-0.000 282 005 923 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 923(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
-0.000 282 005 923(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:

|-0.000 282 005 923| = 0.000 282 005 923


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: 0.000 282 005 923.

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) 0.000 282 005 923 × 2 = 0 + 0.000 564 011 846;
  • 2) 0.000 564 011 846 × 2 = 0 + 0.001 128 023 692;
  • 3) 0.001 128 023 692 × 2 = 0 + 0.002 256 047 384;
  • 4) 0.002 256 047 384 × 2 = 0 + 0.004 512 094 768;
  • 5) 0.004 512 094 768 × 2 = 0 + 0.009 024 189 536;
  • 6) 0.009 024 189 536 × 2 = 0 + 0.018 048 379 072;
  • 7) 0.018 048 379 072 × 2 = 0 + 0.036 096 758 144;
  • 8) 0.036 096 758 144 × 2 = 0 + 0.072 193 516 288;
  • 9) 0.072 193 516 288 × 2 = 0 + 0.144 387 032 576;
  • 10) 0.144 387 032 576 × 2 = 0 + 0.288 774 065 152;
  • 11) 0.288 774 065 152 × 2 = 0 + 0.577 548 130 304;
  • 12) 0.577 548 130 304 × 2 = 1 + 0.155 096 260 608;
  • 13) 0.155 096 260 608 × 2 = 0 + 0.310 192 521 216;
  • 14) 0.310 192 521 216 × 2 = 0 + 0.620 385 042 432;
  • 15) 0.620 385 042 432 × 2 = 1 + 0.240 770 084 864;
  • 16) 0.240 770 084 864 × 2 = 0 + 0.481 540 169 728;
  • 17) 0.481 540 169 728 × 2 = 0 + 0.963 080 339 456;
  • 18) 0.963 080 339 456 × 2 = 1 + 0.926 160 678 912;
  • 19) 0.926 160 678 912 × 2 = 1 + 0.852 321 357 824;
  • 20) 0.852 321 357 824 × 2 = 1 + 0.704 642 715 648;
  • 21) 0.704 642 715 648 × 2 = 1 + 0.409 285 431 296;
  • 22) 0.409 285 431 296 × 2 = 0 + 0.818 570 862 592;
  • 23) 0.818 570 862 592 × 2 = 1 + 0.637 141 725 184;
  • 24) 0.637 141 725 184 × 2 = 1 + 0.274 283 450 368;
  • 25) 0.274 283 450 368 × 2 = 0 + 0.548 566 900 736;
  • 26) 0.548 566 900 736 × 2 = 1 + 0.097 133 801 472;
  • 27) 0.097 133 801 472 × 2 = 0 + 0.194 267 602 944;
  • 28) 0.194 267 602 944 × 2 = 0 + 0.388 535 205 888;
  • 29) 0.388 535 205 888 × 2 = 0 + 0.777 070 411 776;
  • 30) 0.777 070 411 776 × 2 = 1 + 0.554 140 823 552;
  • 31) 0.554 140 823 552 × 2 = 1 + 0.108 281 647 104;
  • 32) 0.108 281 647 104 × 2 = 0 + 0.216 563 294 208;
  • 33) 0.216 563 294 208 × 2 = 0 + 0.433 126 588 416;
  • 34) 0.433 126 588 416 × 2 = 0 + 0.866 253 176 832;
  • 35) 0.866 253 176 832 × 2 = 1 + 0.732 506 353 664;
  • 36) 0.732 506 353 664 × 2 = 1 + 0.465 012 707 328;
  • 37) 0.465 012 707 328 × 2 = 0 + 0.930 025 414 656;
  • 38) 0.930 025 414 656 × 2 = 1 + 0.860 050 829 312;
  • 39) 0.860 050 829 312 × 2 = 1 + 0.720 101 658 624;
  • 40) 0.720 101 658 624 × 2 = 1 + 0.440 203 317 248;
  • 41) 0.440 203 317 248 × 2 = 0 + 0.880 406 634 496;
  • 42) 0.880 406 634 496 × 2 = 1 + 0.760 813 268 992;
  • 43) 0.760 813 268 992 × 2 = 1 + 0.521 626 537 984;
  • 44) 0.521 626 537 984 × 2 = 1 + 0.043 253 075 968;
  • 45) 0.043 253 075 968 × 2 = 0 + 0.086 506 151 936;
  • 46) 0.086 506 151 936 × 2 = 0 + 0.173 012 303 872;
  • 47) 0.173 012 303 872 × 2 = 0 + 0.346 024 607 744;
  • 48) 0.346 024 607 744 × 2 = 0 + 0.692 049 215 488;
  • 49) 0.692 049 215 488 × 2 = 1 + 0.384 098 430 976;
  • 50) 0.384 098 430 976 × 2 = 0 + 0.768 196 861 952;
  • 51) 0.768 196 861 952 × 2 = 1 + 0.536 393 723 904;
  • 52) 0.536 393 723 904 × 2 = 1 + 0.072 787 447 808;
  • 53) 0.072 787 447 808 × 2 = 0 + 0.145 574 895 616;
  • 54) 0.145 574 895 616 × 2 = 0 + 0.291 149 791 232;
  • 55) 0.291 149 791 232 × 2 = 0 + 0.582 299 582 464;
  • 56) 0.582 299 582 464 × 2 = 1 + 0.164 599 164 928;
  • 57) 0.164 599 164 928 × 2 = 0 + 0.329 198 329 856;
  • 58) 0.329 198 329 856 × 2 = 0 + 0.658 396 659 712;
  • 59) 0.658 396 659 712 × 2 = 1 + 0.316 793 319 424;
  • 60) 0.316 793 319 424 × 2 = 0 + 0.633 586 638 848;
  • 61) 0.633 586 638 848 × 2 = 1 + 0.267 173 277 696;
  • 62) 0.267 173 277 696 × 2 = 0 + 0.534 346 555 392;
  • 63) 0.534 346 555 392 × 2 = 1 + 0.068 693 110 784;
  • 64) 0.068 693 110 784 × 2 = 0 + 0.137 386 221 568;

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:


0.000 282 005 923(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0111 0111 0000 1011 0001 0010 1010(2)

6. Positive number before normalization:

0.000 282 005 923(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0111 0111 0000 1011 0001 0010 1010(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:


0.000 282 005 923(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0111 0111 0000 1011 0001 0010 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0111 0111 0000 1011 0001 0010 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0111 0111 0000 1011 0001 0010 1010(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 0011 0111 0111 0000 1011 0001 0010 1010


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 0011 0111 0111 0000 1011 0001 0010 1010 =

0010 0111 1011 0100 0110 0011 0111 0111 0000 1011 0001 0010 1010


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 0011 0111 0111 0000 1011 0001 0010 1010


Decimal number -0.000 282 005 923 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0111 0111 0000 1011 0001 0010 1010

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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