-0.000 282 005 914 464 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 464(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
-0.000 282 005 914 464(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:

|-0.000 282 005 914 464| = 0.000 282 005 914 464


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: 0.000 282 005 914 464.

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) 0.000 282 005 914 464 × 2 = 0 + 0.000 564 011 828 928;
  • 2) 0.000 564 011 828 928 × 2 = 0 + 0.001 128 023 657 856;
  • 3) 0.001 128 023 657 856 × 2 = 0 + 0.002 256 047 315 712;
  • 4) 0.002 256 047 315 712 × 2 = 0 + 0.004 512 094 631 424;
  • 5) 0.004 512 094 631 424 × 2 = 0 + 0.009 024 189 262 848;
  • 6) 0.009 024 189 262 848 × 2 = 0 + 0.018 048 378 525 696;
  • 7) 0.018 048 378 525 696 × 2 = 0 + 0.036 096 757 051 392;
  • 8) 0.036 096 757 051 392 × 2 = 0 + 0.072 193 514 102 784;
  • 9) 0.072 193 514 102 784 × 2 = 0 + 0.144 387 028 205 568;
  • 10) 0.144 387 028 205 568 × 2 = 0 + 0.288 774 056 411 136;
  • 11) 0.288 774 056 411 136 × 2 = 0 + 0.577 548 112 822 272;
  • 12) 0.577 548 112 822 272 × 2 = 1 + 0.155 096 225 644 544;
  • 13) 0.155 096 225 644 544 × 2 = 0 + 0.310 192 451 289 088;
  • 14) 0.310 192 451 289 088 × 2 = 0 + 0.620 384 902 578 176;
  • 15) 0.620 384 902 578 176 × 2 = 1 + 0.240 769 805 156 352;
  • 16) 0.240 769 805 156 352 × 2 = 0 + 0.481 539 610 312 704;
  • 17) 0.481 539 610 312 704 × 2 = 0 + 0.963 079 220 625 408;
  • 18) 0.963 079 220 625 408 × 2 = 1 + 0.926 158 441 250 816;
  • 19) 0.926 158 441 250 816 × 2 = 1 + 0.852 316 882 501 632;
  • 20) 0.852 316 882 501 632 × 2 = 1 + 0.704 633 765 003 264;
  • 21) 0.704 633 765 003 264 × 2 = 1 + 0.409 267 530 006 528;
  • 22) 0.409 267 530 006 528 × 2 = 0 + 0.818 535 060 013 056;
  • 23) 0.818 535 060 013 056 × 2 = 1 + 0.637 070 120 026 112;
  • 24) 0.637 070 120 026 112 × 2 = 1 + 0.274 140 240 052 224;
  • 25) 0.274 140 240 052 224 × 2 = 0 + 0.548 280 480 104 448;
  • 26) 0.548 280 480 104 448 × 2 = 1 + 0.096 560 960 208 896;
  • 27) 0.096 560 960 208 896 × 2 = 0 + 0.193 121 920 417 792;
  • 28) 0.193 121 920 417 792 × 2 = 0 + 0.386 243 840 835 584;
  • 29) 0.386 243 840 835 584 × 2 = 0 + 0.772 487 681 671 168;
  • 30) 0.772 487 681 671 168 × 2 = 1 + 0.544 975 363 342 336;
  • 31) 0.544 975 363 342 336 × 2 = 1 + 0.089 950 726 684 672;
  • 32) 0.089 950 726 684 672 × 2 = 0 + 0.179 901 453 369 344;
  • 33) 0.179 901 453 369 344 × 2 = 0 + 0.359 802 906 738 688;
  • 34) 0.359 802 906 738 688 × 2 = 0 + 0.719 605 813 477 376;
  • 35) 0.719 605 813 477 376 × 2 = 1 + 0.439 211 626 954 752;
  • 36) 0.439 211 626 954 752 × 2 = 0 + 0.878 423 253 909 504;
  • 37) 0.878 423 253 909 504 × 2 = 1 + 0.756 846 507 819 008;
  • 38) 0.756 846 507 819 008 × 2 = 1 + 0.513 693 015 638 016;
  • 39) 0.513 693 015 638 016 × 2 = 1 + 0.027 386 031 276 032;
  • 40) 0.027 386 031 276 032 × 2 = 0 + 0.054 772 062 552 064;
  • 41) 0.054 772 062 552 064 × 2 = 0 + 0.109 544 125 104 128;
  • 42) 0.109 544 125 104 128 × 2 = 0 + 0.219 088 250 208 256;
  • 43) 0.219 088 250 208 256 × 2 = 0 + 0.438 176 500 416 512;
  • 44) 0.438 176 500 416 512 × 2 = 0 + 0.876 353 000 833 024;
  • 45) 0.876 353 000 833 024 × 2 = 1 + 0.752 706 001 666 048;
  • 46) 0.752 706 001 666 048 × 2 = 1 + 0.505 412 003 332 096;
  • 47) 0.505 412 003 332 096 × 2 = 1 + 0.010 824 006 664 192;
  • 48) 0.010 824 006 664 192 × 2 = 0 + 0.021 648 013 328 384;
  • 49) 0.021 648 013 328 384 × 2 = 0 + 0.043 296 026 656 768;
  • 50) 0.043 296 026 656 768 × 2 = 0 + 0.086 592 053 313 536;
  • 51) 0.086 592 053 313 536 × 2 = 0 + 0.173 184 106 627 072;
  • 52) 0.173 184 106 627 072 × 2 = 0 + 0.346 368 213 254 144;
  • 53) 0.346 368 213 254 144 × 2 = 0 + 0.692 736 426 508 288;
  • 54) 0.692 736 426 508 288 × 2 = 1 + 0.385 472 853 016 576;
  • 55) 0.385 472 853 016 576 × 2 = 0 + 0.770 945 706 033 152;
  • 56) 0.770 945 706 033 152 × 2 = 1 + 0.541 891 412 066 304;
  • 57) 0.541 891 412 066 304 × 2 = 1 + 0.083 782 824 132 608;
  • 58) 0.083 782 824 132 608 × 2 = 0 + 0.167 565 648 265 216;
  • 59) 0.167 565 648 265 216 × 2 = 0 + 0.335 131 296 530 432;
  • 60) 0.335 131 296 530 432 × 2 = 0 + 0.670 262 593 060 864;
  • 61) 0.670 262 593 060 864 × 2 = 1 + 0.340 525 186 121 728;
  • 62) 0.340 525 186 121 728 × 2 = 0 + 0.681 050 372 243 456;
  • 63) 0.681 050 372 243 456 × 2 = 1 + 0.362 100 744 486 912;
  • 64) 0.362 100 744 486 912 × 2 = 0 + 0.724 201 488 973 824;

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:


0.000 282 005 914 464(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1110 0000 0101 1000 1010(2)

6. Positive number before normalization:

0.000 282 005 914 464(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1110 0000 0101 1000 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:


0.000 282 005 914 464(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1110 0000 0101 1000 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1110 0000 0101 1000 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 1110 0000 0101 1000 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 1110 0000 0101 1000 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 1110 0000 0101 1000 1010 =

0010 0111 1011 0100 0110 0010 1110 0000 1110 0000 0101 1000 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 1110 0000 0101 1000 1010


Decimal number -0.000 282 005 914 464 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 1110 0000 0101 1000 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