-0.000 282 005 940 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 940 9(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
-0.000 282 005 940 9(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:

|-0.000 282 005 940 9| = 0.000 282 005 940 9


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: 0.000 282 005 940 9.

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) 0.000 282 005 940 9 × 2 = 0 + 0.000 564 011 881 8;
  • 2) 0.000 564 011 881 8 × 2 = 0 + 0.001 128 023 763 6;
  • 3) 0.001 128 023 763 6 × 2 = 0 + 0.002 256 047 527 2;
  • 4) 0.002 256 047 527 2 × 2 = 0 + 0.004 512 095 054 4;
  • 5) 0.004 512 095 054 4 × 2 = 0 + 0.009 024 190 108 8;
  • 6) 0.009 024 190 108 8 × 2 = 0 + 0.018 048 380 217 6;
  • 7) 0.018 048 380 217 6 × 2 = 0 + 0.036 096 760 435 2;
  • 8) 0.036 096 760 435 2 × 2 = 0 + 0.072 193 520 870 4;
  • 9) 0.072 193 520 870 4 × 2 = 0 + 0.144 387 041 740 8;
  • 10) 0.144 387 041 740 8 × 2 = 0 + 0.288 774 083 481 6;
  • 11) 0.288 774 083 481 6 × 2 = 0 + 0.577 548 166 963 2;
  • 12) 0.577 548 166 963 2 × 2 = 1 + 0.155 096 333 926 4;
  • 13) 0.155 096 333 926 4 × 2 = 0 + 0.310 192 667 852 8;
  • 14) 0.310 192 667 852 8 × 2 = 0 + 0.620 385 335 705 6;
  • 15) 0.620 385 335 705 6 × 2 = 1 + 0.240 770 671 411 2;
  • 16) 0.240 770 671 411 2 × 2 = 0 + 0.481 541 342 822 4;
  • 17) 0.481 541 342 822 4 × 2 = 0 + 0.963 082 685 644 8;
  • 18) 0.963 082 685 644 8 × 2 = 1 + 0.926 165 371 289 6;
  • 19) 0.926 165 371 289 6 × 2 = 1 + 0.852 330 742 579 2;
  • 20) 0.852 330 742 579 2 × 2 = 1 + 0.704 661 485 158 4;
  • 21) 0.704 661 485 158 4 × 2 = 1 + 0.409 322 970 316 8;
  • 22) 0.409 322 970 316 8 × 2 = 0 + 0.818 645 940 633 6;
  • 23) 0.818 645 940 633 6 × 2 = 1 + 0.637 291 881 267 2;
  • 24) 0.637 291 881 267 2 × 2 = 1 + 0.274 583 762 534 4;
  • 25) 0.274 583 762 534 4 × 2 = 0 + 0.549 167 525 068 8;
  • 26) 0.549 167 525 068 8 × 2 = 1 + 0.098 335 050 137 6;
  • 27) 0.098 335 050 137 6 × 2 = 0 + 0.196 670 100 275 2;
  • 28) 0.196 670 100 275 2 × 2 = 0 + 0.393 340 200 550 4;
  • 29) 0.393 340 200 550 4 × 2 = 0 + 0.786 680 401 100 8;
  • 30) 0.786 680 401 100 8 × 2 = 1 + 0.573 360 802 201 6;
  • 31) 0.573 360 802 201 6 × 2 = 1 + 0.146 721 604 403 2;
  • 32) 0.146 721 604 403 2 × 2 = 0 + 0.293 443 208 806 4;
  • 33) 0.293 443 208 806 4 × 2 = 0 + 0.586 886 417 612 8;
  • 34) 0.586 886 417 612 8 × 2 = 1 + 0.173 772 835 225 6;
  • 35) 0.173 772 835 225 6 × 2 = 0 + 0.347 545 670 451 2;
  • 36) 0.347 545 670 451 2 × 2 = 0 + 0.695 091 340 902 4;
  • 37) 0.695 091 340 902 4 × 2 = 1 + 0.390 182 681 804 8;
  • 38) 0.390 182 681 804 8 × 2 = 0 + 0.780 365 363 609 6;
  • 39) 0.780 365 363 609 6 × 2 = 1 + 0.560 730 727 219 2;
  • 40) 0.560 730 727 219 2 × 2 = 1 + 0.121 461 454 438 4;
  • 41) 0.121 461 454 438 4 × 2 = 0 + 0.242 922 908 876 8;
  • 42) 0.242 922 908 876 8 × 2 = 0 + 0.485 845 817 753 6;
  • 43) 0.485 845 817 753 6 × 2 = 0 + 0.971 691 635 507 2;
  • 44) 0.971 691 635 507 2 × 2 = 1 + 0.943 383 271 014 4;
  • 45) 0.943 383 271 014 4 × 2 = 1 + 0.886 766 542 028 8;
  • 46) 0.886 766 542 028 8 × 2 = 1 + 0.773 533 084 057 6;
  • 47) 0.773 533 084 057 6 × 2 = 1 + 0.547 066 168 115 2;
  • 48) 0.547 066 168 115 2 × 2 = 1 + 0.094 132 336 230 4;
  • 49) 0.094 132 336 230 4 × 2 = 0 + 0.188 264 672 460 8;
  • 50) 0.188 264 672 460 8 × 2 = 0 + 0.376 529 344 921 6;
  • 51) 0.376 529 344 921 6 × 2 = 0 + 0.753 058 689 843 2;
  • 52) 0.753 058 689 843 2 × 2 = 1 + 0.506 117 379 686 4;
  • 53) 0.506 117 379 686 4 × 2 = 1 + 0.012 234 759 372 8;
  • 54) 0.012 234 759 372 8 × 2 = 0 + 0.024 469 518 745 6;
  • 55) 0.024 469 518 745 6 × 2 = 0 + 0.048 939 037 491 2;
  • 56) 0.048 939 037 491 2 × 2 = 0 + 0.097 878 074 982 4;
  • 57) 0.097 878 074 982 4 × 2 = 0 + 0.195 756 149 964 8;
  • 58) 0.195 756 149 964 8 × 2 = 0 + 0.391 512 299 929 6;
  • 59) 0.391 512 299 929 6 × 2 = 0 + 0.783 024 599 859 2;
  • 60) 0.783 024 599 859 2 × 2 = 1 + 0.566 049 199 718 4;
  • 61) 0.566 049 199 718 4 × 2 = 1 + 0.132 098 399 436 8;
  • 62) 0.132 098 399 436 8 × 2 = 0 + 0.264 196 798 873 6;
  • 63) 0.264 196 798 873 6 × 2 = 0 + 0.528 393 597 747 2;
  • 64) 0.528 393 597 747 2 × 2 = 1 + 0.056 787 195 494 4;

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:


0.000 282 005 940 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1011 0001 1111 0001 1000 0001 1001(2)

6. Positive number before normalization:

0.000 282 005 940 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1011 0001 1111 0001 1000 0001 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:


0.000 282 005 940 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1011 0001 1111 0001 1000 0001 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1011 0001 1111 0001 1000 0001 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0100 1011 0001 1111 0001 1000 0001 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 0100 1011 0001 1111 0001 1000 0001 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 0100 1011 0001 1111 0001 1000 0001 1001 =

0010 0111 1011 0100 0110 0100 1011 0001 1111 0001 1000 0001 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 0100 1011 0001 1111 0001 1000 0001 1001


Decimal number -0.000 282 005 940 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0100 1011 0001 1111 0001 1000 0001 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