-0.000 282 005 914 408 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 408(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
-0.000 282 005 914 408(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:

|-0.000 282 005 914 408| = 0.000 282 005 914 408


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: 0.000 282 005 914 408.

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) 0.000 282 005 914 408 × 2 = 0 + 0.000 564 011 828 816;
  • 2) 0.000 564 011 828 816 × 2 = 0 + 0.001 128 023 657 632;
  • 3) 0.001 128 023 657 632 × 2 = 0 + 0.002 256 047 315 264;
  • 4) 0.002 256 047 315 264 × 2 = 0 + 0.004 512 094 630 528;
  • 5) 0.004 512 094 630 528 × 2 = 0 + 0.009 024 189 261 056;
  • 6) 0.009 024 189 261 056 × 2 = 0 + 0.018 048 378 522 112;
  • 7) 0.018 048 378 522 112 × 2 = 0 + 0.036 096 757 044 224;
  • 8) 0.036 096 757 044 224 × 2 = 0 + 0.072 193 514 088 448;
  • 9) 0.072 193 514 088 448 × 2 = 0 + 0.144 387 028 176 896;
  • 10) 0.144 387 028 176 896 × 2 = 0 + 0.288 774 056 353 792;
  • 11) 0.288 774 056 353 792 × 2 = 0 + 0.577 548 112 707 584;
  • 12) 0.577 548 112 707 584 × 2 = 1 + 0.155 096 225 415 168;
  • 13) 0.155 096 225 415 168 × 2 = 0 + 0.310 192 450 830 336;
  • 14) 0.310 192 450 830 336 × 2 = 0 + 0.620 384 901 660 672;
  • 15) 0.620 384 901 660 672 × 2 = 1 + 0.240 769 803 321 344;
  • 16) 0.240 769 803 321 344 × 2 = 0 + 0.481 539 606 642 688;
  • 17) 0.481 539 606 642 688 × 2 = 0 + 0.963 079 213 285 376;
  • 18) 0.963 079 213 285 376 × 2 = 1 + 0.926 158 426 570 752;
  • 19) 0.926 158 426 570 752 × 2 = 1 + 0.852 316 853 141 504;
  • 20) 0.852 316 853 141 504 × 2 = 1 + 0.704 633 706 283 008;
  • 21) 0.704 633 706 283 008 × 2 = 1 + 0.409 267 412 566 016;
  • 22) 0.409 267 412 566 016 × 2 = 0 + 0.818 534 825 132 032;
  • 23) 0.818 534 825 132 032 × 2 = 1 + 0.637 069 650 264 064;
  • 24) 0.637 069 650 264 064 × 2 = 1 + 0.274 139 300 528 128;
  • 25) 0.274 139 300 528 128 × 2 = 0 + 0.548 278 601 056 256;
  • 26) 0.548 278 601 056 256 × 2 = 1 + 0.096 557 202 112 512;
  • 27) 0.096 557 202 112 512 × 2 = 0 + 0.193 114 404 225 024;
  • 28) 0.193 114 404 225 024 × 2 = 0 + 0.386 228 808 450 048;
  • 29) 0.386 228 808 450 048 × 2 = 0 + 0.772 457 616 900 096;
  • 30) 0.772 457 616 900 096 × 2 = 1 + 0.544 915 233 800 192;
  • 31) 0.544 915 233 800 192 × 2 = 1 + 0.089 830 467 600 384;
  • 32) 0.089 830 467 600 384 × 2 = 0 + 0.179 660 935 200 768;
  • 33) 0.179 660 935 200 768 × 2 = 0 + 0.359 321 870 401 536;
  • 34) 0.359 321 870 401 536 × 2 = 0 + 0.718 643 740 803 072;
  • 35) 0.718 643 740 803 072 × 2 = 1 + 0.437 287 481 606 144;
  • 36) 0.437 287 481 606 144 × 2 = 0 + 0.874 574 963 212 288;
  • 37) 0.874 574 963 212 288 × 2 = 1 + 0.749 149 926 424 576;
  • 38) 0.749 149 926 424 576 × 2 = 1 + 0.498 299 852 849 152;
  • 39) 0.498 299 852 849 152 × 2 = 0 + 0.996 599 705 698 304;
  • 40) 0.996 599 705 698 304 × 2 = 1 + 0.993 199 411 396 608;
  • 41) 0.993 199 411 396 608 × 2 = 1 + 0.986 398 822 793 216;
  • 42) 0.986 398 822 793 216 × 2 = 1 + 0.972 797 645 586 432;
  • 43) 0.972 797 645 586 432 × 2 = 1 + 0.945 595 291 172 864;
  • 44) 0.945 595 291 172 864 × 2 = 1 + 0.891 190 582 345 728;
  • 45) 0.891 190 582 345 728 × 2 = 1 + 0.782 381 164 691 456;
  • 46) 0.782 381 164 691 456 × 2 = 1 + 0.564 762 329 382 912;
  • 47) 0.564 762 329 382 912 × 2 = 1 + 0.129 524 658 765 824;
  • 48) 0.129 524 658 765 824 × 2 = 0 + 0.259 049 317 531 648;
  • 49) 0.259 049 317 531 648 × 2 = 0 + 0.518 098 635 063 296;
  • 50) 0.518 098 635 063 296 × 2 = 1 + 0.036 197 270 126 592;
  • 51) 0.036 197 270 126 592 × 2 = 0 + 0.072 394 540 253 184;
  • 52) 0.072 394 540 253 184 × 2 = 0 + 0.144 789 080 506 368;
  • 53) 0.144 789 080 506 368 × 2 = 0 + 0.289 578 161 012 736;
  • 54) 0.289 578 161 012 736 × 2 = 0 + 0.579 156 322 025 472;
  • 55) 0.579 156 322 025 472 × 2 = 1 + 0.158 312 644 050 944;
  • 56) 0.158 312 644 050 944 × 2 = 0 + 0.316 625 288 101 888;
  • 57) 0.316 625 288 101 888 × 2 = 0 + 0.633 250 576 203 776;
  • 58) 0.633 250 576 203 776 × 2 = 1 + 0.266 501 152 407 552;
  • 59) 0.266 501 152 407 552 × 2 = 0 + 0.533 002 304 815 104;
  • 60) 0.533 002 304 815 104 × 2 = 1 + 0.066 004 609 630 208;
  • 61) 0.066 004 609 630 208 × 2 = 0 + 0.132 009 219 260 416;
  • 62) 0.132 009 219 260 416 × 2 = 0 + 0.264 018 438 520 832;
  • 63) 0.264 018 438 520 832 × 2 = 0 + 0.528 036 877 041 664;
  • 64) 0.528 036 877 041 664 × 2 = 1 + 0.056 073 754 083 328;

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:


0.000 282 005 914 408(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1110 0100 0010 0101 0001(2)

6. Positive number before normalization:

0.000 282 005 914 408(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1110 0100 0010 0101 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:


0.000 282 005 914 408(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1110 0100 0010 0101 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1110 0100 0010 0101 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1110 0100 0010 0101 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 1101 1111 1110 0100 0010 0101 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 1101 1111 1110 0100 0010 0101 0001 =

0010 0111 1011 0100 0110 0010 1101 1111 1110 0100 0010 0101 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 1101 1111 1110 0100 0010 0101 0001


Decimal number -0.000 282 005 914 408 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1110 0100 0010 0101 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