-0.000 282 005 914 376 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 376(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
-0.000 282 005 914 376(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:

|-0.000 282 005 914 376| = 0.000 282 005 914 376


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: 0.000 282 005 914 376.

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) 0.000 282 005 914 376 × 2 = 0 + 0.000 564 011 828 752;
  • 2) 0.000 564 011 828 752 × 2 = 0 + 0.001 128 023 657 504;
  • 3) 0.001 128 023 657 504 × 2 = 0 + 0.002 256 047 315 008;
  • 4) 0.002 256 047 315 008 × 2 = 0 + 0.004 512 094 630 016;
  • 5) 0.004 512 094 630 016 × 2 = 0 + 0.009 024 189 260 032;
  • 6) 0.009 024 189 260 032 × 2 = 0 + 0.018 048 378 520 064;
  • 7) 0.018 048 378 520 064 × 2 = 0 + 0.036 096 757 040 128;
  • 8) 0.036 096 757 040 128 × 2 = 0 + 0.072 193 514 080 256;
  • 9) 0.072 193 514 080 256 × 2 = 0 + 0.144 387 028 160 512;
  • 10) 0.144 387 028 160 512 × 2 = 0 + 0.288 774 056 321 024;
  • 11) 0.288 774 056 321 024 × 2 = 0 + 0.577 548 112 642 048;
  • 12) 0.577 548 112 642 048 × 2 = 1 + 0.155 096 225 284 096;
  • 13) 0.155 096 225 284 096 × 2 = 0 + 0.310 192 450 568 192;
  • 14) 0.310 192 450 568 192 × 2 = 0 + 0.620 384 901 136 384;
  • 15) 0.620 384 901 136 384 × 2 = 1 + 0.240 769 802 272 768;
  • 16) 0.240 769 802 272 768 × 2 = 0 + 0.481 539 604 545 536;
  • 17) 0.481 539 604 545 536 × 2 = 0 + 0.963 079 209 091 072;
  • 18) 0.963 079 209 091 072 × 2 = 1 + 0.926 158 418 182 144;
  • 19) 0.926 158 418 182 144 × 2 = 1 + 0.852 316 836 364 288;
  • 20) 0.852 316 836 364 288 × 2 = 1 + 0.704 633 672 728 576;
  • 21) 0.704 633 672 728 576 × 2 = 1 + 0.409 267 345 457 152;
  • 22) 0.409 267 345 457 152 × 2 = 0 + 0.818 534 690 914 304;
  • 23) 0.818 534 690 914 304 × 2 = 1 + 0.637 069 381 828 608;
  • 24) 0.637 069 381 828 608 × 2 = 1 + 0.274 138 763 657 216;
  • 25) 0.274 138 763 657 216 × 2 = 0 + 0.548 277 527 314 432;
  • 26) 0.548 277 527 314 432 × 2 = 1 + 0.096 555 054 628 864;
  • 27) 0.096 555 054 628 864 × 2 = 0 + 0.193 110 109 257 728;
  • 28) 0.193 110 109 257 728 × 2 = 0 + 0.386 220 218 515 456;
  • 29) 0.386 220 218 515 456 × 2 = 0 + 0.772 440 437 030 912;
  • 30) 0.772 440 437 030 912 × 2 = 1 + 0.544 880 874 061 824;
  • 31) 0.544 880 874 061 824 × 2 = 1 + 0.089 761 748 123 648;
  • 32) 0.089 761 748 123 648 × 2 = 0 + 0.179 523 496 247 296;
  • 33) 0.179 523 496 247 296 × 2 = 0 + 0.359 046 992 494 592;
  • 34) 0.359 046 992 494 592 × 2 = 0 + 0.718 093 984 989 184;
  • 35) 0.718 093 984 989 184 × 2 = 1 + 0.436 187 969 978 368;
  • 36) 0.436 187 969 978 368 × 2 = 0 + 0.872 375 939 956 736;
  • 37) 0.872 375 939 956 736 × 2 = 1 + 0.744 751 879 913 472;
  • 38) 0.744 751 879 913 472 × 2 = 1 + 0.489 503 759 826 944;
  • 39) 0.489 503 759 826 944 × 2 = 0 + 0.979 007 519 653 888;
  • 40) 0.979 007 519 653 888 × 2 = 1 + 0.958 015 039 307 776;
  • 41) 0.958 015 039 307 776 × 2 = 1 + 0.916 030 078 615 552;
  • 42) 0.916 030 078 615 552 × 2 = 1 + 0.832 060 157 231 104;
  • 43) 0.832 060 157 231 104 × 2 = 1 + 0.664 120 314 462 208;
  • 44) 0.664 120 314 462 208 × 2 = 1 + 0.328 240 628 924 416;
  • 45) 0.328 240 628 924 416 × 2 = 0 + 0.656 481 257 848 832;
  • 46) 0.656 481 257 848 832 × 2 = 1 + 0.312 962 515 697 664;
  • 47) 0.312 962 515 697 664 × 2 = 0 + 0.625 925 031 395 328;
  • 48) 0.625 925 031 395 328 × 2 = 1 + 0.251 850 062 790 656;
  • 49) 0.251 850 062 790 656 × 2 = 0 + 0.503 700 125 581 312;
  • 50) 0.503 700 125 581 312 × 2 = 1 + 0.007 400 251 162 624;
  • 51) 0.007 400 251 162 624 × 2 = 0 + 0.014 800 502 325 248;
  • 52) 0.014 800 502 325 248 × 2 = 0 + 0.029 601 004 650 496;
  • 53) 0.029 601 004 650 496 × 2 = 0 + 0.059 202 009 300 992;
  • 54) 0.059 202 009 300 992 × 2 = 0 + 0.118 404 018 601 984;
  • 55) 0.118 404 018 601 984 × 2 = 0 + 0.236 808 037 203 968;
  • 56) 0.236 808 037 203 968 × 2 = 0 + 0.473 616 074 407 936;
  • 57) 0.473 616 074 407 936 × 2 = 0 + 0.947 232 148 815 872;
  • 58) 0.947 232 148 815 872 × 2 = 1 + 0.894 464 297 631 744;
  • 59) 0.894 464 297 631 744 × 2 = 1 + 0.788 928 595 263 488;
  • 60) 0.788 928 595 263 488 × 2 = 1 + 0.577 857 190 526 976;
  • 61) 0.577 857 190 526 976 × 2 = 1 + 0.155 714 381 053 952;
  • 62) 0.155 714 381 053 952 × 2 = 0 + 0.311 428 762 107 904;
  • 63) 0.311 428 762 107 904 × 2 = 0 + 0.622 857 524 215 808;
  • 64) 0.622 857 524 215 808 × 2 = 1 + 0.245 715 048 431 616;

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:


0.000 282 005 914 376(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0101 0100 0000 0111 1001(2)

6. Positive number before normalization:

0.000 282 005 914 376(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0101 0100 0000 0111 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:


0.000 282 005 914 376(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0101 0100 0000 0111 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 0101 0100 0000 0111 1001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 0101 0100 0000 0111 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 0010 1101 1111 0101 0100 0000 0111 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 0010 1101 1111 0101 0100 0000 0111 1001 =

0010 0111 1011 0100 0110 0010 1101 1111 0101 0100 0000 0111 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 0010 1101 1111 0101 0100 0000 0111 1001


Decimal number -0.000 282 005 914 376 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 0101 0100 0000 0111 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