-0.100 000 000 005 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.100 000 000 005 4(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
-0.100 000 000 005 4(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:

|-0.100 000 000 005 4| = 0.100 000 000 005 4


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: 0.100 000 000 005 4.

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) 0.100 000 000 005 4 × 2 = 0 + 0.200 000 000 010 8;
  • 2) 0.200 000 000 010 8 × 2 = 0 + 0.400 000 000 021 6;
  • 3) 0.400 000 000 021 6 × 2 = 0 + 0.800 000 000 043 2;
  • 4) 0.800 000 000 043 2 × 2 = 1 + 0.600 000 000 086 4;
  • 5) 0.600 000 000 086 4 × 2 = 1 + 0.200 000 000 172 8;
  • 6) 0.200 000 000 172 8 × 2 = 0 + 0.400 000 000 345 6;
  • 7) 0.400 000 000 345 6 × 2 = 0 + 0.800 000 000 691 2;
  • 8) 0.800 000 000 691 2 × 2 = 1 + 0.600 000 001 382 4;
  • 9) 0.600 000 001 382 4 × 2 = 1 + 0.200 000 002 764 8;
  • 10) 0.200 000 002 764 8 × 2 = 0 + 0.400 000 005 529 6;
  • 11) 0.400 000 005 529 6 × 2 = 0 + 0.800 000 011 059 2;
  • 12) 0.800 000 011 059 2 × 2 = 1 + 0.600 000 022 118 4;
  • 13) 0.600 000 022 118 4 × 2 = 1 + 0.200 000 044 236 8;
  • 14) 0.200 000 044 236 8 × 2 = 0 + 0.400 000 088 473 6;
  • 15) 0.400 000 088 473 6 × 2 = 0 + 0.800 000 176 947 2;
  • 16) 0.800 000 176 947 2 × 2 = 1 + 0.600 000 353 894 4;
  • 17) 0.600 000 353 894 4 × 2 = 1 + 0.200 000 707 788 8;
  • 18) 0.200 000 707 788 8 × 2 = 0 + 0.400 001 415 577 6;
  • 19) 0.400 001 415 577 6 × 2 = 0 + 0.800 002 831 155 2;
  • 20) 0.800 002 831 155 2 × 2 = 1 + 0.600 005 662 310 4;
  • 21) 0.600 005 662 310 4 × 2 = 1 + 0.200 011 324 620 8;
  • 22) 0.200 011 324 620 8 × 2 = 0 + 0.400 022 649 241 6;
  • 23) 0.400 022 649 241 6 × 2 = 0 + 0.800 045 298 483 2;
  • 24) 0.800 045 298 483 2 × 2 = 1 + 0.600 090 596 966 4;
  • 25) 0.600 090 596 966 4 × 2 = 1 + 0.200 181 193 932 8;
  • 26) 0.200 181 193 932 8 × 2 = 0 + 0.400 362 387 865 6;
  • 27) 0.400 362 387 865 6 × 2 = 0 + 0.800 724 775 731 2;
  • 28) 0.800 724 775 731 2 × 2 = 1 + 0.601 449 551 462 4;
  • 29) 0.601 449 551 462 4 × 2 = 1 + 0.202 899 102 924 8;
  • 30) 0.202 899 102 924 8 × 2 = 0 + 0.405 798 205 849 6;
  • 31) 0.405 798 205 849 6 × 2 = 0 + 0.811 596 411 699 2;
  • 32) 0.811 596 411 699 2 × 2 = 1 + 0.623 192 823 398 4;
  • 33) 0.623 192 823 398 4 × 2 = 1 + 0.246 385 646 796 8;
  • 34) 0.246 385 646 796 8 × 2 = 0 + 0.492 771 293 593 6;
  • 35) 0.492 771 293 593 6 × 2 = 0 + 0.985 542 587 187 2;
  • 36) 0.985 542 587 187 2 × 2 = 1 + 0.971 085 174 374 4;
  • 37) 0.971 085 174 374 4 × 2 = 1 + 0.942 170 348 748 8;
  • 38) 0.942 170 348 748 8 × 2 = 1 + 0.884 340 697 497 6;
  • 39) 0.884 340 697 497 6 × 2 = 1 + 0.768 681 394 995 2;
  • 40) 0.768 681 394 995 2 × 2 = 1 + 0.537 362 789 990 4;
  • 41) 0.537 362 789 990 4 × 2 = 1 + 0.074 725 579 980 8;
  • 42) 0.074 725 579 980 8 × 2 = 0 + 0.149 451 159 961 6;
  • 43) 0.149 451 159 961 6 × 2 = 0 + 0.298 902 319 923 2;
  • 44) 0.298 902 319 923 2 × 2 = 0 + 0.597 804 639 846 4;
  • 45) 0.597 804 639 846 4 × 2 = 1 + 0.195 609 279 692 8;
  • 46) 0.195 609 279 692 8 × 2 = 0 + 0.391 218 559 385 6;
  • 47) 0.391 218 559 385 6 × 2 = 0 + 0.782 437 118 771 2;
  • 48) 0.782 437 118 771 2 × 2 = 1 + 0.564 874 237 542 4;
  • 49) 0.564 874 237 542 4 × 2 = 1 + 0.129 748 475 084 8;
  • 50) 0.129 748 475 084 8 × 2 = 0 + 0.259 496 950 169 6;
  • 51) 0.259 496 950 169 6 × 2 = 0 + 0.518 993 900 339 2;
  • 52) 0.518 993 900 339 2 × 2 = 1 + 0.037 987 800 678 4;
  • 53) 0.037 987 800 678 4 × 2 = 0 + 0.075 975 601 356 8;
  • 54) 0.075 975 601 356 8 × 2 = 0 + 0.151 951 202 713 6;
  • 55) 0.151 951 202 713 6 × 2 = 0 + 0.303 902 405 427 2;
  • 56) 0.303 902 405 427 2 × 2 = 0 + 0.607 804 810 854 4;

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:


0.100 000 000 005 4(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000(2)

6. Positive number before normalization:

0.100 000 000 005 4(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000(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:


0.100 000 000 005 4(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000(2) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000(2) × 20 =

1.1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000(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):
1.1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000


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. 1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000 =

1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000


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) =
1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000


Decimal number -0.100 000 000 005 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1001 1001 1001 1001 1001 1001 1001 1001 1111 1000 1001 1001 0000

» Convert decimal -0.100 000 000 014 1 to 64 bit double precision IEEE 754 binary floating point representation standard

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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