-0.000 000 101 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 101 9(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
-0.000 000 101 9(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:

|-0.000 000 101 9| = 0.000 000 101 9


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: 0.000 000 101 9.

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) 0.000 000 101 9 × 2 = 0 + 0.000 000 203 8;
  • 2) 0.000 000 203 8 × 2 = 0 + 0.000 000 407 6;
  • 3) 0.000 000 407 6 × 2 = 0 + 0.000 000 815 2;
  • 4) 0.000 000 815 2 × 2 = 0 + 0.000 001 630 4;
  • 5) 0.000 001 630 4 × 2 = 0 + 0.000 003 260 8;
  • 6) 0.000 003 260 8 × 2 = 0 + 0.000 006 521 6;
  • 7) 0.000 006 521 6 × 2 = 0 + 0.000 013 043 2;
  • 8) 0.000 013 043 2 × 2 = 0 + 0.000 026 086 4;
  • 9) 0.000 026 086 4 × 2 = 0 + 0.000 052 172 8;
  • 10) 0.000 052 172 8 × 2 = 0 + 0.000 104 345 6;
  • 11) 0.000 104 345 6 × 2 = 0 + 0.000 208 691 2;
  • 12) 0.000 208 691 2 × 2 = 0 + 0.000 417 382 4;
  • 13) 0.000 417 382 4 × 2 = 0 + 0.000 834 764 8;
  • 14) 0.000 834 764 8 × 2 = 0 + 0.001 669 529 6;
  • 15) 0.001 669 529 6 × 2 = 0 + 0.003 339 059 2;
  • 16) 0.003 339 059 2 × 2 = 0 + 0.006 678 118 4;
  • 17) 0.006 678 118 4 × 2 = 0 + 0.013 356 236 8;
  • 18) 0.013 356 236 8 × 2 = 0 + 0.026 712 473 6;
  • 19) 0.026 712 473 6 × 2 = 0 + 0.053 424 947 2;
  • 20) 0.053 424 947 2 × 2 = 0 + 0.106 849 894 4;
  • 21) 0.106 849 894 4 × 2 = 0 + 0.213 699 788 8;
  • 22) 0.213 699 788 8 × 2 = 0 + 0.427 399 577 6;
  • 23) 0.427 399 577 6 × 2 = 0 + 0.854 799 155 2;
  • 24) 0.854 799 155 2 × 2 = 1 + 0.709 598 310 4;
  • 25) 0.709 598 310 4 × 2 = 1 + 0.419 196 620 8;
  • 26) 0.419 196 620 8 × 2 = 0 + 0.838 393 241 6;
  • 27) 0.838 393 241 6 × 2 = 1 + 0.676 786 483 2;
  • 28) 0.676 786 483 2 × 2 = 1 + 0.353 572 966 4;
  • 29) 0.353 572 966 4 × 2 = 0 + 0.707 145 932 8;
  • 30) 0.707 145 932 8 × 2 = 1 + 0.414 291 865 6;
  • 31) 0.414 291 865 6 × 2 = 0 + 0.828 583 731 2;
  • 32) 0.828 583 731 2 × 2 = 1 + 0.657 167 462 4;
  • 33) 0.657 167 462 4 × 2 = 1 + 0.314 334 924 8;
  • 34) 0.314 334 924 8 × 2 = 0 + 0.628 669 849 6;
  • 35) 0.628 669 849 6 × 2 = 1 + 0.257 339 699 2;
  • 36) 0.257 339 699 2 × 2 = 0 + 0.514 679 398 4;
  • 37) 0.514 679 398 4 × 2 = 1 + 0.029 358 796 8;
  • 38) 0.029 358 796 8 × 2 = 0 + 0.058 717 593 6;
  • 39) 0.058 717 593 6 × 2 = 0 + 0.117 435 187 2;
  • 40) 0.117 435 187 2 × 2 = 0 + 0.234 870 374 4;
  • 41) 0.234 870 374 4 × 2 = 0 + 0.469 740 748 8;
  • 42) 0.469 740 748 8 × 2 = 0 + 0.939 481 497 6;
  • 43) 0.939 481 497 6 × 2 = 1 + 0.878 962 995 2;
  • 44) 0.878 962 995 2 × 2 = 1 + 0.757 925 990 4;
  • 45) 0.757 925 990 4 × 2 = 1 + 0.515 851 980 8;
  • 46) 0.515 851 980 8 × 2 = 1 + 0.031 703 961 6;
  • 47) 0.031 703 961 6 × 2 = 0 + 0.063 407 923 2;
  • 48) 0.063 407 923 2 × 2 = 0 + 0.126 815 846 4;
  • 49) 0.126 815 846 4 × 2 = 0 + 0.253 631 692 8;
  • 50) 0.253 631 692 8 × 2 = 0 + 0.507 263 385 6;
