-0.000 282 005 628 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 628(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
-0.000 282 005 628(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:

|-0.000 282 005 628| = 0.000 282 005 628


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: 0.000 282 005 628.

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) 0.000 282 005 628 × 2 = 0 + 0.000 564 011 256;
  • 2) 0.000 564 011 256 × 2 = 0 + 0.001 128 022 512;
  • 3) 0.001 128 022 512 × 2 = 0 + 0.002 256 045 024;
  • 4) 0.002 256 045 024 × 2 = 0 + 0.004 512 090 048;
  • 5) 0.004 512 090 048 × 2 = 0 + 0.009 024 180 096;
  • 6) 0.009 024 180 096 × 2 = 0 + 0.018 048 360 192;
  • 7) 0.018 048 360 192 × 2 = 0 + 0.036 096 720 384;
  • 8) 0.036 096 720 384 × 2 = 0 + 0.072 193 440 768;
  • 9) 0.072 193 440 768 × 2 = 0 + 0.144 386 881 536;
  • 10) 0.144 386 881 536 × 2 = 0 + 0.288 773 763 072;
  • 11) 0.288 773 763 072 × 2 = 0 + 0.577 547 526 144;
  • 12) 0.577 547 526 144 × 2 = 1 + 0.155 095 052 288;
  • 13) 0.155 095 052 288 × 2 = 0 + 0.310 190 104 576;
  • 14) 0.310 190 104 576 × 2 = 0 + 0.620 380 209 152;
  • 15) 0.620 380 209 152 × 2 = 1 + 0.240 760 418 304;
  • 16) 0.240 760 418 304 × 2 = 0 + 0.481 520 836 608;
  • 17) 0.481 520 836 608 × 2 = 0 + 0.963 041 673 216;
  • 18) 0.963 041 673 216 × 2 = 1 + 0.926 083 346 432;
  • 19) 0.926 083 346 432 × 2 = 1 + 0.852 166 692 864;
  • 20) 0.852 166 692 864 × 2 = 1 + 0.704 333 385 728;
  • 21) 0.704 333 385 728 × 2 = 1 + 0.408 666 771 456;
  • 22) 0.408 666 771 456 × 2 = 0 + 0.817 333 542 912;
  • 23) 0.817 333 542 912 × 2 = 1 + 0.634 667 085 824;
  • 24) 0.634 667 085 824 × 2 = 1 + 0.269 334 171 648;
  • 25) 0.269 334 171 648 × 2 = 0 + 0.538 668 343 296;
  • 26) 0.538 668 343 296 × 2 = 1 + 0.077 336 686 592;
  • 27) 0.077 336 686 592 × 2 = 0 + 0.154 673 373 184;
  • 28) 0.154 673 373 184 × 2 = 0 + 0.309 346 746 368;
  • 29) 0.309 346 746 368 × 2 = 0 + 0.618 693 492 736;
  • 30) 0.618 693 492 736 × 2 = 1 + 0.237 386 985 472;
  • 31) 0.237 386 985 472 × 2 = 0 + 0.474 773 970 944;
  • 32) 0.474 773 970 944 × 2 = 0 + 0.949 547 941 888;
  • 33) 0.949 547 941 888 × 2 = 1 + 0.899 095 883 776;
  • 34) 0.899 095 883 776 × 2 = 1 + 0.798 191 767 552;
  • 35) 0.798 191 767 552 × 2 = 1 + 0.596 383 535 104;
  • 36) 0.596 383 535 104 × 2 = 1 + 0.192 767 070 208;
  • 37) 0.192 767 070 208 × 2 = 0 + 0.385 534 140 416;
  • 38) 0.385 534 140 416 × 2 = 0 + 0.771 068 280 832;
  • 39) 0.771 068 280 832 × 2 = 1 + 0.542 136 561 664;
  • 40) 0.542 136 561 664 × 2 = 1 + 0.084 273 123 328;
  • 41) 0.084 273 123 328 × 2 = 0 + 0.168 546 246 656;
  • 42) 0.168 546 246 656 × 2 = 0 + 0.337 092 493 312;
  • 43) 0.337 092 493 312 × 2 = 0 + 0.674 184 986 624;
  • 44) 0.674 184 986 624 × 2 = 1 + 0.348 369 973 248;
  • 45) 0.348 369 973 248 × 2 = 0 + 0.696 739 946 496;
  • 46) 0.696 739 946 496 × 2 = 1 + 0.393 479 892 992;
  • 47) 0.393 479 892 992 × 2 = 0 + 0.786 959 785 984;
  • 48) 0.786 959 785 984 × 2 = 1 + 0.573 919 571 968;
  • 49) 0.573 919 571 968 × 2 = 1 + 0.147 839 143 936;
  • 50) 0.147 839 143 936 × 2 = 0 + 0.295 678 287 872;
  • 51) 0.295 678 287 872 × 2 = 0 + 0.591 356 575 744;
  • 52) 0.591 356 575 744 × 2 = 1 + 0.182 713 151 488;
  • 53) 0.182 713 151 488 × 2 = 0 + 0.365 426 302 976;
  • 54) 0.365 426 302 976 × 2 = 0 + 0.730 852 605 952;
  • 55) 0.730 852 605 952 × 2 = 1 + 0.461 705 211 904;
  • 56) 0.461 705 211 904 × 2 = 0 + 0.923 410 423 808;
  • 57) 0.923 410 423 808 × 2 = 1 + 0.846 820 847 616;
  • 58) 0.846 820 847 616 × 2 = 1 + 0.693 641 695 232;
  • 59) 0.693 641 695 232 × 2 = 1 + 0.387 283 390 464;
  • 60) 0.387 283 390 464 × 2 = 0 + 0.774 566 780 928;
  • 61) 0.774 566 780 928 × 2 = 1 + 0.549 133 561 856;
  • 62) 0.549 133 561 856 × 2 = 1 + 0.098 267 123 712;
  • 63) 0.098 267 123 712 × 2 = 0 + 0.196 534 247 424;
  • 64) 0.196 534 247 424 × 2 = 0 + 0.393 068 494 848;

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:


0.000 282 005 628(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1111 0011 0001 0101 1001 0010 1110 1100(2)

6. Positive number before normalization:

0.000 282 005 628(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1111 0011 0001 0101 1001 0010 1110 1100(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:


0.000 282 005 628(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1111 0011 0001 0101 1001 0010 1110 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1111 0011 0001 0101 1001 0010 1110 1100(2) × 20 =

1.0010 0111 1011 0100 0100 1111 0011 0001 0101 1001 0010 1110 1100(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 0100 1111 0011 0001 0101 1001 0010 1110 1100


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 0100 1111 0011 0001 0101 1001 0010 1110 1100 =

0010 0111 1011 0100 0100 1111 0011 0001 0101 1001 0010 1110 1100


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 0100 1111 0011 0001 0101 1001 0010 1110 1100


Decimal number -0.000 282 005 628 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 1111 0011 0001 0101 1001 0010 1110 1100

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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