × 2 = 0 + 0.004 512 094 632 672;
  • 5) 0.004 512 094 632 672 × 2 = 0 + 0.009 024 189 265 344;
  • 6) 0.009 024 189 265 344 × 2 = 0 + 0.018 048 378 530 688;
  • 7) 0.018 048 378 530 688 × 2 = 0 + 0.036 096 757 061 376;
  • 8) 0.036 096 757 061 376 × 2 = 0 + 0.072 193 514 122 752;
  • 9) 0.072 193 514 122 752 × 2 = 0 + 0.144 387 028 245 504;
  • 10) 0.144 387 028 245 504 × 2 = 0 + 0.288 774 056 491 008;
  • 11) 0.288 774 056 491 008 × 2 = 0 + 0.577 548 112 982 016;
  • 12) 0.577 548 112 982 016 × 2 = 1 + 0.155 096 225 964 032;
  • 13) 0.155 096 225 964 032 × 2 = 0 + 0.310 192 451 928 064;
  • 14) 0.310 192 451 928 064 × 2 = 0 + 0.620 384 903 856 128;
  • 15) 0.620 384 903 856 128 × 2 = 1 + 0.240 769 807 712 256;
  • 16) 0.240 769 807 712 256 × 2 = 0 + 0.481 539 615 424 512;
  • 17) 0.481 539 615 424 512 × 2 = 0 + 0.963 079 230 849 024;
  • 18) 0.963 079 230 849 024 × 2 = 1 + 0.926 158 461 698 048;
  • 19) 0.926 158 461 698 048 × 2 = 1 + 0.852 316 923 396 096;
  • 20) 0.852 316 923 396 096 × 2 = 1 + 0.704 633 846 792 192;
  • 21) 0.704 633 846 792 192 × 2 = 1 + 0.409 267 693 584 384;
  • 22) 0.409 267 693 584 384 × 2 = 0 + 0.818 535 387 168 768;
  • 23) 0.818 535 387 168 768 × 2 = 1 + 0.637 070 774 337 536;
  • 24) 0.637 070 774 337 536 × 2 = 1 + 0.274 141 548 675 072;
  • 25) 0.274 141 548 675 072 × 2 = 0 + 0.548 283 097 350 144;
  • 26) 0.548 283 097 350 144 × 2 = 1 + 0.096 566 194 700 288;
  • 27) 0.096 566 194 700 288 × 2 = 0 + 0.193 132 389 400 576;
  • 28) 0.193 132 389 400 576 × 2 = 0 + 0.386 264 778 801 152;
  • 29) 0.386 264 778 801 152 × 2 = 0 + 0.772 529 557 602 304;
  • 30) 0.772 529 557 602 304 × 2 = 1 + 0.545 059 115 204 608;
  • 31) 0.545 059 115 204 608 × 2 = 1 + 0.090 118 230 409 216;
  • 32) 0.090 118 230 409 216 × 2 = 0 + 0.180 236 460 818 432;
  • 33) 0.180 236 460 818 432 × 2 = 0 + 0.360 472 921 636 864;
  • 34) 0.360 472 921 636 864 × 2 = 0 + 0.720 945 843 273 728;
  • 35) 0.720 945 843 273 728 × 2 = 1 + 0.441 891 686 547 456;
  • 36) 0.441 891 686 547 456 × 2 = 0 + 0.883 783 373 094 912;
  • 37) 0.883 783 373 094 912 × 2 = 1 + 0.767 566 746 189 824;
  • 38) 0.767 566 746 189 824 × 2 = 1 + 0.535 133 492 379 648;
  • 39) 0.535 133 492 379 648 × 2 = 1 + 0.070 266 984 759 296;
  • 40) 0.070 266 984 759 296 × 2 = 0 + 0.140 533 969 518 592;
  • 41) 0.140 533 969 518 592 × 2 = 0 + 0.281 067 939 037 184;
  • 42) 0.281 067 939 037 184 × 2 = 0 + 0.562 135 878 074 368;
  • 43) 0.562 135 878 074 368 × 2 = 1 + 0.124 271 756 148 736;
  • 44) 0.124 271 756 148 736 × 2 = 0 + 0.248 543 512 297 472;
  • 45) 0.248 543 512 297 472 × 2 = 0 + 0.497 087 024 594 944;
  • 46) 0.497 087 024 594 944 × 2 = 0 + 0.994 174 049 189 888;
  • 47) 0.994 174 049 189 888 × 2 = 1 + 0.988 348 098 379 776;
  • 48) 0.988 348 098 379 776 × 2 = 1 + 0.976 696 196 759 552;
  • 49) 0.976 696 196 759 552 × 2 = 1 + 0.953 392 393 519 104;
  • 50) 0.953 392 393 519 104 × 2 = 1 + 0.906 784 787 038 208;
  • 51) 0.906 784 787 038 208 × 2 = 1 + 0.813 569 574 076 416;
  • 52) 0.813 569 574 076 416 × 2 = 1 + 0.627 139 148 152 832;
  • 53) 0.627 139 148 152 832 × 2 = 1 + 0.254 278 296 305 664;
  • 54) 0.254 278 296 305 664 × 2 = 0 + 0.508 556 592 611 328;
  • 55) 0.508 556 592 611 328 × 2 = 1 + 0.017 113 185 222 656;
  • 56) 0.017 113 185 222 656 × 2 = 0 + 0.034 226 370 445 312;
  • 57) 0.034 226 370 445 312 × 2 = 0 + 0.068 452 740 890 624;
  • 58) 0.068 452 740 890 624 × 2 = 0 + 0.136 905 481 781 248;
  • 59) 0.136 905 481 781 248 × 2 = 0 + 0.273 810 963 562 496;
  • 60) 0.273 810 963 562 496 × 2 = 0 + 0.547 621 927 124 992;
  • 61) 0.547 621 927 124 992 × 2 = 1 + 0.095 243 854 249 984;
  • 62) 0.095 243 854 249 984 × 2 = 0 + 0.190 487 708 499 968;
  • 63) 0.190 487 708 499 968 × 2 = 0 + 0.380 975 416 999 936;
  • 64) 0.380 975 416 999 936 × 2 = 0 + 0.761 950 833 999 872;

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:


0.000 282 005 914 542(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0011 1111 1010 0000 1000(2)

6. Positive number before normalization:

0.000 282 005 914 542(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0011 1111 1010 0000 1000(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:


0.000 282 005 914 542(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0011 1111 1010 0000 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 0011 1111 1010 0000 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0010 0011 1111 1010 0000 1000(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 0011 1111 1010 0000 1000


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 0011 1111 1010 0000 1000 =

0010 0111 1011 0100 0110 0010 1110 0010 0011 1111 1010 0000 1000


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 0011 1111 1010 0000 1000


Decimal number -0.000 282 005 914 542 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0010 0011 1111 1010 0000 1000

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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