-0.008 788 423 613 77 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 613 77(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
-0.008 788 423 613 77(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:

|-0.008 788 423 613 77| = 0.008 788 423 613 77


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: 0.008 788 423 613 77.

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) 0.008 788 423 613 77 × 2 = 0 + 0.017 576 847 227 54;
  • 2) 0.017 576 847 227 54 × 2 = 0 + 0.035 153 694 455 08;
  • 3) 0.035 153 694 455 08 × 2 = 0 + 0.070 307 388 910 16;
  • 4) 0.070 307 388 910 16 × 2 = 0 + 0.140 614 777 820 32;
  • 5) 0.140 614 777 820 32 × 2 = 0 + 0.281 229 555 640 64;
  • 6) 0.281 229 555 640 64 × 2 = 0 + 0.562 459 111 281 28;
  • 7) 0.562 459 111 281 28 × 2 = 1 + 0.124 918 222 562 56;
  • 8) 0.124 918 222 562 56 × 2 = 0 + 0.249 836 445 125 12;
  • 9) 0.249 836 445 125 12 × 2 = 0 + 0.499 672 890 250 24;
  • 10) 0.499 672 890 250 24 × 2 = 0 + 0.999 345 780 500 48;
  • 11) 0.999 345 780 500 48 × 2 = 1 + 0.998 691 561 000 96;
  • 12) 0.998 691 561 000 96 × 2 = 1 + 0.997 383 122 001 92;
  • 13) 0.997 383 122 001 92 × 2 = 1 + 0.994 766 244 003 84;
  • 14) 0.994 766 244 003 84 × 2 = 1 + 0.989 532 488 007 68;
  • 15) 0.989 532 488 007 68 × 2 = 1 + 0.979 064 976 015 36;
  • 16) 0.979 064 976 015 36 × 2 = 1 + 0.958 129 952 030 72;
  • 17) 0.958 129 952 030 72 × 2 = 1 + 0.916 259 904 061 44;
  • 18) 0.916 259 904 061 44 × 2 = 1 + 0.832 519 808 122 88;
  • 19) 0.832 519 808 122 88 × 2 = 1 + 0.665 039 616 245 76;
  • 20) 0.665 039 616 245 76 × 2 = 1 + 0.330 079 232 491 52;
  • 21) 0.330 079 232 491 52 × 2 = 0 + 0.660 158 464 983 04;
  • 22) 0.660 158 464 983 04 × 2 = 1 + 0.320 316 929 966 08;
  • 23) 0.320 316 929 966 08 × 2 = 0 + 0.640 633 859 932 16;
  • 24) 0.640 633 859 932 16 × 2 = 1 + 0.281 267 719 864 32;
  • 25) 0.281 267 719 864 32 × 2 = 0 + 0.562 535 439 728 64;
  • 26) 0.562 535 439 728 64 × 2 = 1 + 0.125 070 879 457 28;
  • 27) 0.125 070 879 457 28 × 2 = 0 + 0.250 141 758 914 56;
  • 28) 0.250 141 758 914 56 × 2 = 0 + 0.500 283 517 829 12;
  • 29) 0.500 283 517 829 12 × 2 = 1 + 0.000 567 035 658 24;
  • 30) 0.000 567 035 658 24 × 2 = 0 + 0.001 134 071 316 48;
  • 31) 0.001 134 071 316 48 × 2 = 0 + 0.002 268 142 632 96;
  • 32) 0.002 268 142 632 96 × 2 = 0 + 0.004 536 285 265 92;
  • 33) 0.004 536 285 265 92 × 2 = 0 + 0.009 072 570 531 84;
  • 34) 0.009 072 570 531 84 × 2 = 0 + 0.018 145 141 063 68;
  • 35) 0.018 145 141 063 68 × 2 = 0 + 0.036 290 282 127 36;
  • 36) 0.036 290 282 127 36 × 2 = 0 + 0.072 580 564 254 72;
  • 37) 0.072 580 564 254 72 × 2 = 0 + 0.145 161 128 509 44;
  • 38) 0.145 161 128 509 44 × 2 = 0 + 0.290 322 257 018 88;
  • 39) 0.290 322 257 018 88 × 2 = 0 + 0.580 644 514 037 76;
  • 40) 0.580 644 514 037 76 × 2 = 1 + 0.161 289 028 075 52;
  • 41) 0.161 289 028 075 52 × 2 = 0 + 0.322 578 056 151 04;
  • 42) 0.322 578 056 151 04 × 2 = 0 + 0.645 156 112 302 08;
  • 43) 0.645 156 112 302 08 × 2 = 1 + 0.290 312 224 604 16;
  • 44) 0.290 312 224 604 16 × 2 = 0 + 0.580 624 449 208 32;
  • 45) 0.580 624 449 208 32 × 2 = 1 + 0.161 248 898 416 64;
  • 46) 0.161 248 898 416 64 × 2 = 0 + 0.322 497 796 833 28;
  • 47) 0.322 497 796 833 28 × 2 = 0 + 0.644 995 593 666 56;
  • 48) 0.644 995 593 666 56 × 2 = 1 + 0.289 991 187 333 12;
  • 49) 0.289 991 187 333 12 × 2 = 0 + 0.579 982 374 666 24;
  • 50) 0.579 982 374 666 24 × 2 = 1 + 0.159 964 749 332 48;
  • 51) 0.159 964 749 332 48 × 2 = 0 + 0.319 929 498 664 96;
  • 52) 0.319 929 498 664 96 × 2 = 0 + 0.639 858 997 329 92;
  • 53) 0.639 858 997 329 92 × 2 = 1 + 0.279 717 994 659 84;
  • 54) 0.279 717 994 659 84 × 2 = 0 + 0.559 435 989 319 68;
  • 55) 0.559 435 989 319 68 × 2 = 1 + 0.118 871 978 639 36;
  • 56) 0.118 871 978 639 36 × 2 = 0 + 0.237 743 957 278 72;
  • 57) 0.237 743 957 278 72 × 2 = 0 + 0.475 487 914 557 44;
  • 58) 0.475 487 914 557 44 × 2 = 0 + 0.950 975 829 114 88;
  • 59) 0.950 975 829 114 88 × 2 = 1 + 0.901 951 658 229 76;

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:


0.008 788 423 613 77(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0010 1001 0100 1010 001(2)

6. Positive number before normalization:

0.008 788 423 613 77(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0010 1001 0100 1010 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 613 77(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0010 1001 0100 1010 001(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0010 1001 0100 1010 001(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 1001 0100 1010 0101 0001(2) × 2-7


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): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0100 0000 0000 1001 0100 1010 0101 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 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) =

1016(10) =

011 1111 1000(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. 0001 1111 1111 1010 1010 0100 0000 0000 1001 0100 1010 0101 0001 =

0001 1111 1111 1010 1010 0100 0000 0000 1001 0100 1010 0101 0001


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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0000 1001 0100 1010 0101 0001


Decimal number -0.008 788 423 613 77 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 1001 0100 1010 0101 0001

Share

» Convert decimal -0.008 788 423 613 84 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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