-0.000 282 005 914 439 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 439(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
-0.000 282 005 914 439(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:

|-0.000 282 005 914 439| = 0.000 282 005 914 439


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: 0.000 282 005 914 439.

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) 0.000 282 005 914 439 × 2 = 0 + 0.000 564 011 828 878;
  • 2) 0.000 564 011 828 878 × 2 = 0 + 0.001 128 023 657 756;
  • 3) 0.001 128 023 657 756 × 2 = 0 + 0.002 256 047 315 512;
  • 4) 0.002 256 047 315 512 × 2 = 0 + 0.004 512 094 631 024;
  • 5) 0.004 512 094 631 024 × 2 = 0 + 0.009 024 189 262 048;
  • 6) 0.009 024 189 262 048 × 2 = 0 + 0.018 048 378 524 096;
  • 7) 0.018 048 378 524 096 × 2 = 0 + 0.036 096 757 048 192;
  • 8) 0.036 096 757 048 192 × 2 = 0 + 0.072 193 514 096 384;
  • 9) 0.072 193 514 096 384 × 2 = 0 + 0.144 387 028 192 768;
  • 10) 0.144 387 028 192 768 × 2 = 0 + 0.288 774 056 385 536;
  • 11) 0.288 774 056 385 536 × 2 = 0 + 0.577 548 112 771 072;
  • 12) 0.577 548 112 771 072 × 2 = 1 + 0.155 096 225 542 144;
  • 13) 0.155 096 225 542 144 × 2 = 0 + 0.310 192 451 084 288;
  • 14) 0.310 192 451 084 288 × 2 = 0 + 0.620 384 902 168 576;
  • 15) 0.620 384 902 168 576 × 2 = 1 + 0.240 769 804 337 152;
  • 16) 0.240 769 804 337 152 × 2 = 0 + 0.481 539 608 674 304;
  • 17) 0.481 539 608 674 304 × 2 = 0 + 0.963 079 217 348 608;
  • 18) 0.963 079 217 348 608 × 2 = 1 + 0.926 158 434 697 216;
  • 19) 0.926 158 434 697 216 × 2 = 1 + 0.852 316 869 394 432;
  • 20) 0.852 316 869 394 432 × 2 = 1 + 0.704 633 738 788 864;
  • 21) 0.704 633 738 788 864 × 2 = 1 + 0.409 267 477 577 728;
  • 22) 0.409 267 477 577 728 × 2 = 0 + 0.818 534 955 155 456;
  • 23) 0.818 534 955 155 456 × 2 = 1 + 0.637 069 910 310 912;
  • 24) 0.637 069 910 310 912 × 2 = 1 + 0.274 139 820 621 824;
  • 25) 0.274 139 820 621 824 × 2 = 0 + 0.548 279 641 243 648;
  • 26) 0.548 279 641 243 648 × 2 = 1 + 0.096 559 282 487 296;
  • 27) 0.096 559 282 487 296 × 2 = 0 + 0.193 118 564 974 592;
  • 28) 0.193 118 564 974 592 × 2 = 0 + 0.386 237 129 949 184;
  • 29) 0.386 237 129 949 184 × 2 = 0 + 0.772 474 259 898 368;
  • 30) 0.772 474 259 898 368 × 2 = 1 + 0.544 948 519 796 736;
  • 31) 0.544 948 519 796 736 × 2 = 1 + 0.089 897 039 593 472;
  • 32) 0.089 897 039 593 472 × 2 = 0 + 0.179 794 079 186 944;
  • 33) 0.179 794 079 186 944 × 2 = 0 + 0.359 588 158 373 888;
  • 34) 0.359 588 158 373 888 × 2 = 0 + 0.719 176 316 747 776;
  • 35) 0.719 176 316 747 776 × 2 = 1 + 0.438 352 633 495 552;
  • 36) 0.438 352 633 495 552 × 2 = 0 + 0.876 705 266 991 104;
  • 37) 0.876 705 266 991 104 × 2 = 1 + 0.753 410 533 982 208;
  • 38) 0.753 410 533 982 208 × 2 = 1 + 0.506 821 067 964 416;
  • 39) 0.506 821 067 964 416 × 2 = 1 + 0.013 642 135 928 832;
  • 40) 0.013 642 135 928 832 × 2 = 0 + 0.027 284 271 857 664;
  • 41) 0.027 284 271 857 664 × 2 = 0 + 0.054 568 543 715 328;
  • 42) 0.054 568 543 715 328 × 2 = 0 + 0.109 137 087 430 656;
  • 43) 0.109 137 087 430 656 × 2 = 0 + 0.218 274 174 861 312;
  • 44) 0.218 274 174 861 312 × 2 = 0 + 0.436 548 349 722 624;
  • 45) 0.436 548 349 722 624 × 2 = 0 + 0.873 096 699 445 248;
  • 46) 0.873 096 699 445 248 × 2 = 1 + 0.746 193 398 890 496;
  • 47) 0.746 193 398 890 496 × 2 = 1 + 0.492 386 797 780 992;
  • 48) 0.492 386 797 780 992 × 2 = 0 + 0.984 773 595 561 984;
  • 49) 0.984 773 595 561 984 × 2 = 1 + 0.969 547 191 123 968;
  • 50) 0.969 547 191 123 968 × 2 = 1 + 0.939 094 382 247 936;
  • 51) 0.939 094 382 247 936 × 2 = 1 + 0.878 188 764 495 872;
  • 52) 0.878 188 764 495 872 × 2 = 1 + 0.756 377 528 991 744;
  • 53) 0.756 377 528 991 744 × 2 = 1 + 0.512 755 057 983 488;
  • 54) 0.512 755 057 983 488 × 2 = 1 + 0.025 510 115 966 976;
  • 55) 0.025 510 115 966 976 × 2 = 0 + 0.051 020 231 933 952;
  • 56) 0.051 020 231 933 952 × 2 = 0 + 0.102 040 463 867 904;
  • 57) 0.102 040 463 867 904 × 2 = 0 + 0.204 080 927 735 808;
  • 58) 0.204 080 927 735 808 × 2 = 0 + 0.408 161 855 471 616;
  • 59) 0.408 161 855 471 616 × 2 = 0 + 0.816 323 710 943 232;
  • 60) 0.816 323 710 943 232 × 2 = 1 + 0.632 647 421 886 464;
  • 61) 0.632 647 421 886 464 × 2 = 1 + 0.265 294 843 772 928;
  • 62) 0.265 294 843 772 928 × 2 = 0 + 0.530 589 687 545 856;
  • 63) 0.530 589 687 545 856 × 2 = 1 + 0.061 179 375 091 712;
  • 64) 0.061 179 375 091 712 × 2 = 0 + 0.122 358 750 183 424;

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:


0.000 282 005 914 439(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0110 1111 1100 0001 1010(2)

6. Positive number before normalization:

0.000 282 005 914 439(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0110 1111 1100 0001 1010(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:


0.000 282 005 914 439(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0110 1111 1100 0001 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0110 1111 1100 0001 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 0110 1111 1100 0001 1010(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 0110 1111 1100 0001 1010


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 0110 1111 1100 0001 1010 =

0010 0111 1011 0100 0110 0010 1110 0000 0110 1111 1100 0001 1010


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 0110 1111 1100 0001 1010


Decimal number -0.000 282 005 914 439 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 0110 1111 1100 0001 1010

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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