-0.000 282 005 796 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 796(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
-0.000 282 005 796(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:

|-0.000 282 005 796| = 0.000 282 005 796


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: 0.000 282 005 796.

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) 0.000 282 005 796 × 2 = 0 + 0.000 564 011 592;
  • 2) 0.000 564 011 592 × 2 = 0 + 0.001 128 023 184;
  • 3) 0.001 128 023 184 × 2 = 0 + 0.002 256 046 368;
  • 4) 0.002 256 046 368 × 2 = 0 + 0.004 512 092 736;
  • 5) 0.004 512 092 736 × 2 = 0 + 0.009 024 185 472;
  • 6) 0.009 024 185 472 × 2 = 0 + 0.018 048 370 944;
  • 7) 0.018 048 370 944 × 2 = 0 + 0.036 096 741 888;
  • 8) 0.036 096 741 888 × 2 = 0 + 0.072 193 483 776;
  • 9) 0.072 193 483 776 × 2 = 0 + 0.144 386 967 552;
  • 10) 0.144 386 967 552 × 2 = 0 + 0.288 773 935 104;
  • 11) 0.288 773 935 104 × 2 = 0 + 0.577 547 870 208;
  • 12) 0.577 547 870 208 × 2 = 1 + 0.155 095 740 416;
  • 13) 0.155 095 740 416 × 2 = 0 + 0.310 191 480 832;
  • 14) 0.310 191 480 832 × 2 = 0 + 0.620 382 961 664;
  • 15) 0.620 382 961 664 × 2 = 1 + 0.240 765 923 328;
  • 16) 0.240 765 923 328 × 2 = 0 + 0.481 531 846 656;
  • 17) 0.481 531 846 656 × 2 = 0 + 0.963 063 693 312;
  • 18) 0.963 063 693 312 × 2 = 1 + 0.926 127 386 624;
  • 19) 0.926 127 386 624 × 2 = 1 + 0.852 254 773 248;
  • 20) 0.852 254 773 248 × 2 = 1 + 0.704 509 546 496;
  • 21) 0.704 509 546 496 × 2 = 1 + 0.409 019 092 992;
  • 22) 0.409 019 092 992 × 2 = 0 + 0.818 038 185 984;
  • 23) 0.818 038 185 984 × 2 = 1 + 0.636 076 371 968;
  • 24) 0.636 076 371 968 × 2 = 1 + 0.272 152 743 936;
  • 25) 0.272 152 743 936 × 2 = 0 + 0.544 305 487 872;
  • 26) 0.544 305 487 872 × 2 = 1 + 0.088 610 975 744;
  • 27) 0.088 610 975 744 × 2 = 0 + 0.177 221 951 488;
  • 28) 0.177 221 951 488 × 2 = 0 + 0.354 443 902 976;
  • 29) 0.354 443 902 976 × 2 = 0 + 0.708 887 805 952;
  • 30) 0.708 887 805 952 × 2 = 1 + 0.417 775 611 904;
  • 31) 0.417 775 611 904 × 2 = 0 + 0.835 551 223 808;
  • 32) 0.835 551 223 808 × 2 = 1 + 0.671 102 447 616;
  • 33) 0.671 102 447 616 × 2 = 1 + 0.342 204 895 232;
  • 34) 0.342 204 895 232 × 2 = 0 + 0.684 409 790 464;
  • 35) 0.684 409 790 464 × 2 = 1 + 0.368 819 580 928;
  • 36) 0.368 819 580 928 × 2 = 0 + 0.737 639 161 856;
  • 37) 0.737 639 161 856 × 2 = 1 + 0.475 278 323 712;
  • 38) 0.475 278 323 712 × 2 = 0 + 0.950 556 647 424;
  • 39) 0.950 556 647 424 × 2 = 1 + 0.901 113 294 848;
  • 40) 0.901 113 294 848 × 2 = 1 + 0.802 226 589 696;
  • 41) 0.802 226 589 696 × 2 = 1 + 0.604 453 179 392;
  • 42) 0.604 453 179 392 × 2 = 1 + 0.208 906 358 784;
  • 43) 0.208 906 358 784 × 2 = 0 + 0.417 812 717 568;
  • 44) 0.417 812 717 568 × 2 = 0 + 0.835 625 435 136;
  • 45) 0.835 625 435 136 × 2 = 1 + 0.671 250 870 272;
  • 46) 0.671 250 870 272 × 2 = 1 + 0.342 501 740 544;
  • 47) 0.342 501 740 544 × 2 = 0 + 0.685 003 481 088;
  • 48) 0.685 003 481 088 × 2 = 1 + 0.370 006 962 176;
  • 49) 0.370 006 962 176 × 2 = 0 + 0.740 013 924 352;
  • 50) 0.740 013 924 352 × 2 = 1 + 0.480 027 848 704;
  • 51) 0.480 027 848 704 × 2 = 0 + 0.960 055 697 408;
  • 52) 0.960 055 697 408 × 2 = 1 + 0.920 111 394 816;
  • 53) 0.920 111 394 816 × 2 = 1 + 0.840 222 789 632;
  • 54) 0.840 222 789 632 × 2 = 1 + 0.680 445 579 264;
  • 55) 0.680 445 579 264 × 2 = 1 + 0.360 891 158 528;
  • 56) 0.360 891 158 528 × 2 = 0 + 0.721 782 317 056;
  • 57) 0.721 782 317 056 × 2 = 1 + 0.443 564 634 112;
  • 58) 0.443 564 634 112 × 2 = 0 + 0.887 129 268 224;
  • 59) 0.887 129 268 224 × 2 = 1 + 0.774 258 536 448;
  • 60) 0.774 258 536 448 × 2 = 1 + 0.548 517 072 896;
  • 61) 0.548 517 072 896 × 2 = 1 + 0.097 034 145 792;
  • 62) 0.097 034 145 792 × 2 = 0 + 0.194 068 291 584;
  • 63) 0.194 068 291 584 × 2 = 0 + 0.388 136 583 168;
  • 64) 0.388 136 583 168 × 2 = 0 + 0.776 273 166 336;

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:


0.000 282 005 796(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1011 1100 1101 0101 1110 1011 1000(2)

6. Positive number before normalization:

0.000 282 005 796(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1011 1100 1101 0101 1110 1011 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:


0.000 282 005 796(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1011 1100 1101 0101 1110 1011 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1011 1100 1101 0101 1110 1011 1000(2) × 20 =

1.0010 0111 1011 0100 0101 1010 1011 1100 1101 0101 1110 1011 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 0101 1010 1011 1100 1101 0101 1110 1011 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 0101 1010 1011 1100 1101 0101 1110 1011 1000 =

0010 0111 1011 0100 0101 1010 1011 1100 1101 0101 1110 1011 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 0101 1010 1011 1100 1101 0101 1110 1011 1000


Decimal number -0.000 282 005 796 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1010 1011 1100 1101 0101 1110 1011 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