-0.000 282 005 914 506 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 506(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
-0.000 282 005 914 506(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:

|-0.000 282 005 914 506| = 0.000 282 005 914 506


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: 0.000 282 005 914 506.

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) 0.000 282 005 914 506 × 2 = 0 + 0.000 564 011 829 012;
  • 2) 0.000 564 011 829 012 × 2 = 0 + 0.001 128 023 658 024;
  • 3) 0.001 128 023 658 024 × 2 = 0 + 0.002 256 047 316 048;
  • 4) 0.002 256 047 316 048 × 2 = 0 + 0.004 512 094 632 096;
  • 5) 0.004 512 094 632 096 × 2 = 0 + 0.009 024 189 264 192;
  • 6) 0.009 024 189 264 192 × 2 = 0 + 0.018 048 378 528 384;
  • 7) 0.018 048 378 528 384 × 2 = 0 + 0.036 096 757 056 768;
  • 8) 0.036 096 757 056 768 × 2 = 0 + 0.072 193 514 113 536;
  • 9) 0.072 193 514 113 536 × 2 = 0 + 0.144 387 028 227 072;
  • 10) 0.144 387 028 227 072 × 2 = 0 + 0.288 774 056 454 144;
  • 11) 0.288 774 056 454 144 × 2 = 0 + 0.577 548 112 908 288;
  • 12) 0.577 548 112 908 288 × 2 = 1 + 0.155 096 225 816 576;
  • 13) 0.155 096 225 816 576 × 2 = 0 + 0.310 192 451 633 152;
  • 14) 0.310 192 451 633 152 × 2 = 0 + 0.620 384 903 266 304;
  • 15) 0.620 384 903 266 304 × 2 = 1 + 0.240 769 806 532 608;
  • 16) 0.240 769 806 532 608 × 2 = 0 + 0.481 539 613 065 216;
  • 17) 0.481 539 613 065 216 × 2 = 0 + 0.963 079 226 130 432;
  • 18) 0.963 079 226 130 432 × 2 = 1 + 0.926 158 452 260 864;
  • 19) 0.926 158 452 260 864 × 2 = 1 + 0.852 316 904 521 728;
  • 20) 0.852 316 904 521 728 × 2 = 1 + 0.704 633 809 043 456;
  • 21) 0.704 633 809 043 456 × 2 = 1 + 0.409 267 618 086 912;
  • 22) 0.409 267 618 086 912 × 2 = 0 + 0.818 535 236 173 824;
  • 23) 0.818 535 236 173 824 × 2 = 1 + 0.637 070 472 347 648;
  • 24) 0.637 070 472 347 648 × 2 = 1 + 0.274 140 944 695 296;
  • 25) 0.274 140 944 695 296 × 2 = 0 + 0.548 281 889 390 592;
  • 26) 0.548 281 889 390 592 × 2 = 1 + 0.096 563 778 781 184;
  • 27) 0.096 563 778 781 184 × 2 = 0 + 0.193 127 557 562 368;
  • 28) 0.193 127 557 562 368 × 2 = 0 + 0.386 255 115 124 736;
  • 29) 0.386 255 115 124 736 × 2 = 0 + 0.772 510 230 249 472;
  • 30) 0.772 510 230 249 472 × 2 = 1 + 0.545 020 460 498 944;
  • 31) 0.545 020 460 498 944 × 2 = 1 + 0.090 040 920 997 888;
  • 32) 0.090 040 920 997 888 × 2 = 0 + 0.180 081 841 995 776;
  • 33) 0.180 081 841 995 776 × 2 = 0 + 0.360 163 683 991 552;
  • 34) 0.360 163 683 991 552 × 2 = 0 + 0.720 327 367 983 104;
  • 35) 0.720 327 367 983 104 × 2 = 1 + 0.440 654 735 966 208;
  • 36) 0.440 654 735 966 208 × 2 = 0 + 0.881 309 471 932 416;
  • 37) 0.881 309 471 932 416 × 2 = 1 + 0.762 618 943 864 832;
  • 38) 0.762 618 943 864 832 × 2 = 1 + 0.525 237 887 729 664;
  • 39) 0.525 237 887 729 664 × 2 = 1 + 0.050 475 775 459 328;
  • 40) 0.050 475 775 459 328 × 2 = 0 + 0.100 951 550 918 656;
  • 41) 0.100 951 550 918 656 × 2 = 0 + 0.201 903 101 837 312;
  • 42) 0.201 903 101 837 312 × 2 = 0 + 0.403 806 203 674 624;
  • 43) 0.403 806 203 674 624 × 2 = 0 + 0.807 612 407 349 248;
  • 44) 0.807 612 407 349 248 × 2 = 1 + 0.615 224 814 698 496;
  • 45) 0.615 224 814 698 496 × 2 = 1 + 0.230 449 629 396 992;
  • 46) 0.230 449 629 396 992 × 2 = 0 + 0.460 899 258 793 984;
  • 47) 0.460 899 258 793 984 × 2 = 0 + 0.921 798 517 587 968;
  • 48) 0.921 798 517 587 968 × 2 = 1 + 0.843 597 035 175 936;
  • 49) 0.843 597 035 175 936 × 2 = 1 + 0.687 194 070 351 872;
  • 50) 0.687 194 070 351 872 × 2 = 1 + 0.374 388 140 703 744;
  • 51) 0.374 388 140 703 744 × 2 = 0 + 0.748 776 281 407 488;
  • 52) 0.748 776 281 407 488 × 2 = 1 + 0.497 552 562 814 976;
  • 53) 0.497 552 562 814 976 × 2 = 0 + 0.995 105 125 629 952;
  • 54) 0.995 105 125 629 952 × 2 = 1 + 0.990 210 251 259 904;
  • 55) 0.990 210 251 259 904 × 2 = 1 + 0.980 420 502 519 808;
  • 56) 0.980 420 502 519 808 × 2 = 1 + 0.960 841 005 039 616;
  • 57) 0.960 841 005 039 616 × 2 = 1 + 0.921 682 010 079 232;
  • 58) 0.921 682 010 079 232 × 2 = 1 + 0.843 364 020 158 464;
  • 59) 0.843 364 020 158 464 × 2 = 1 + 0.686 728 040 316 928;
  • 60) 0.686 728 040 316 928 × 2 = 1 + 0.373 456 080 633 856;
  • 61) 0.373 456 080 633 856 × 2 = 0 + 0.746 912 161 267 712;
  • 62) 0.746 912 161 267 712 × 2 = 1 + 0.493 824 322 535 424;
  • 63) 0.493 824 322 535 424 × 2 = 0 + 0.987 648 645 070 848;
  • 64) 0.987 648 645 070 848 × 2 = 1 + 0.975 297 290 141 696;

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:


0.000 282 005 914 506(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1001 1101 0111 1111 0101(2)

6. Positive number before normalization:

0.000 282 005 914 506(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1001 1101 0111 1111 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:


0.000 282 005 914 506(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1001 1101 0111 1111 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1001 1101 0111 1111 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 1001 1101 0111 1111 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 0001 1001 1101 0111 1111 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 0001 1001 1101 0111 1111 0101 =

0010 0111 1011 0100 0110 0010 1110 0001 1001 1101 0111 1111 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 0001 1001 1101 0111 1111 0101


Decimal number -0.000 282 005 914 506 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 1001 1101 0111 1111 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