-0.000 282 005 914 421 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 421(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
-0.000 282 005 914 421(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:

|-0.000 282 005 914 421| = 0.000 282 005 914 421


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: 0.000 282 005 914 421.

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) 0.000 282 005 914 421 × 2 = 0 + 0.000 564 011 828 842;
  • 2) 0.000 564 011 828 842 × 2 = 0 + 0.001 128 023 657 684;
  • 3) 0.001 128 023 657 684 × 2 = 0 + 0.002 256 047 315 368;
  • 4) 0.002 256 047 315 368 × 2 = 0 + 0.004 512 094 630 736;
  • 5) 0.004 512 094 630 736 × 2 = 0 + 0.009 024 189 261 472;
  • 6) 0.009 024 189 261 472 × 2 = 0 + 0.018 048 378 522 944;
  • 7) 0.018 048 378 522 944 × 2 = 0 + 0.036 096 757 045 888;
  • 8) 0.036 096 757 045 888 × 2 = 0 + 0.072 193 514 091 776;
  • 9) 0.072 193 514 091 776 × 2 = 0 + 0.144 387 028 183 552;
  • 10) 0.144 387 028 183 552 × 2 = 0 + 0.288 774 056 367 104;
  • 11) 0.288 774 056 367 104 × 2 = 0 + 0.577 548 112 734 208;
  • 12) 0.577 548 112 734 208 × 2 = 1 + 0.155 096 225 468 416;
  • 13) 0.155 096 225 468 416 × 2 = 0 + 0.310 192 450 936 832;
  • 14) 0.310 192 450 936 832 × 2 = 0 + 0.620 384 901 873 664;
  • 15) 0.620 384 901 873 664 × 2 = 1 + 0.240 769 803 747 328;
  • 16) 0.240 769 803 747 328 × 2 = 0 + 0.481 539 607 494 656;
  • 17) 0.481 539 607 494 656 × 2 = 0 + 0.963 079 214 989 312;
  • 18) 0.963 079 214 989 312 × 2 = 1 + 0.926 158 429 978 624;
  • 19) 0.926 158 429 978 624 × 2 = 1 + 0.852 316 859 957 248;
  • 20) 0.852 316 859 957 248 × 2 = 1 + 0.704 633 719 914 496;
  • 21) 0.704 633 719 914 496 × 2 = 1 + 0.409 267 439 828 992;
  • 22) 0.409 267 439 828 992 × 2 = 0 + 0.818 534 879 657 984;
  • 23) 0.818 534 879 657 984 × 2 = 1 + 0.637 069 759 315 968;
  • 24) 0.637 069 759 315 968 × 2 = 1 + 0.274 139 518 631 936;
  • 25) 0.274 139 518 631 936 × 2 = 0 + 0.548 279 037 263 872;
  • 26) 0.548 279 037 263 872 × 2 = 1 + 0.096 558 074 527 744;
  • 27) 0.096 558 074 527 744 × 2 = 0 + 0.193 116 149 055 488;
  • 28) 0.193 116 149 055 488 × 2 = 0 + 0.386 232 298 110 976;
  • 29) 0.386 232 298 110 976 × 2 = 0 + 0.772 464 596 221 952;
  • 30) 0.772 464 596 221 952 × 2 = 1 + 0.544 929 192 443 904;
  • 31) 0.544 929 192 443 904 × 2 = 1 + 0.089 858 384 887 808;
  • 32) 0.089 858 384 887 808 × 2 = 0 + 0.179 716 769 775 616;
  • 33) 0.179 716 769 775 616 × 2 = 0 + 0.359 433 539 551 232;
  • 34) 0.359 433 539 551 232 × 2 = 0 + 0.718 867 079 102 464;
  • 35) 0.718 867 079 102 464 × 2 = 1 + 0.437 734 158 204 928;
  • 36) 0.437 734 158 204 928 × 2 = 0 + 0.875 468 316 409 856;
  • 37) 0.875 468 316 409 856 × 2 = 1 + 0.750 936 632 819 712;
  • 38) 0.750 936 632 819 712 × 2 = 1 + 0.501 873 265 639 424;
  • 39) 0.501 873 265 639 424 × 2 = 1 + 0.003 746 531 278 848;
  • 40) 0.003 746 531 278 848 × 2 = 0 + 0.007 493 062 557 696;
  • 41) 0.007 493 062 557 696 × 2 = 0 + 0.014 986 125 115 392;
  • 42) 0.014 986 125 115 392 × 2 = 0 + 0.029 972 250 230 784;
  • 43) 0.029 972 250 230 784 × 2 = 0 + 0.059 944 500 461 568;
  • 44) 0.059 944 500 461 568 × 2 = 0 + 0.119 889 000 923 136;
  • 45) 0.119 889 000 923 136 × 2 = 0 + 0.239 778 001 846 272;
  • 46) 0.239 778 001 846 272 × 2 = 0 + 0.479 556 003 692 544;
  • 47) 0.479 556 003 692 544 × 2 = 0 + 0.959 112 007 385 088;
  • 48) 0.959 112 007 385 088 × 2 = 1 + 0.918 224 014 770 176;
  • 49) 0.918 224 014 770 176 × 2 = 1 + 0.836 448 029 540 352;
  • 50) 0.836 448 029 540 352 × 2 = 1 + 0.672 896 059 080 704;
  • 51) 0.672 896 059 080 704 × 2 = 1 + 0.345 792 118 161 408;
  • 52) 0.345 792 118 161 408 × 2 = 0 + 0.691 584 236 322 816;
  • 53) 0.691 584 236 322 816 × 2 = 1 + 0.383 168 472 645 632;
  • 54) 0.383 168 472 645 632 × 2 = 0 + 0.766 336 945 291 264;
  • 55) 0.766 336 945 291 264 × 2 = 1 + 0.532 673 890 582 528;
  • 56) 0.532 673 890 582 528 × 2 = 1 + 0.065 347 781 165 056;
  • 57) 0.065 347 781 165 056 × 2 = 0 + 0.130 695 562 330 112;
  • 58) 0.130 695 562 330 112 × 2 = 0 + 0.261 391 124 660 224;
  • 59) 0.261 391 124 660 224 × 2 = 0 + 0.522 782 249 320 448;
  • 60) 0.522 782 249 320 448 × 2 = 1 + 0.045 564 498 640 896;
  • 61) 0.045 564 498 640 896 × 2 = 0 + 0.091 128 997 281 792;
  • 62) 0.091 128 997 281 792 × 2 = 0 + 0.182 257 994 563 584;
  • 63) 0.182 257 994 563 584 × 2 = 0 + 0.364 515 989 127 168;
  • 64) 0.364 515 989 127 168 × 2 = 0 + 0.729 031 978 254 336;

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:


0.000 282 005 914 421(10) =

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

6. Positive number before normalization:

0.000 282 005 914 421(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 0001 1110 1011 0001 0000(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:


0.000 282 005 914 421(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1110 0000 0001 1110 1011 0001 0000(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 0000 0001 1110 1011 0001 0000


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 0000 0001 1110 1011 0001 0000 =

0010 0111 1011 0100 0110 0010 1110 0000 0001 1110 1011 0001 0000


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 0000 0001 1110 1011 0001 0000


Decimal number -0.000 282 005 914 421 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 0001 1110 1011 0001 0000

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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