-0.000 282 005 914 516 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 516(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
-0.000 282 005 914 516(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:

|-0.000 282 005 914 516| = 0.000 282 005 914 516


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: 0.000 282 005 914 516.

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) 0.000 282 005 914 516 × 2 = 0 + 0.000 564 011 829 032;
  • 2) 0.000 564 011 829 032 × 2 = 0 + 0.001 128 023 658 064;
  • 3) 0.001 128 023 658 064 × 2 = 0 + 0.002 256 047 316 128;
  • 4) 0.002 256 047 316 128 × 2 = 0 + 0.004 512 094 632 256;
  • 5) 0.004 512 094 632 256 × 2 = 0 + 0.009 024 189 264 512;
  • 6) 0.009 024 189 264 512 × 2 = 0 + 0.018 048 378 529 024;
  • 7) 0.018 048 378 529 024 × 2 = 0 + 0.036 096 757 058 048;
  • 8) 0.036 096 757 058 048 × 2 = 0 + 0.072 193 514 116 096;
  • 9) 0.072 193 514 116 096 × 2 = 0 + 0.144 387 028 232 192;
  • 10) 0.144 387 028 232 192 × 2 = 0 + 0.288 774 056 464 384;
  • 11) 0.288 774 056 464 384 × 2 = 0 + 0.577 548 112 928 768;
  • 12) 0.577 548 112 928 768 × 2 = 1 + 0.155 096 225 857 536;
  • 13) 0.155 096 225 857 536 × 2 = 0 + 0.310 192 451 715 072;
  • 14) 0.310 192 451 715 072 × 2 = 0 + 0.620 384 903 430 144;
  • 15) 0.620 384 903 430 144 × 2 = 1 + 0.240 769 806 860 288;
  • 16) 0.240 769 806 860 288 × 2 = 0 + 0.481 539 613 720 576;
  • 17) 0.481 539 613 720 576 × 2 = 0 + 0.963 079 227 441 152;
  • 18) 0.963 079 227 441 152 × 2 = 1 + 0.926 158 454 882 304;
  • 19) 0.926 158 454 882 304 × 2 = 1 + 0.852 316 909 764 608;
  • 20) 0.852 316 909 764 608 × 2 = 1 + 0.704 633 819 529 216;
  • 21) 0.704 633 819 529 216 × 2 = 1 + 0.409 267 639 058 432;
  • 22) 0.409 267 639 058 432 × 2 = 0 + 0.818 535 278 116 864;
  • 23) 0.818 535 278 116 864 × 2 = 1 + 0.637 070 556 233 728;
  • 24) 0.637 070 556 233 728 × 2 = 1 + 0.274 141 112 467 456;
  • 25) 0.274 141 112 467 456 × 2 = 0 + 0.548 282 224 934 912;
  • 26) 0.548 282 224 934 912 × 2 = 1 + 0.096 564 449 869 824;
  • 27) 0.096 564 449 869 824 × 2 = 0 + 0.193 128 899 739 648;
  • 28) 0.193 128 899 739 648 × 2 = 0 + 0.386 257 799 479 296;
  • 29) 0.386 257 799 479 296 × 2 = 0 + 0.772 515 598 958 592;
  • 30) 0.772 515 598 958 592 × 2 = 1 + 0.545 031 197 917 184;
  • 31) 0.545 031 197 917 184 × 2 = 1 + 0.090 062 395 834 368;
  • 32) 0.090 062 395 834 368 × 2 = 0 + 0.180 124 791 668 736;
  • 33) 0.180 124 791 668 736 × 2 = 0 + 0.360 249 583 337 472;
  • 34) 0.360 249 583 337 472 × 2 = 0 + 0.720 499 166 674 944;
  • 35) 0.720 499 166 674 944 × 2 = 1 + 0.440 998 333 349 888;
  • 36) 0.440 998 333 349 888 × 2 = 0 + 0.881 996 666 699 776;
  • 37) 0.881 996 666 699 776 × 2 = 1 + 0.763 993 333 399 552;
  • 38) 0.763 993 333 399 552 × 2 = 1 + 0.527 986 666 799 104;
  • 39) 0.527 986 666 799 104 × 2 = 1 + 0.055 973 333 598 208;
  • 40) 0.055 973 333 598 208 × 2 = 0 + 0.111 946 667 196 416;
  • 41) 0.111 946 667 196 416 × 2 = 0 + 0.223 893 334 392 832;
  • 42) 0.223 893 334 392 832 × 2 = 0 + 0.447 786 668 785 664;
  • 43) 0.447 786 668 785 664 × 2 = 0 + 0.895 573 337 571 328;
  • 44) 0.895 573 337 571 328 × 2 = 1 + 0.791 146 675 142 656;
  • 45) 0.791 146 675 142 656 × 2 = 1 + 0.582 293 350 285 312;
  • 46) 0.582 293 350 285 312 × 2 = 1 + 0.164 586 700 570 624;
  • 47) 0.164 586 700 570 624 × 2 = 0 + 0.329 173 401 141 248;
  • 48) 0.329 173 401 141 248 × 2 = 0 + 0.658 346 802 282 496;
  • 49) 0.658 346 802 282 496 × 2 = 1 + 0.316 693 604 564 992;
  • 50) 0.316 693 604 564 992 × 2 = 0 + 0.633 387 209 129 984;
  • 51) 0.633 387 209 129 984 × 2 = 1 + 0.266 774 418 259 968;
  • 52) 0.266 774 418 259 968 × 2 = 0 + 0.533 548 836 519 936;
  • 53) 0.533 548 836 519 936 × 2 = 1 + 0.067 097 673 039 872;
  • 54) 0.067 097 673 039 872 × 2 = 0 + 0.134 195 346 079 744;
  • 55) 0.134 195 346 079 744 × 2 = 0 + 0.268 390 692 159 488;
  • 56) 0.268 390 692 159 488 × 2 = 0 + 0.536 781 384 318 976;
  • 57) 0.536 781 384 318 976 × 2 = 1 + 0.073 562 768 637 952;
  • 58) 0.073 562 768 637 952 × 2 = 0 + 0.147 125 537 275 904;
  • 59) 0.147 125 537 275 904 × 2 = 0 + 0.294 251 074 551 808;
  • 60) 0.294 251 074 551 808 × 2 = 0 + 0.588 502 149 103 616;
  • 61) 0.588 502 149 103 616 × 2 = 1 + 0.177 004 298 207 232;
  • 62) 0.177 004 298 207 232 × 2 = 0 + 0.354 008 596 414 464;
  • 63) 0.354 008 596 414 464 × 2 = 0 + 0.708 017 192 828 928;
  • 64) 0.708 017 192 828 928 × 2 = 1 + 0.416 034 385 657 856;

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:


0.000 282 005 914 516(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1100 1010 1000 1000 1001(2)

6. Positive number before normalization:

0.000 282 005 914 516(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1100 1010 1000 1000 1001(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:


0.000 282 005 914 516(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1100 1010 1000 1000 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1100 1010 1000 1000 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 1100 1010 1000 1000 1001(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 1100 1010 1000 1000 1001


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 1100 1010 1000 1000 1001 =

0010 0111 1011 0100 0110 0010 1110 0001 1100 1010 1000 1000 1001


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 1100 1010 1000 1000 1001


Decimal number -0.000 282 005 914 516 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 1100 1010 1000 1000 1001

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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