-0.000 007 820 538 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 820 538(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
-0.000 007 820 538(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:

|-0.000 007 820 538| = 0.000 007 820 538


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: 0.000 007 820 538.

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) 0.000 007 820 538 × 2 = 0 + 0.000 015 641 076;
  • 2) 0.000 015 641 076 × 2 = 0 + 0.000 031 282 152;
  • 3) 0.000 031 282 152 × 2 = 0 + 0.000 062 564 304;
  • 4) 0.000 062 564 304 × 2 = 0 + 0.000 125 128 608;
  • 5) 0.000 125 128 608 × 2 = 0 + 0.000 250 257 216;
  • 6) 0.000 250 257 216 × 2 = 0 + 0.000 500 514 432;
  • 7) 0.000 500 514 432 × 2 = 0 + 0.001 001 028 864;
  • 8) 0.001 001 028 864 × 2 = 0 + 0.002 002 057 728;
  • 9) 0.002 002 057 728 × 2 = 0 + 0.004 004 115 456;
  • 10) 0.004 004 115 456 × 2 = 0 + 0.008 008 230 912;
  • 11) 0.008 008 230 912 × 2 = 0 + 0.016 016 461 824;
  • 12) 0.016 016 461 824 × 2 = 0 + 0.032 032 923 648;
  • 13) 0.032 032 923 648 × 2 = 0 + 0.064 065 847 296;
  • 14) 0.064 065 847 296 × 2 = 0 + 0.128 131 694 592;
  • 15) 0.128 131 694 592 × 2 = 0 + 0.256 263 389 184;
  • 16) 0.256 263 389 184 × 2 = 0 + 0.512 526 778 368;
  • 17) 0.512 526 778 368 × 2 = 1 + 0.025 053 556 736;
  • 18) 0.025 053 556 736 × 2 = 0 + 0.050 107 113 472;
  • 19) 0.050 107 113 472 × 2 = 0 + 0.100 214 226 944;
  • 20) 0.100 214 226 944 × 2 = 0 + 0.200 428 453 888;
  • 21) 0.200 428 453 888 × 2 = 0 + 0.400 856 907 776;
  • 22) 0.400 856 907 776 × 2 = 0 + 0.801 713 815 552;
  • 23) 0.801 713 815 552 × 2 = 1 + 0.603 427 631 104;
  • 24) 0.603 427 631 104 × 2 = 1 + 0.206 855 262 208;
  • 25) 0.206 855 262 208 × 2 = 0 + 0.413 710 524 416;
  • 26) 0.413 710 524 416 × 2 = 0 + 0.827 421 048 832;
  • 27) 0.827 421 048 832 × 2 = 1 + 0.654 842 097 664;
  • 28) 0.654 842 097 664 × 2 = 1 + 0.309 684 195 328;
  • 29) 0.309 684 195 328 × 2 = 0 + 0.619 368 390 656;
  • 30) 0.619 368 390 656 × 2 = 1 + 0.238 736 781 312;
  • 31) 0.238 736 781 312 × 2 = 0 + 0.477 473 562 624;
  • 32) 0.477 473 562 624 × 2 = 0 + 0.954 947 125 248;
  • 33) 0.954 947 125 248 × 2 = 1 + 0.909 894 250 496;
  • 34) 0.909 894 250 496 × 2 = 1 + 0.819 788 500 992;
  • 35) 0.819 788 500 992 × 2 = 1 + 0.639 577 001 984;
  • 36) 0.639 577 001 984 × 2 = 1 + 0.279 154 003 968;
  • 37) 0.279 154 003 968 × 2 = 0 + 0.558 308 007 936;
  • 38) 0.558 308 007 936 × 2 = 1 + 0.116 616 015 872;
  • 39) 0.116 616 015 872 × 2 = 0 + 0.233 232 031 744;
  • 40) 0.233 232 031 744 × 2 = 0 + 0.466 464 063 488;
  • 41) 0.466 464 063 488 × 2 = 0 + 0.932 928 126 976;
  • 42) 0.932 928 126 976 × 2 = 1 + 0.865 856 253 952;
  • 43) 0.865 856 253 952 × 2 = 1 + 0.731 712 507 904;
  • 44) 0.731 712 507 904 × 2 = 1 + 0.463 425 015 808;
