-0.750 000 128 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.750 000 128 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
-0.750 000 128 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:

|-0.750 000 128 7| = 0.750 000 128 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: 0.750 000 128 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) 0.750 000 128 7 × 2 = 1 + 0.500 000 257 4;
  • 2) 0.500 000 257 4 × 2 = 1 + 0.000 000 514 8;
  • 3) 0.000 000 514 8 × 2 = 0 + 0.000 001 029 6;
  • 4) 0.000 001 029 6 × 2 = 0 + 0.000 002 059 2;
  • 5) 0.000 002 059 2 × 2 = 0 + 0.000 004 118 4;
  • 6) 0.000 004 118 4 × 2 = 0 + 0.000 008 236 8;
  • 7) 0.000 008 236 8 × 2 = 0 + 0.000 016 473 6;
  • 8) 0.000 016 473 6 × 2 = 0 + 0.000 032 947 2;
  • 9) 0.000 032 947 2 × 2 = 0 + 0.000 065 894 4;
  • 10) 0.000 065 894 4 × 2 = 0 + 0.000 131 788 8;
  • 11) 0.000 131 788 8 × 2 = 0 + 0.000 263 577 6;
  • 12) 0.000 263 577 6 × 2 = 0 + 0.000 527 155 2;
  • 13) 0.000 527 155 2 × 2 = 0 + 0.001 054 310 4;
  • 14) 0.001 054 310 4 × 2 = 0 + 0.002 108 620 8;
  • 15) 0.002 108 620 8 × 2 = 0 + 0.004 217 241 6;
  • 16) 0.004 217 241 6 × 2 = 0 + 0.008 434 483 2;
  • 17) 0.008 434 483 2 × 2 = 0 + 0.016 868 966 4;
  • 18) 0.016 868 966 4 × 2 = 0 + 0.033 737 932 8;
  • 19) 0.033 737 932 8 × 2 = 0 + 0.067 475 865 6;
  • 20) 0.067 475 865 6 × 2 = 0 + 0.134 951 731 2;
  • 21) 0.134 951 731 2 × 2 = 0 + 0.269 903 462 4;
  • 22) 0.269 903 462 4 × 2 = 0 + 0.539 806 924 8;
  • 23) 0.539 806 924 8 × 2 = 1 + 0.079 613 849 6;
  • 24) 0.079 613 849 6 × 2 = 0 + 0.159 227 699 2;
  • 25) 0.159 227 699 2 × 2 = 0 + 0.318 455 398 4;
  • 26) 0.318 455 398 4 × 2 = 0 + 0.636 910 796 8;
  • 27) 0.636 910 796 8 × 2 = 1 + 0.273 821 593 6;
  • 28) 0.273 821 593 6 × 2 = 0 + 0.547 643 187 2;
  • 29) 0.547 643 187 2 × 2 = 1 + 0.095 286 374 4;
  • 30) 0.095 286 374 4 × 2 = 0 + 0.190 572 748 8;
  • 31) 0.190 572 748 8 × 2 = 0 + 0.381 145 497 6;
  • 32) 0.381 145 497 6 × 2 = 0 + 0.762 290 995 2;
  • 33) 0.762 290 995 2 × 2 = 1 + 0.524 581 990 4;
  • 34) 0.524 581 990 4 × 2 = 1 + 0.049 163 980 8;
  • 35) 0.049 163 980 8 × 2 = 0 + 0.098 327 961 6;
  • 36) 0.098 327 961 6 × 2 = 0 + 0.196 655 923 2;
  • 37) 0.196 655 923 2 × 2 = 0 + 0.393 311 846 4;
  • 38) 0.393 311 846 4 × 2 = 0 + 0.786 623 692 8;
  • 39) 0.786 623 692 8 × 2 = 1 + 0.573 247 385 6;
  • 40) 0.573 247 385 6 × 2 = 1 + 0.146 494 771 2;
  • 41) 0.146 494 771 2 × 2 = 0 + 0.292 989 542 4;
  • 42) 0.292 989 542 4 × 2 = 0 + 0.585 979 084 8;
  • 43) 0.585 979 084 8 × 2 = 1 + 0.171 958 169 6;
  • 44) 0.171 958 169 6 × 2 = 0 + 0.343 916 339 2;
  • 45) 0.343 916 339 2 × 2 = 0 + 0.687 832 678 4;
  • 46) 0.687 832 678 4 × 2 = 1 + 0.375 665 356 8;
  • 47) 0.375 665 356 8 × 2 = 0 + 0.751 330 713 6;
  • 48) 0.751 330 713 6 × 2 = 1 + 0.502 661 427 2;
  • 49) 0.502 661 427 2 × 2 = 1 + 0.005 322 854 4;
  • 50) 0.005 322 854 4 × 2 = 0 + 0.010 645 708 8;
  • 51) 0.010 645 708 8 × 2 = 0 + 0.021 291 417 6;
  • 52) 0.021 291 417 6 × 2 = 0 + 0.042 582 835 2;
  • 53) 0.042 582 835 2 × 2 = 0 + 0.085 165 670 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:


0.750 000 128 7(10) =

0.1100 0000 0000 0000 0000 0010 0010 1000 1100 0011 0010 0101 1000 0(2)

6. Positive number before normalization:

0.750 000 128 7(10) =

0.1100 0000 0000 0000 0000 0010 0010 1000 1100 0011 0010 0101 1000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.750 000 128 7(10) =

0.1100 0000 0000 0000 0000 0010 0010 1000 1100 0011 0010 0101 1000 0(2) =

0.1100 0000 0000 0000 0000 0010 0010 1000 1100 0011 0010 0101 1000 0(2) × 20 =

1.1000 0000 0000 0000 0000 0100 0101 0001 1000 0110 0100 1011 0000(2) × 2-1


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): -1

Mantissa (not normalized):
1.1000 0000 0000 0000 0000 0100 0101 0001 1000 0110 0100 1011 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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) =

1022(10) =

011 1111 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. 1000 0000 0000 0000 0000 0100 0101 0001 1000 0110 0100 1011 0000 =

1000 0000 0000 0000 0000 0100 0101 0001 1000 0110 0100 1011 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 1110

Mantissa (52 bits) =
1000 0000 0000 0000 0000 0100 0101 0001 1000 0110 0100 1011 0000


Decimal number -0.750 000 128 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1110 - 1000 0000 0000 0000 0000 0100 0101 0001 1000 0110 0100 1011 0000

» Convert decimal -0.750 000 125 2 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