-0.105 000 000 007 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.105 000 000 007 7(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
-0.105 000 000 007 7(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:

|-0.105 000 000 007 7| = 0.105 000 000 007 7


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: 0.105 000 000 007 7.

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) 0.105 000 000 007 7 × 2 = 0 + 0.210 000 000 015 4;
  • 2) 0.210 000 000 015 4 × 2 = 0 + 0.420 000 000 030 8;
  • 3) 0.420 000 000 030 8 × 2 = 0 + 0.840 000 000 061 6;
  • 4) 0.840 000 000 061 6 × 2 = 1 + 0.680 000 000 123 2;
  • 5) 0.680 000 000 123 2 × 2 = 1 + 0.360 000 000 246 4;
  • 6) 0.360 000 000 246 4 × 2 = 0 + 0.720 000 000 492 8;
  • 7) 0.720 000 000 492 8 × 2 = 1 + 0.440 000 000 985 6;
  • 8) 0.440 000 000 985 6 × 2 = 0 + 0.880 000 001 971 2;
  • 9) 0.880 000 001 971 2 × 2 = 1 + 0.760 000 003 942 4;
  • 10) 0.760 000 003 942 4 × 2 = 1 + 0.520 000 007 884 8;
  • 11) 0.520 000 007 884 8 × 2 = 1 + 0.040 000 015 769 6;
  • 12) 0.040 000 015 769 6 × 2 = 0 + 0.080 000 031 539 2;
  • 13) 0.080 000 031 539 2 × 2 = 0 + 0.160 000 063 078 4;
  • 14) 0.160 000 063 078 4 × 2 = 0 + 0.320 000 126 156 8;
  • 15) 0.320 000 126 156 8 × 2 = 0 + 0.640 000 252 313 6;
  • 16) 0.640 000 252 313 6 × 2 = 1 + 0.280 000 504 627 2;
  • 17) 0.280 000 504 627 2 × 2 = 0 + 0.560 001 009 254 4;
  • 18) 0.560 001 009 254 4 × 2 = 1 + 0.120 002 018 508 8;
  • 19) 0.120 002 018 508 8 × 2 = 0 + 0.240 004 037 017 6;
  • 20) 0.240 004 037 017 6 × 2 = 0 + 0.480 008 074 035 2;
  • 21) 0.480 008 074 035 2 × 2 = 0 + 0.960 016 148 070 4;
  • 22) 0.960 016 148 070 4 × 2 = 1 + 0.920 032 296 140 8;
  • 23) 0.920 032 296 140 8 × 2 = 1 + 0.840 064 592 281 6;
  • 24) 0.840 064 592 281 6 × 2 = 1 + 0.680 129 184 563 2;
  • 25) 0.680 129 184 563 2 × 2 = 1 + 0.360 258 369 126 4;
  • 26) 0.360 258 369 126 4 × 2 = 0 + 0.720 516 738 252 8;
  • 27) 0.720 516 738 252 8 × 2 = 1 + 0.441 033 476 505 6;
  • 28) 0.441 033 476 505 6 × 2 = 0 + 0.882 066 953 011 2;
  • 29) 0.882 066 953 011 2 × 2 = 1 + 0.764 133 906 022 4;
  • 30) 0.764 133 906 022 4 × 2 = 1 + 0.528 267 812 044 8;
  • 31) 0.528 267 812 044 8 × 2 = 1 + 0.056 535 624 089 6;
  • 32) 0.056 535 624 089 6 × 2 = 0 + 0.113 071 248 179 2;
  • 33) 0.113 071 248 179 2 × 2 = 0 + 0.226 142 496 358 4;
  • 34) 0.226 142 496 358 4 × 2 = 0 + 0.452 284 992 716 8;
  • 35) 0.452 284 992 716 8 × 2 = 0 + 0.904 569 985 433 6;
  • 36) 0.904 569 985 433 6 × 2 = 1 + 0.809 139 970 867 2;
  • 37) 0.809 139 970 867 2 × 2 = 1 + 0.618 279 941 734 4;
  • 38) 0.618 279 941 734 4 × 2 = 1 + 0.236 559 883 468 8;
  • 39) 0.236 559 883 468 8 × 2 = 0 + 0.473 119 766 937 6;
  • 40) 0.473 119 766 937 6 × 2 = 0 + 0.946 239 533 875 2;
  • 41) 0.946 239 533 875 2 × 2 = 1 + 0.892 479 067 750 4;
  • 42) 0.892 479 067 750 4 × 2 = 1 + 0.784 958 135 500 8;
  • 43) 0.784 958 135 500 8 × 2 = 1 + 0.569 916 271 001 6;
  • 44) 0.569 916 271 001 6 × 2 = 1 + 0.139 832 542 003 2;
  • 45) 0.139 832 542 003 2 × 2 = 0 + 0.279 665 084 006 4;
  • 46) 0.279 665 084 006 4 × 2 = 0 + 0.559 330 168 012 8;
  • 47) 0.559 330 168 012 8 × 2 = 1 + 0.118 660 336 025 6;
  • 48) 0.118 660 336 025 6 × 2 = 0 + 0.237 320 672 051 2;
  • 49) 0.237 320 672 051 2 × 2 = 0 + 0.474 641 344 102 4;
  • 50) 0.474 641 344 102 4 × 2 = 0 + 0.949 282 688 204 8;
  • 51) 0.949 282 688 204 8 × 2 = 1 + 0.898 565 376 409 6;
  • 52) 0.898 565 376 409 6 × 2 = 1 + 0.797 130 752 819 2;
  • 53) 0.797 130 752 819 2 × 2 = 1 + 0.594 261 505 638 4;
  • 54) 0.594 261 505 638 4 × 2 = 1 + 0.188 523 011 276 8;
  • 55) 0.188 523 011 276 8 × 2 = 0 + 0.377 046 022 553 6;
  • 56) 0.377 046 022 553 6 × 2 = 0 + 0.754 092 045 107 2;

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:


0.105 000 000 007 7(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100(2)

6. Positive number before normalization:

0.105 000 000 007 7(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.105 000 000 007 7(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100(2) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100(2) × 20 =

1.1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100(2) × 2-4


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):
1.1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100 =

1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 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 1011

Mantissa (52 bits) =
1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 1100


Decimal number -0.105 000 000 007 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 0111 1010 1110 0001 1100 1111 0010 0011 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