  • 51) 0.507 263 385 6 × 2 = 1 + 0.014 526 771 2;
  • 52) 0.014 526 771 2 × 2 = 0 + 0.029 053 542 4;
  • 53) 0.029 053 542 4 × 2 = 0 + 0.058 107 084 8;
  • 54) 0.058 107 084 8 × 2 = 0 + 0.116 214 169 6;
  • 55) 0.116 214 169 6 × 2 = 0 + 0.232 428 339 2;
  • 56) 0.232 428 339 2 × 2 = 0 + 0.464 856 678 4;
  • 57) 0.464 856 678 4 × 2 = 0 + 0.929 713 356 8;
  • 58) 0.929 713 356 8 × 2 = 1 + 0.859 426 713 6;
  • 59) 0.859 426 713 6 × 2 = 1 + 0.718 853 427 2;
  • 60) 0.718 853 427 2 × 2 = 1 + 0.437 706 854 4;
  • 61) 0.437 706 854 4 × 2 = 0 + 0.875 413 708 8;
  • 62) 0.875 413 708 8 × 2 = 1 + 0.750 827 417 6;
  • 63) 0.750 827 417 6 × 2 = 1 + 0.501 654 835 2;
  • 64) 0.501 654 835 2 × 2 = 1 + 0.003 309 670 4;
  • 65) 0.003 309 670 4 × 2 = 0 + 0.006 619 340 8;
  • 66) 0.006 619 340 8 × 2 = 0 + 0.013 238 681 6;
  • 67) 0.013 238 681 6 × 2 = 0 + 0.026 477 363 2;
  • 68) 0.026 477 363 2 × 2 = 0 + 0.052 954 726 4;
  • 69) 0.052 954 726 4 × 2 = 0 + 0.105 909 452 8;
  • 70) 0.105 909 452 8 × 2 = 0 + 0.211 818 905 6;
  • 71) 0.211 818 905 6 × 2 = 0 + 0.423 637 811 2;
  • 72) 0.423 637 811 2 × 2 = 0 + 0.847 275 622 4;
  • 73) 0.847 275 622 4 × 2 = 1 + 0.694 551 244 8;
  • 74) 0.694 551 244 8 × 2 = 1 + 0.389 102 489 6;
  • 75) 0.389 102 489 6 × 2 = 0 + 0.778 204 979 2;
  • 76) 0.778 204 979 2 × 2 = 1 + 0.556 409 958 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:


0.000 000 101 9(10) =

0.0000 0000 0000 0000 0000 0001 1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101(2)

6. Positive number before normalization:

0.000 000 101 9(10) =

0.0000 0000 0000 0000 0000 0001 1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 101 9(10) =

0.0000 0000 0000 0000 0000 0001 1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101(2) =

0.0000 0000 0000 0000 0000 0001 1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101(2) × 20 =

1.1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101(2) × 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): -24

Mantissa (not normalized):
1.1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(11-1) - 1 =

(-24 + 1 023)(10) =

999(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;
  • 999 ÷ 2 = 499 + 1;
  • 499 ÷ 2 = 249 + 1;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 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) =

999(10) =

011 1110 0111(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. 1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101 =

1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101


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 0111

Mantissa (52 bits) =
1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101


Decimal number -0.000 000 101 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0111 - 1011 0101 1010 1000 0011 1100 0010 0000 0111 0111 0000 0000 1101

» Convert decimal -0.000 000 103 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