-0.008 788 423 612 711 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 711 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
-0.008 788 423 612 711 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:

|-0.008 788 423 612 711 4| = 0.008 788 423 612 711 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: 0.008 788 423 612 711 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) 0.008 788 423 612 711 4 × 2 = 0 + 0.017 576 847 225 422 8;
  • 2) 0.017 576 847 225 422 8 × 2 = 0 + 0.035 153 694 450 845 6;
  • 3) 0.035 153 694 450 845 6 × 2 = 0 + 0.070 307 388 901 691 2;
  • 4) 0.070 307 388 901 691 2 × 2 = 0 + 0.140 614 777 803 382 4;
  • 5) 0.140 614 777 803 382 4 × 2 = 0 + 0.281 229 555 606 764 8;
  • 6) 0.281 229 555 606 764 8 × 2 = 0 + 0.562 459 111 213 529 6;
  • 7) 0.562 459 111 213 529 6 × 2 = 1 + 0.124 918 222 427 059 2;
  • 8) 0.124 918 222 427 059 2 × 2 = 0 + 0.249 836 444 854 118 4;
  • 9) 0.249 836 444 854 118 4 × 2 = 0 + 0.499 672 889 708 236 8;
  • 10) 0.499 672 889 708 236 8 × 2 = 0 + 0.999 345 779 416 473 6;
  • 11) 0.999 345 779 416 473 6 × 2 = 1 + 0.998 691 558 832 947 2;
  • 12) 0.998 691 558 832 947 2 × 2 = 1 + 0.997 383 117 665 894 4;
  • 13) 0.997 383 117 665 894 4 × 2 = 1 + 0.994 766 235 331 788 8;
  • 14) 0.994 766 235 331 788 8 × 2 = 1 + 0.989 532 470 663 577 6;
  • 15) 0.989 532 470 663 577 6 × 2 = 1 + 0.979 064 941 327 155 2;
  • 16) 0.979 064 941 327 155 2 × 2 = 1 + 0.958 129 882 654 310 4;
  • 17) 0.958 129 882 654 310 4 × 2 = 1 + 0.916 259 765 308 620 8;
  • 18) 0.916 259 765 308 620 8 × 2 = 1 + 0.832 519 530 617 241 6;
  • 19) 0.832 519 530 617 241 6 × 2 = 1 + 0.665 039 061 234 483 2;
  • 20) 0.665 039 061 234 483 2 × 2 = 1 + 0.330 078 122 468 966 4;
  • 21) 0.330 078 122 468 966 4 × 2 = 0 + 0.660 156 244 937 932 8;
  • 22) 0.660 156 244 937 932 8 × 2 = 1 + 0.320 312 489 875 865 6;
  • 23) 0.320 312 489 875 865 6 × 2 = 0 + 0.640 624 979 751 731 2;
  • 24) 0.640 624 979 751 731 2 × 2 = 1 + 0.281 249 959 503 462 4;
  • 25) 0.281 249 959 503 462 4 × 2 = 0 + 0.562 499 919 006 924 8;
  • 26) 0.562 499 919 006 924 8 × 2 = 1 + 0.124 999 838 013 849 6;
  • 27) 0.124 999 838 013 849 6 × 2 = 0 + 0.249 999 676 027 699 2;
  • 28) 0.249 999 676 027 699 2 × 2 = 0 + 0.499 999 352 055 398 4;
  • 29) 0.499 999 352 055 398 4 × 2 = 0 + 0.999 998 704 110 796 8;
  • 30) 0.999 998 704 110 796 8 × 2 = 1 + 0.999 997 408 221 593 6;
  • 31) 0.999 997 408 221 593 6 × 2 = 1 + 0.999 994 816 443 187 2;
  • 32) 0.999 994 816 443 187 2 × 2 = 1 + 0.999 989 632 886 374 4;
  • 33) 0.999 989 632 886 374 4 × 2 = 1 + 0.999 979 265 772 748 8;
  • 34) 0.999 979 265 772 748 8 × 2 = 1 + 0.999 958 531 545 497 6;
  • 35) 0.999 958 531 545 497 6 × 2 = 1 + 0.999 917 063 090 995 2;
  • 36) 0.999 917 063 090 995 2 × 2 = 1 + 0.999 834 126 181 990 4;