  • 45) 0.463 425 015 808 × 2 = 0 + 0.926 850 031 616;
  • 46) 0.926 850 031 616 × 2 = 1 + 0.853 700 063 232;
  • 47) 0.853 700 063 232 × 2 = 1 + 0.707 400 126 464;
  • 48) 0.707 400 126 464 × 2 = 1 + 0.414 800 252 928;
  • 49) 0.414 800 252 928 × 2 = 0 + 0.829 600 505 856;
  • 50) 0.829 600 505 856 × 2 = 1 + 0.659 201 011 712;
  • 51) 0.659 201 011 712 × 2 = 1 + 0.318 402 023 424;
  • 52) 0.318 402 023 424 × 2 = 0 + 0.636 804 046 848;
  • 53) 0.636 804 046 848 × 2 = 1 + 0.273 608 093 696;
  • 54) 0.273 608 093 696 × 2 = 0 + 0.547 216 187 392;
  • 55) 0.547 216 187 392 × 2 = 1 + 0.094 432 374 784;
  • 56) 0.094 432 374 784 × 2 = 0 + 0.188 864 749 568;
  • 57) 0.188 864 749 568 × 2 = 0 + 0.377 729 499 136;
  • 58) 0.377 729 499 136 × 2 = 0 + 0.755 458 998 272;
  • 59) 0.755 458 998 272 × 2 = 1 + 0.510 917 996 544;
  • 60) 0.510 917 996 544 × 2 = 1 + 0.021 835 993 088;
  • 61) 0.021 835 993 088 × 2 = 0 + 0.043 671 986 176;
  • 62) 0.043 671 986 176 × 2 = 0 + 0.087 343 972 352;
  • 63) 0.087 343 972 352 × 2 = 0 + 0.174 687 944 704;
  • 64) 0.174 687 944 704 × 2 = 0 + 0.349 375 889 408;
  • 65) 0.349 375 889 408 × 2 = 0 + 0.698 751 778 816;
  • 66) 0.698 751 778 816 × 2 = 1 + 0.397 503 557 632;
  • 67) 0.397 503 557 632 × 2 = 0 + 0.795 007 115 264;
  • 68) 0.795 007 115 264 × 2 = 1 + 0.590 014 230 528;
  • 69) 0.590 014 230 528 × 2 = 1 + 0.180 028 461 056;

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:


0.000 007 820 538(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1111 0100 0111 0111 0110 1010 0011 0000 0101 1(2)

6. Positive number before normalization:

0.000 007 820 538(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1111 0100 0111 0111 0110 1010 0011 0000 0101 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 820 538(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1111 0100 0111 0111 0110 1010 0011 0000 0101 1(2) =

0.0000 0000 0000 0000 1000 0011 0011 0100 1111 0100 0111 0111 0110 1010 0011 0000 0101 1(2) × 20 =

1.0000 0110 0110 1001 1110 1000 1110 1110 1101 0100 0110 0000 1011(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0110 0110 1001 1110 1000 1110 1110 1101 0100 0110 0000 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1006(10) =

011 1110 1110(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. 0000 0110 0110 1001 1110 1000 1110 1110 1101 0100 0110 0000 1011 =

0000 0110 0110 1001 1110 1000 1110 1110 1101 0100 0110 0000 1011


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 1110 1110

Mantissa (52 bits) =
0000 0110 0110 1001 1110 1000 1110 1110 1101 0100 0110 0000 1011


Decimal number -0.000 007 820 538 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 0110 1001 1110 1000 1110 1110 1101 0100 0110 0000 1011

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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