-0.000 282 005 608 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 608(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
-0.000 282 005 608(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:

|-0.000 282 005 608| = 0.000 282 005 608


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: 0.000 282 005 608.

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) 0.000 282 005 608 × 2 = 0 + 0.000 564 011 216;
  • 2) 0.000 564 011 216 × 2 = 0 + 0.001 128 022 432;
  • 3) 0.001 128 022 432 × 2 = 0 + 0.002 256 044 864;
  • 4) 0.002 256 044 864 × 2 = 0 + 0.004 512 089 728;
  • 5) 0.004 512 089 728 × 2 = 0 + 0.009 024 179 456;
  • 6) 0.009 024 179 456 × 2 = 0 + 0.018 048 358 912;
  • 7) 0.018 048 358 912 × 2 = 0 + 0.036 096 717 824;
  • 8) 0.036 096 717 824 × 2 = 0 + 0.072 193 435 648;
  • 9) 0.072 193 435 648 × 2 = 0 + 0.144 386 871 296;
  • 10) 0.144 386 871 296 × 2 = 0 + 0.288 773 742 592;
  • 11) 0.288 773 742 592 × 2 = 0 + 0.577 547 485 184;
  • 12) 0.577 547 485 184 × 2 = 1 + 0.155 094 970 368;
  • 13) 0.155 094 970 368 × 2 = 0 + 0.310 189 940 736;
  • 14) 0.310 189 940 736 × 2 = 0 + 0.620 379 881 472;
  • 15) 0.620 379 881 472 × 2 = 1 + 0.240 759 762 944;
  • 16) 0.240 759 762 944 × 2 = 0 + 0.481 519 525 888;
  • 17) 0.481 519 525 888 × 2 = 0 + 0.963 039 051 776;
  • 18) 0.963 039 051 776 × 2 = 1 + 0.926 078 103 552;
  • 19) 0.926 078 103 552 × 2 = 1 + 0.852 156 207 104;
  • 20) 0.852 156 207 104 × 2 = 1 + 0.704 312 414 208;
  • 21) 0.704 312 414 208 × 2 = 1 + 0.408 624 828 416;
  • 22) 0.408 624 828 416 × 2 = 0 + 0.817 249 656 832;
  • 23) 0.817 249 656 832 × 2 = 1 + 0.634 499 313 664;
  • 24) 0.634 499 313 664 × 2 = 1 + 0.268 998 627 328;
  • 25) 0.268 998 627 328 × 2 = 0 + 0.537 997 254 656;
  • 26) 0.537 997 254 656 × 2 = 1 + 0.075 994 509 312;
  • 27) 0.075 994 509 312 × 2 = 0 + 0.151 989 018 624;
  • 28) 0.151 989 018 624 × 2 = 0 + 0.303 978 037 248;
  • 29) 0.303 978 037 248 × 2 = 0 + 0.607 956 074 496;
  • 30) 0.607 956 074 496 × 2 = 1 + 0.215 912 148 992;
  • 31) 0.215 912 148 992 × 2 = 0 + 0.431 824 297 984;
  • 32) 0.431 824 297 984 × 2 = 0 + 0.863 648 595 968;
  • 33) 0.863 648 595 968 × 2 = 1 + 0.727 297 191 936;
  • 34) 0.727 297 191 936 × 2 = 1 + 0.454 594 383 872;
  • 35) 0.454 594 383 872 × 2 = 0 + 0.909 188 767 744;
  • 36) 0.909 188 767 744 × 2 = 1 + 0.818 377 535 488;
  • 37) 0.818 377 535 488 × 2 = 1 + 0.636 755 070 976;
  • 38) 0.636 755 070 976 × 2 = 1 + 0.273 510 141 952;
  • 39) 0.273 510 141 952 × 2 = 0 + 0.547 020 283 904;
  • 40) 0.547 020 283 904 × 2 = 1 + 0.094 040 567 808;
  • 41) 0.094 040 567 808 × 2 = 0 + 0.188 081 135 616;
  • 42) 0.188 081 135 616 × 2 = 0 + 0.376 162 271 232;
  • 43) 0.376 162 271 232 × 2 = 0 + 0.752 324 542 464;
  • 44) 0.752 324 542 464 × 2 = 1 + 0.504 649 084 928;
  • 45) 0.504 649 084 928 × 2 = 1 + 0.009 298 169 856;
  • 46) 0.009 298 169 856 × 2 = 0 + 0.018 596 339 712;
  • 47) 0.018 596 339 712 × 2 = 0 + 0.037 192 679 424;
  • 48) 0.037 192 679 424 × 2 = 0 + 0.074 385 358 848;
  • 49) 0.074 385 358 848 × 2 = 0 + 0.148 770 717 696;
  • 50) 0.148 770 717 696 × 2 = 0 + 0.297 541 435 392;
  • 51) 0.297 541 435 392 × 2 = 0 + 0.595 082 870 784;
  • 52) 0.595 082 870 784 × 2 = 1 + 0.190 165 741 568;
  • 53) 0.190 165 741 568 × 2 = 0 + 0.380 331 483 136;
  • 54) 0.380 331 483 136 × 2 = 0 + 0.760 662 966 272;
  • 55) 0.760 662 966 272 × 2 = 1 + 0.521 325 932 544;
  • 56) 0.521 325 932 544 × 2 = 1 + 0.042 651 865 088;
  • 57) 0.042 651 865 088 × 2 = 0 + 0.085 303 730 176;
  • 58) 0.085 303 730 176 × 2 = 0 + 0.170 607 460 352;
  • 59) 0.170 607 460 352 × 2 = 0 + 0.341 214 920 704;
  • 60) 0.341 214 920 704 × 2 = 0 + 0.682 429 841 408;
  • 61) 0.682 429 841 408 × 2 = 1 + 0.364 859 682 816;
  • 62) 0.364 859 682 816 × 2 = 0 + 0.729 719 365 632;
  • 63) 0.729 719 365 632 × 2 = 1 + 0.459 438 731 264;
  • 64) 0.459 438 731 264 × 2 = 0 + 0.918 877 462 528;

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:


0.000 282 005 608(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1101 0001 1000 0001 0011 0000 1010(2)

6. Positive number before normalization:

0.000 282 005 608(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1101 0001 1000 0001 0011 0000 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:


0.000 282 005 608(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1101 0001 1000 0001 0011 0000 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1101 1101 0001 1000 0001 0011 0000 1010(2) × 20 =

1.0010 0111 1011 0100 0100 1101 1101 0001 1000 0001 0011 0000 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 0100 1101 1101 0001 1000 0001 0011 0000 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 0100 1101 1101 0001 1000 0001 0011 0000 1010 =

0010 0111 1011 0100 0100 1101 1101 0001 1000 0001 0011 0000 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 0100 1101 1101 0001 1000 0001 0011 0000 1010


Decimal number -0.000 282 005 608 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 1101 1101 0001 1000 0001 0011 0000 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