  • 37) 0.999 834 126 181 990 4 × 2 = 1 + 0.999 668 252 363 980 8;
  • 38) 0.999 668 252 363 980 8 × 2 = 1 + 0.999 336 504 727 961 6;
  • 39) 0.999 336 504 727 961 6 × 2 = 1 + 0.998 673 009 455 923 2;
  • 40) 0.998 673 009 455 923 2 × 2 = 1 + 0.997 346 018 911 846 4;
  • 41) 0.997 346 018 911 846 4 × 2 = 1 + 0.994 692 037 823 692 8;
  • 42) 0.994 692 037 823 692 8 × 2 = 1 + 0.989 384 075 647 385 6;
  • 43) 0.989 384 075 647 385 6 × 2 = 1 + 0.978 768 151 294 771 2;
  • 44) 0.978 768 151 294 771 2 × 2 = 1 + 0.957 536 302 589 542 4;
  • 45) 0.957 536 302 589 542 4 × 2 = 1 + 0.915 072 605 179 084 8;
  • 46) 0.915 072 605 179 084 8 × 2 = 1 + 0.830 145 210 358 169 6;
  • 47) 0.830 145 210 358 169 6 × 2 = 1 + 0.660 290 420 716 339 2;
  • 48) 0.660 290 420 716 339 2 × 2 = 1 + 0.320 580 841 432 678 4;
  • 49) 0.320 580 841 432 678 4 × 2 = 0 + 0.641 161 682 865 356 8;
  • 50) 0.641 161 682 865 356 8 × 2 = 1 + 0.282 323 365 730 713 6;
  • 51) 0.282 323 365 730 713 6 × 2 = 0 + 0.564 646 731 461 427 2;
  • 52) 0.564 646 731 461 427 2 × 2 = 1 + 0.129 293 462 922 854 4;
  • 53) 0.129 293 462 922 854 4 × 2 = 0 + 0.258 586 925 845 708 8;
  • 54) 0.258 586 925 845 708 8 × 2 = 0 + 0.517 173 851 691 417 6;
  • 55) 0.517 173 851 691 417 6 × 2 = 1 + 0.034 347 703 382 835 2;
  • 56) 0.034 347 703 382 835 2 × 2 = 0 + 0.068 695 406 765 670 4;
  • 57) 0.068 695 406 765 670 4 × 2 = 0 + 0.137 390 813 531 340 8;
  • 58) 0.137 390 813 531 340 8 × 2 = 0 + 0.274 781 627 062 681 6;
  • 59) 0.274 781 627 062 681 6 × 2 = 0 + 0.549 563 254 125 363 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:


0.008 788 423 612 711 4(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1111 0101 0010 000(2)

6. Positive number before normalization:

0.008 788 423 612 711 4(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1111 0101 0010 000(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:


0.008 788 423 612 711 4(10) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1111 0101 0010 000(2) =

0.0000 0010 0011 1111 1111 0101 0100 0111 1111 1111 1111 1111 0101 0010 000(2) × 20 =

1.0001 1111 1111 1010 1010 0011 1111 1111 1111 1111 1010 1001 0000(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 0011 1111 1111 1111 1111 1010 1001 0000


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 0011 1111 1111 1111 1111 1010 1001 0000 =

0001 1111 1111 1010 1010 0011 1111 1111 1111 1111 1010 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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0011 1111 1111 1111 1111 1010 1001 0000


Decimal number -0.008 788 423 612 711 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0011 1111 1111 1111 1111 1010 1001 0000

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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