-0.000 282 005 914 494 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 494(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
-0.000 282 005 914 494(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:

|-0.000 282 005 914 494| = 0.000 282 005 914 494


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: 0.000 282 005 914 494.

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) 0.000 282 005 914 494 × 2 = 0 + 0.000 564 011 828 988;
  • 2) 0.000 564 011 828 988 × 2 = 0 + 0.001 128 023 657 976;
  • 3) 0.001 128 023 657 976 × 2 = 0 + 0.002 256 047 315 952;
  • 4) 0.002 256 047 315 952 × 2 = 0 + 0.004 512 094 631 904;
  • 5) 0.004 512 094 631 904 × 2 = 0 + 0.009 024 189 263 808;
  • 6) 0.009 024 189 263 808 × 2 = 0 + 0.018 048 378 527 616;
  • 7) 0.018 048 378 527 616 × 2 = 0 + 0.036 096 757 055 232;
  • 8) 0.036 096 757 055 232 × 2 = 0 + 0.072 193 514 110 464;
  • 9) 0.072 193 514 110 464 × 2 = 0 + 0.144 387 028 220 928;
  • 10) 0.144 387 028 220 928 × 2 = 0 + 0.288 774 056 441 856;
  • 11) 0.288 774 056 441 856 × 2 = 0 + 0.577 548 112 883 712;
  • 12) 0.577 548 112 883 712 × 2 = 1 + 0.155 096 225 767 424;
  • 13) 0.155 096 225 767 424 × 2 = 0 + 0.310 192 451 534 848;
  • 14) 0.310 192 451 534 848 × 2 = 0 + 0.620 384 903 069 696;
  • 15) 0.620 384 903 069 696 × 2 = 1 + 0.240 769 806 139 392;
  • 16) 0.240 769 806 139 392 × 2 = 0 + 0.481 539 612 278 784;
  • 17) 0.481 539 612 278 784 × 2 = 0 + 0.963 079 224 557 568;
  • 18) 0.963 079 224 557 568 × 2 = 1 + 0.926 158 449 115 136;
  • 19) 0.926 158 449 115 136 × 2 = 1 + 0.852 316 898 230 272;
  • 20) 0.852 316 898 230 272 × 2 = 1 + 0.704 633 796 460 544;
  • 21) 0.704 633 796 460 544 × 2 = 1 + 0.409 267 592 921 088;
  • 22) 0.409 267 592 921 088 × 2 = 0 + 0.818 535 185 842 176;
  • 23) 0.818 535 185 842 176 × 2 = 1 + 0.637 070 371 684 352;
  • 24) 0.637 070 371 684 352 × 2 = 1 + 0.274 140 743 368 704;
  • 25) 0.274 140 743 368 704 × 2 = 0 + 0.548 281 486 737 408;
  • 26) 0.548 281 486 737 408 × 2 = 1 + 0.096 562 973 474 816;
  • 27) 0.096 562 973 474 816 × 2 = 0 + 0.193 125 946 949 632;
  • 28) 0.193 125 946 949 632 × 2 = 0 + 0.386 251 893 899 264;
  • 29) 0.386 251 893 899 264 × 2 = 0 + 0.772 503 787 798 528;
  • 30) 0.772 503 787 798 528 × 2 = 1 + 0.545 007 575 597 056;
  • 31) 0.545 007 575 597 056 × 2 = 1 + 0.090 015 151 194 112;
  • 32) 0.090 015 151 194 112 × 2 = 0 + 0.180 030 302 388 224;
  • 33) 0.180 030 302 388 224 × 2 = 0 + 0.360 060 604 776 448;
  • 34) 0.360 060 604 776 448 × 2 = 0 + 0.720 121 209 552 896;
  • 35) 0.720 121 209 552 896 × 2 = 1 + 0.440 242 419 105 792;
  • 36) 0.440 242 419 105 792 × 2 = 0 + 0.880 484 838 211 584;
  • 37) 0.880 484 838 211 584 × 2 = 1 + 0.760 969 676 423 168;
  • 38) 0.760 969 676 423 168 × 2 = 1 + 0.521 939 352 846 336;
  • 39) 0.521 939 352 846 336 × 2 = 1 + 0.043 878 705 692 672;
  • 40) 0.043 878 705 692 672 × 2 = 0 + 0.087 757 411 385 344;
  • 41) 0.087 757 411 385 344 × 2 = 0 + 0.175 514 822 770 688;
  • 42) 0.175 514 822 770 688 × 2 = 0 + 0.351 029 645 541 376;
  • 43) 0.351 029 645 541 376 × 2 = 0 + 0.702 059 291 082 752;
  • 44) 0.702 059 291 082 752 × 2 = 1 + 0.404 118 582 165 504;
  • 45) 0.404 118 582 165 504 × 2 = 0 + 0.808 237 164 331 008;
  • 46) 0.808 237 164 331 008 × 2 = 1 + 0.616 474 328 662 016;
  • 47) 0.616 474 328 662 016 × 2 = 1 + 0.232 948 657 324 032;
  • 48) 0.232 948 657 324 032 × 2 = 0 + 0.465 897 314 648 064;
  • 49) 0.465 897 314 648 064 × 2 = 0 + 0.931 794 629 296 128;
  • 50) 0.931 794 629 296 128 × 2 = 1 + 0.863 589 258 592 256;
  • 51) 0.863 589 258 592 256 × 2 = 1 + 0.727 178 517 184 512;
  • 52) 0.727 178 517 184 512 × 2 = 1 + 0.454 357 034 369 024;
  • 53) 0.454 357 034 369 024 × 2 = 0 + 0.908 714 068 738 048;
  • 54) 0.908 714 068 738 048 × 2 = 1 + 0.817 428 137 476 096;
  • 55) 0.817 428 137 476 096 × 2 = 1 + 0.634 856 274 952 192;
  • 56) 0.634 856 274 952 192 × 2 = 1 + 0.269 712 549 904 384;
  • 57) 0.269 712 549 904 384 × 2 = 0 + 0.539 425 099 808 768;
  • 58) 0.539 425 099 808 768 × 2 = 1 + 0.078 850 199 617 536;
  • 59) 0.078 850 199 617 536 × 2 = 0 + 0.157 700 399 235 072;
  • 60) 0.157 700 399 235 072 × 2 = 0 + 0.315 400 798 470 144;
  • 61) 0.315 400 798 470 144 × 2 = 0 + 0.630 801 596 940 288;
  • 62) 0.630 801 596 940 288 × 2 = 1 + 0.261 603 193 880 576;
  • 63) 0.261 603 193 880 576 × 2 = 0 + 0.523 206 387 761 152;
  • 64) 0.523 206 387 761 152 × 2 = 1 + 0.046 412 775 522 304;

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:


0.000 282 005 914 494(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0111 0111 0100 0101(2)

6. Positive number before normalization:

0.000 282 005 914 494(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0111 0111 0100 0101(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:


0.000 282 005 914 494(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0111 0111 0100 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0110 0111 0111 0100 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 0110 0111 0111 0100 0101(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 0110 0111 0111 0100 0101


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 0110 0111 0111 0100 0101 =

0010 0111 1011 0100 0110 0010 1110 0001 0110 0111 0111 0100 0101


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 0110 0111 0111 0100 0101


Decimal number -0.000 282 005 914 494 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 0110 0111 0111 0100 0101

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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