-0.000 282 005 914 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 5(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 5(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 5| = 0.000 282 005 914 5


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 5.

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 5 × 2 = 0 + 0.000 564 011 829;
  • 2) 0.000 564 011 829 × 2 = 0 + 0.001 128 023 658;
  • 3) 0.001 128 023 658 × 2 = 0 + 0.002 256 047 316;
  • 4) 0.002 256 047 316 × 2 = 0 + 0.004 512 094 632;
  • 5) 0.004 512 094 632 × 2 = 0 + 0.009 024 189 264;
  • 6) 0.009 024 189 264 × 2 = 0 + 0.018 048 378 528;
  • 7) 0.018 048 378 528 × 2 = 0 + 0.036 096 757 056;
  • 8) 0.036 096 757 056 × 2 = 0 + 0.072 193 514 112;
  • 9) 0.072 193 514 112 × 2 = 0 + 0.144 387 028 224;
  • 10) 0.144 387 028 224 × 2 = 0 + 0.288 774 056 448;
  • 11) 0.288 774 056 448 × 2 = 0 + 0.577 548 112 896;
  • 12) 0.577 548 112 896 × 2 = 1 + 0.155 096 225 792;
  • 13) 0.155 096 225 792 × 2 = 0 + 0.310 192 451 584;
  • 14) 0.310 192 451 584 × 2 = 0 + 0.620 384 903 168;
  • 15) 0.620 384 903 168 × 2 = 1 + 0.240 769 806 336;
  • 16) 0.240 769 806 336 × 2 = 0 + 0.481 539 612 672;
  • 17) 0.481 539 612 672 × 2 = 0 + 0.963 079 225 344;
  • 18) 0.963 079 225 344 × 2 = 1 + 0.926 158 450 688;
  • 19) 0.926 158 450 688 × 2 = 1 + 0.852 316 901 376;
  • 20) 0.852 316 901 376 × 2 = 1 + 0.704 633 802 752;
  • 21) 0.704 633 802 752 × 2 = 1 + 0.409 267 605 504;
  • 22) 0.409 267 605 504 × 2 = 0 + 0.818 535 211 008;
  • 23) 0.818 535 211 008 × 2 = 1 + 0.637 070 422 016;
  • 24) 0.637 070 422 016 × 2 = 1 + 0.274 140 844 032;
  • 25) 0.274 140 844 032 × 2 = 0 + 0.548 281 688 064;
  • 26) 0.548 281 688 064 × 2 = 1 + 0.096 563 376 128;
  • 27) 0.096 563 376 128 × 2 = 0 + 0.193 126 752 256;
  • 28) 0.193 126 752 256 × 2 = 0 + 0.386 253 504 512;
  • 29) 0.386 253 504 512 × 2 = 0 + 0.772 507 009 024;
  • 30) 0.772 507 009 024 × 2 = 1 + 0.545 014 018 048;
  • 31) 0.545 014 018 048 × 2 = 1 + 0.090 028 036 096;
  • 32) 0.090 028 036 096 × 2 = 0 + 0.180 056 072 192;
  • 33) 0.180 056 072 192 × 2 = 0 + 0.360 112 144 384;
  • 34) 0.360 112 144 384 × 2 = 0 + 0.720 224 288 768;
  • 35) 0.720 224 288 768 × 2 = 1 + 0.440 448 577 536;
  • 36) 0.440 448 577 536 × 2 = 0 + 0.880 897 155 072;
  • 37) 0.880 897 155 072 × 2 = 1 + 0.761 794 310 144;
  • 38) 0.761 794 310 144 × 2 = 1 + 0.523 588 620 288;
  • 39) 0.523 588 620 288 × 2 = 1 + 0.047 177 240 576;
  • 40) 0.047 177 240 576 × 2 = 0 + 0.094 354 481 152;
  • 41) 0.094 354 481 152 × 2 = 0 + 0.188 708 962 304;
  • 42) 0.188 708 962 304 × 2 = 0 + 0.377 417 924 608;
  • 43) 0.377 417 924 608 × 2 = 0 + 0.754 835 849 216;
  • 44) 0.754 835 849 216 × 2 = 1 + 0.509 671 698 432;
  • 45) 0.509 671 698 432 × 2 = 1 + 0.019 343 396 864;
  • 46) 0.019 343 396 864 × 2 = 0 + 0.038 686 793 728;
  • 47) 0.038 686 793 728 × 2 = 0 + 0.077 373 587 456;
  • 48) 0.077 373 587 456 × 2 = 0 + 0.154 747 174 912;
  • 49) 0.154 747 174 912 × 2 = 0 + 0.309 494 349 824;
  • 50) 0.309 494 349 824 × 2 = 0 + 0.618 988 699 648;
  • 51) 0.618 988 699 648 × 2 = 1 + 0.237 977 399 296;
  • 52) 0.237 977 399 296 × 2 = 0 + 0.475 954 798 592;
  • 53) 0.475 954 798 592 × 2 = 0 + 0.951 909 597 184;
  • 54) 0.951 909 597 184 × 2 = 1 + 0.903 819 194 368;
  • 55) 0.903 819 194 368 × 2 = 1 + 0.807 638 388 736;
  • 56) 0.807 638 388 736 × 2 = 1 + 0.615 276 777 472;
  • 57) 0.615 276 777 472 × 2 = 1 + 0.230 553 554 944;
  • 58) 0.230 553 554 944 × 2 = 0 + 0.461 107 109 888;
  • 59) 0.461 107 109 888 × 2 = 0 + 0.922 214 219 776;
  • 60) 0.922 214 219 776 × 2 = 1 + 0.844 428 439 552;
  • 61) 0.844 428 439 552 × 2 = 1 + 0.688 856 879 104;
  • 62) 0.688 856 879 104 × 2 = 1 + 0.377 713 758 208;
  • 63) 0.377 713 758 208 × 2 = 0 + 0.755 427 516 416;
  • 64) 0.755 427 516 416 × 2 = 1 + 0.510 855 032 832;

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 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1000 0010 0111 1001 1101(2)

6. Positive number before normalization:

0.000 282 005 914 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1000 0010 0111 1001 1101(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 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1000 0010 0111 1001 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1000 0010 0111 1001 1101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 1000 0010 0111 1001 1101(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 1000 0010 0111 1001 1101


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 1000 0010 0111 1001 1101 =

0010 0111 1011 0100 0110 0010 1110 0001 1000 0010 0111 1001 1101


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 1000 0010 0111 1001 1101


Decimal number -0.000 282 005 914 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 1000 0010 0111 1001 1101

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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