-0.000 282 005 794 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 794(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
-0.000 282 005 794(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:

|-0.000 282 005 794| = 0.000 282 005 794


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: 0.000 282 005 794.

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) 0.000 282 005 794 × 2 = 0 + 0.000 564 011 588;
  • 2) 0.000 564 011 588 × 2 = 0 + 0.001 128 023 176;
  • 3) 0.001 128 023 176 × 2 = 0 + 0.002 256 046 352;
  • 4) 0.002 256 046 352 × 2 = 0 + 0.004 512 092 704;
  • 5) 0.004 512 092 704 × 2 = 0 + 0.009 024 185 408;
  • 6) 0.009 024 185 408 × 2 = 0 + 0.018 048 370 816;
  • 7) 0.018 048 370 816 × 2 = 0 + 0.036 096 741 632;
  • 8) 0.036 096 741 632 × 2 = 0 + 0.072 193 483 264;
  • 9) 0.072 193 483 264 × 2 = 0 + 0.144 386 966 528;
  • 10) 0.144 386 966 528 × 2 = 0 + 0.288 773 933 056;
  • 11) 0.288 773 933 056 × 2 = 0 + 0.577 547 866 112;
  • 12) 0.577 547 866 112 × 2 = 1 + 0.155 095 732 224;
  • 13) 0.155 095 732 224 × 2 = 0 + 0.310 191 464 448;
  • 14) 0.310 191 464 448 × 2 = 0 + 0.620 382 928 896;
  • 15) 0.620 382 928 896 × 2 = 1 + 0.240 765 857 792;
  • 16) 0.240 765 857 792 × 2 = 0 + 0.481 531 715 584;
  • 17) 0.481 531 715 584 × 2 = 0 + 0.963 063 431 168;
  • 18) 0.963 063 431 168 × 2 = 1 + 0.926 126 862 336;
  • 19) 0.926 126 862 336 × 2 = 1 + 0.852 253 724 672;
  • 20) 0.852 253 724 672 × 2 = 1 + 0.704 507 449 344;
  • 21) 0.704 507 449 344 × 2 = 1 + 0.409 014 898 688;
  • 22) 0.409 014 898 688 × 2 = 0 + 0.818 029 797 376;
  • 23) 0.818 029 797 376 × 2 = 1 + 0.636 059 594 752;
  • 24) 0.636 059 594 752 × 2 = 1 + 0.272 119 189 504;
  • 25) 0.272 119 189 504 × 2 = 0 + 0.544 238 379 008;
  • 26) 0.544 238 379 008 × 2 = 1 + 0.088 476 758 016;
  • 27) 0.088 476 758 016 × 2 = 0 + 0.176 953 516 032;
  • 28) 0.176 953 516 032 × 2 = 0 + 0.353 907 032 064;
  • 29) 0.353 907 032 064 × 2 = 0 + 0.707 814 064 128;
  • 30) 0.707 814 064 128 × 2 = 1 + 0.415 628 128 256;
  • 31) 0.415 628 128 256 × 2 = 0 + 0.831 256 256 512;
  • 32) 0.831 256 256 512 × 2 = 1 + 0.662 512 513 024;
  • 33) 0.662 512 513 024 × 2 = 1 + 0.325 025 026 048;
  • 34) 0.325 025 026 048 × 2 = 0 + 0.650 050 052 096;
  • 35) 0.650 050 052 096 × 2 = 1 + 0.300 100 104 192;
  • 36) 0.300 100 104 192 × 2 = 0 + 0.600 200 208 384;
  • 37) 0.600 200 208 384 × 2 = 1 + 0.200 400 416 768;
  • 38) 0.200 400 416 768 × 2 = 0 + 0.400 800 833 536;
  • 39) 0.400 800 833 536 × 2 = 0 + 0.801 601 667 072;
  • 40) 0.801 601 667 072 × 2 = 1 + 0.603 203 334 144;
  • 41) 0.603 203 334 144 × 2 = 1 + 0.206 406 668 288;
  • 42) 0.206 406 668 288 × 2 = 0 + 0.412 813 336 576;
  • 43) 0.412 813 336 576 × 2 = 0 + 0.825 626 673 152;
  • 44) 0.825 626 673 152 × 2 = 1 + 0.651 253 346 304;
  • 45) 0.651 253 346 304 × 2 = 1 + 0.302 506 692 608;
  • 46) 0.302 506 692 608 × 2 = 0 + 0.605 013 385 216;
  • 47) 0.605 013 385 216 × 2 = 1 + 0.210 026 770 432;
  • 48) 0.210 026 770 432 × 2 = 0 + 0.420 053 540 864;
  • 49) 0.420 053 540 864 × 2 = 0 + 0.840 107 081 728;
  • 50) 0.840 107 081 728 × 2 = 1 + 0.680 214 163 456;
  • 51) 0.680 214 163 456 × 2 = 1 + 0.360 428 326 912;
  • 52) 0.360 428 326 912 × 2 = 0 + 0.720 856 653 824;
  • 53) 0.720 856 653 824 × 2 = 1 + 0.441 713 307 648;
  • 54) 0.441 713 307 648 × 2 = 0 + 0.883 426 615 296;
  • 55) 0.883 426 615 296 × 2 = 1 + 0.766 853 230 592;
  • 56) 0.766 853 230 592 × 2 = 1 + 0.533 706 461 184;
  • 57) 0.533 706 461 184 × 2 = 1 + 0.067 412 922 368;
  • 58) 0.067 412 922 368 × 2 = 0 + 0.134 825 844 736;
  • 59) 0.134 825 844 736 × 2 = 0 + 0.269 651 689 472;
  • 60) 0.269 651 689 472 × 2 = 0 + 0.539 303 378 944;
  • 61) 0.539 303 378 944 × 2 = 1 + 0.078 606 757 888;
  • 62) 0.078 606 757 888 × 2 = 0 + 0.157 213 515 776;
  • 63) 0.157 213 515 776 × 2 = 0 + 0.314 427 031 552;
  • 64) 0.314 427 031 552 × 2 = 0 + 0.628 854 063 104;

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:


0.000 282 005 794(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1001 1001 1010 0110 1011 1000 1000(2)

6. Positive number before normalization:

0.000 282 005 794(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1001 1001 1010 0110 1011 1000 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:


0.000 282 005 794(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1001 1001 1010 0110 1011 1000 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1010 1001 1001 1010 0110 1011 1000 1000(2) × 20 =

1.0010 0111 1011 0100 0101 1010 1001 1001 1010 0110 1011 1000 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 1001 1001 1010 0110 1011 1000 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 1001 1001 1010 0110 1011 1000 1000 =

0010 0111 1011 0100 0101 1010 1001 1001 1010 0110 1011 1000 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 1001 1001 1010 0110 1011 1000 1000


Decimal number -0.000 282 005 794 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1010 1001 1001 1010 0110 1011 1000 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