-0.000 282 005 978 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 978(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
-0.000 282 005 978(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:

|-0.000 282 005 978| = 0.000 282 005 978


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: 0.000 282 005 978.

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) 0.000 282 005 978 × 2 = 0 + 0.000 564 011 956;
  • 2) 0.000 564 011 956 × 2 = 0 + 0.001 128 023 912;
  • 3) 0.001 128 023 912 × 2 = 0 + 0.002 256 047 824;
  • 4) 0.002 256 047 824 × 2 = 0 + 0.004 512 095 648;
  • 5) 0.004 512 095 648 × 2 = 0 + 0.009 024 191 296;
  • 6) 0.009 024 191 296 × 2 = 0 + 0.018 048 382 592;
  • 7) 0.018 048 382 592 × 2 = 0 + 0.036 096 765 184;
  • 8) 0.036 096 765 184 × 2 = 0 + 0.072 193 530 368;
  • 9) 0.072 193 530 368 × 2 = 0 + 0.144 387 060 736;
  • 10) 0.144 387 060 736 × 2 = 0 + 0.288 774 121 472;
  • 11) 0.288 774 121 472 × 2 = 0 + 0.577 548 242 944;
  • 12) 0.577 548 242 944 × 2 = 1 + 0.155 096 485 888;
  • 13) 0.155 096 485 888 × 2 = 0 + 0.310 192 971 776;
  • 14) 0.310 192 971 776 × 2 = 0 + 0.620 385 943 552;
  • 15) 0.620 385 943 552 × 2 = 1 + 0.240 771 887 104;
  • 16) 0.240 771 887 104 × 2 = 0 + 0.481 543 774 208;
  • 17) 0.481 543 774 208 × 2 = 0 + 0.963 087 548 416;
  • 18) 0.963 087 548 416 × 2 = 1 + 0.926 175 096 832;
  • 19) 0.926 175 096 832 × 2 = 1 + 0.852 350 193 664;
  • 20) 0.852 350 193 664 × 2 = 1 + 0.704 700 387 328;
  • 21) 0.704 700 387 328 × 2 = 1 + 0.409 400 774 656;
  • 22) 0.409 400 774 656 × 2 = 0 + 0.818 801 549 312;
  • 23) 0.818 801 549 312 × 2 = 1 + 0.637 603 098 624;
  • 24) 0.637 603 098 624 × 2 = 1 + 0.275 206 197 248;
  • 25) 0.275 206 197 248 × 2 = 0 + 0.550 412 394 496;
  • 26) 0.550 412 394 496 × 2 = 1 + 0.100 824 788 992;
  • 27) 0.100 824 788 992 × 2 = 0 + 0.201 649 577 984;
  • 28) 0.201 649 577 984 × 2 = 0 + 0.403 299 155 968;
  • 29) 0.403 299 155 968 × 2 = 0 + 0.806 598 311 936;
  • 30) 0.806 598 311 936 × 2 = 1 + 0.613 196 623 872;
  • 31) 0.613 196 623 872 × 2 = 1 + 0.226 393 247 744;
  • 32) 0.226 393 247 744 × 2 = 0 + 0.452 786 495 488;
  • 33) 0.452 786 495 488 × 2 = 0 + 0.905 572 990 976;
  • 34) 0.905 572 990 976 × 2 = 1 + 0.811 145 981 952;
  • 35) 0.811 145 981 952 × 2 = 1 + 0.622 291 963 904;
  • 36) 0.622 291 963 904 × 2 = 1 + 0.244 583 927 808;
  • 37) 0.244 583 927 808 × 2 = 0 + 0.489 167 855 616;
  • 38) 0.489 167 855 616 × 2 = 0 + 0.978 335 711 232;
  • 39) 0.978 335 711 232 × 2 = 1 + 0.956 671 422 464;
  • 40) 0.956 671 422 464 × 2 = 1 + 0.913 342 844 928;
  • 41) 0.913 342 844 928 × 2 = 1 + 0.826 685 689 856;
  • 42) 0.826 685 689 856 × 2 = 1 + 0.653 371 379 712;
  • 43) 0.653 371 379 712 × 2 = 1 + 0.306 742 759 424;
  • 44) 0.306 742 759 424 × 2 = 0 + 0.613 485 518 848;
  • 45) 0.613 485 518 848 × 2 = 1 + 0.226 971 037 696;
  • 46) 0.226 971 037 696 × 2 = 0 + 0.453 942 075 392;
  • 47) 0.453 942 075 392 × 2 = 0 + 0.907 884 150 784;
  • 48) 0.907 884 150 784 × 2 = 1 + 0.815 768 301 568;
  • 49) 0.815 768 301 568 × 2 = 1 + 0.631 536 603 136;
  • 50) 0.631 536 603 136 × 2 = 1 + 0.263 073 206 272;
  • 51) 0.263 073 206 272 × 2 = 0 + 0.526 146 412 544;
  • 52) 0.526 146 412 544 × 2 = 1 + 0.052 292 825 088;
  • 53) 0.052 292 825 088 × 2 = 0 + 0.104 585 650 176;
  • 54) 0.104 585 650 176 × 2 = 0 + 0.209 171 300 352;
  • 55) 0.209 171 300 352 × 2 = 0 + 0.418 342 600 704;
  • 56) 0.418 342 600 704 × 2 = 0 + 0.836 685 201 408;
  • 57) 0.836 685 201 408 × 2 = 1 + 0.673 370 402 816;
  • 58) 0.673 370 402 816 × 2 = 1 + 0.346 740 805 632;
  • 59) 0.346 740 805 632 × 2 = 0 + 0.693 481 611 264;
  • 60) 0.693 481 611 264 × 2 = 1 + 0.386 963 222 528;
  • 61) 0.386 963 222 528 × 2 = 0 + 0.773 926 445 056;
  • 62) 0.773 926 445 056 × 2 = 1 + 0.547 852 890 112;
  • 63) 0.547 852 890 112 × 2 = 1 + 0.095 705 780 224;
  • 64) 0.095 705 780 224 × 2 = 0 + 0.191 411 560 448;

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:


0.000 282 005 978(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 0011 1110 1001 1101 0000 1101 0110(2)

6. Positive number before normalization:

0.000 282 005 978(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 0011 1110 1001 1101 0000 1101 0110(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:


0.000 282 005 978(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 0011 1110 1001 1101 0000 1101 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0111 0011 1110 1001 1101 0000 1101 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0111 0011 1110 1001 1101 0000 1101 0110(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 0111 0011 1110 1001 1101 0000 1101 0110


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 0111 0011 1110 1001 1101 0000 1101 0110 =

0010 0111 1011 0100 0110 0111 0011 1110 1001 1101 0000 1101 0110


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 0111 0011 1110 1001 1101 0000 1101 0110


Decimal number -0.000 282 005 978 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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