-0.000 282 005 914 459 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 459(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
-0.000 282 005 914 459(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:

|-0.000 282 005 914 459| = 0.000 282 005 914 459


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: 0.000 282 005 914 459.

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) 0.000 282 005 914 459 × 2 = 0 + 0.000 564 011 828 918;
  • 2) 0.000 564 011 828 918 × 2 = 0 + 0.001 128 023 657 836;
  • 3) 0.001 128 023 657 836 × 2 = 0 + 0.002 256 047 315 672;
  • 4) 0.002 256 047 315 672 × 2 = 0 + 0.004 512 094 631 344;
  • 5) 0.004 512 094 631 344 × 2 = 0 + 0.009 024 189 262 688;
  • 6) 0.009 024 189 262 688 × 2 = 0 + 0.018 048 378 525 376;
  • 7) 0.018 048 378 525 376 × 2 = 0 + 0.036 096 757 050 752;
  • 8) 0.036 096 757 050 752 × 2 = 0 + 0.072 193 514 101 504;
  • 9) 0.072 193 514 101 504 × 2 = 0 + 0.144 387 028 203 008;
  • 10) 0.144 387 028 203 008 × 2 = 0 + 0.288 774 056 406 016;
  • 11) 0.288 774 056 406 016 × 2 = 0 + 0.577 548 112 812 032;
  • 12) 0.577 548 112 812 032 × 2 = 1 + 0.155 096 225 624 064;
  • 13) 0.155 096 225 624 064 × 2 = 0 + 0.310 192 451 248 128;
  • 14) 0.310 192 451 248 128 × 2 = 0 + 0.620 384 902 496 256;
  • 15) 0.620 384 902 496 256 × 2 = 1 + 0.240 769 804 992 512;
  • 16) 0.240 769 804 992 512 × 2 = 0 + 0.481 539 609 985 024;
  • 17) 0.481 539 609 985 024 × 2 = 0 + 0.963 079 219 970 048;
  • 18) 0.963 079 219 970 048 × 2 = 1 + 0.926 158 439 940 096;
  • 19) 0.926 158 439 940 096 × 2 = 1 + 0.852 316 879 880 192;
  • 20) 0.852 316 879 880 192 × 2 = 1 + 0.704 633 759 760 384;
  • 21) 0.704 633 759 760 384 × 2 = 1 + 0.409 267 519 520 768;
  • 22) 0.409 267 519 520 768 × 2 = 0 + 0.818 535 039 041 536;
  • 23) 0.818 535 039 041 536 × 2 = 1 + 0.637 070 078 083 072;
  • 24) 0.637 070 078 083 072 × 2 = 1 + 0.274 140 156 166 144;
  • 25) 0.274 140 156 166 144 × 2 = 0 + 0.548 280 312 332 288;
  • 26) 0.548 280 312 332 288 × 2 = 1 + 0.096 560 624 664 576;
  • 27) 0.096 560 624 664 576 × 2 = 0 + 0.193 121 249 329 152;
  • 28) 0.193 121 249 329 152 × 2 = 0 + 0.386 242 498 658 304;
  • 29) 0.386 242 498 658 304 × 2 = 0 + 0.772 484 997 316 608;
  • 30) 0.772 484 997 316 608 × 2 = 1 + 0.544 969 994 633 216;
  • 31) 0.544 969 994 633 216 × 2 = 1 + 0.089 939 989 266 432;
  • 32) 0.089 939 989 266 432 × 2 = 0 + 0.179 879 978 532 864;
  • 33) 0.179 879 978 532 864 × 2 = 0 + 0.359 759 957 065 728;
  • 34) 0.359 759 957 065 728 × 2 = 0 + 0.719 519 914 131 456;
  • 35) 0.719 519 914 131 456 × 2 = 1 + 0.439 039 828 262 912;
  • 36) 0.439 039 828 262 912 × 2 = 0 + 0.878 079 656 525 824;
  • 37) 0.878 079 656 525 824 × 2 = 1 + 0.756 159 313 051 648;
  • 38) 0.756 159 313 051 648 × 2 = 1 + 0.512 318 626 103 296;
  • 39) 0.512 318 626 103 296 × 2 = 1 + 0.024 637 252 206 592;
  • 40) 0.024 637 252 206 592 × 2 = 0 + 0.049 274 504 413 184;
  • 41) 0.049 274 504 413 184 × 2 = 0 + 0.098 549 008 826 368;
  • 42) 0.098 549 008 826 368 × 2 = 0 + 0.197 098 017 652 736;
  • 43) 0.197 098 017 652 736 × 2 = 0 + 0.394 196 035 305 472;
  • 44) 0.394 196 035 305 472 × 2 = 0 + 0.788 392 070 610 944;
  • 45) 0.788 392 070 610 944 × 2 = 1 + 0.576 784 141 221 888;
  • 46) 0.576 784 141 221 888 × 2 = 1 + 0.153 568 282 443 776;
  • 47) 0.153 568 282 443 776 × 2 = 0 + 0.307 136 564 887 552;
  • 48) 0.307 136 564 887 552 × 2 = 0 + 0.614 273 129 775 104;
  • 49) 0.614 273 129 775 104 × 2 = 1 + 0.228 546 259 550 208;
  • 50) 0.228 546 259 550 208 × 2 = 0 + 0.457 092 519 100 416;
  • 51) 0.457 092 519 100 416 × 2 = 0 + 0.914 185 038 200 832;
  • 52) 0.914 185 038 200 832 × 2 = 1 + 0.828 370 076 401 664;
  • 53) 0.828 370 076 401 664 × 2 = 1 + 0.656 740 152 803 328;
  • 54) 0.656 740 152 803 328 × 2 = 1 + 0.313 480 305 606 656;
  • 55) 0.313 480 305 606 656 × 2 = 0 + 0.626 960 611 213 312;
  • 56) 0.626 960 611 213 312 × 2 = 1 + 0.253 921 222 426 624;
  • 57) 0.253 921 222 426 624 × 2 = 0 + 0.507 842 444 853 248;
  • 58) 0.507 842 444 853 248 × 2 = 1 + 0.015 684 889 706 496;
  • 59) 0.015 684 889 706 496 × 2 = 0 + 0.031 369 779 412 992;
  • 60) 0.031 369 779 412 992 × 2 = 0 + 0.062 739 558 825 984;
  • 61) 0.062 739 558 825 984 × 2 = 0 + 0.125 479 117 651 968;
  • 62) 0.125 479 117 651 968 × 2 = 0 + 0.250 958 235 303 936;
  • 63) 0.250 958 235 303 936 × 2 = 0 + 0.501 916 470 607 872;
  • 64) 0.501 916 470 607 872 × 2 = 1 + 0.003 832 941 215 744;

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:


0.000 282 005 914 459(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1100 1001 1101 0100 0001(2)

6. Positive number before normalization:

0.000 282 005 914 459(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1100 1001 1101 0100 0001(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:


0.000 282 005 914 459(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1100 1001 1101 0100 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1100 1001 1101 0100 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 1100 1001 1101 0100 0001(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 1100 1001 1101 0100 0001


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 1100 1001 1101 0100 0001 =

0010 0111 1011 0100 0110 0010 1110 0000 1100 1001 1101 0100 0001


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 1100 1001 1101 0100 0001


Decimal number -0.000 282 005 914 459 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 1100 1001 1101 0100 0001

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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