-69.814 999 999 999 997 726 263 251 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -69.814 999 999 999 997 726 263 251(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
-69.814 999 999 999 997 726 263 251(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:

|-69.814 999 999 999 997 726 263 251| = 69.814 999 999 999 997 726 263 251


2. First, convert to binary (in base 2) the integer part: 69.
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;
  • 69 ÷ 2 = 34 + 1;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

69(10) =

100 0101(2)


4. Convert to binary (base 2) the fractional part: 0.814 999 999 999 997 726 263 251.

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) 0.814 999 999 999 997 726 263 251 × 2 = 1 + 0.629 999 999 999 995 452 526 502;
  • 2) 0.629 999 999 999 995 452 526 502 × 2 = 1 + 0.259 999 999 999 990 905 053 004;
  • 3) 0.259 999 999 999 990 905 053 004 × 2 = 0 + 0.519 999 999 999 981 810 106 008;
  • 4) 0.519 999 999 999 981 810 106 008 × 2 = 1 + 0.039 999 999 999 963 620 212 016;
  • 5) 0.039 999 999 999 963 620 212 016 × 2 = 0 + 0.079 999 999 999 927 240 424 032;
  • 6) 0.079 999 999 999 927 240 424 032 × 2 = 0 + 0.159 999 999 999 854 480 848 064;
  • 7) 0.159 999 999 999 854 480 848 064 × 2 = 0 + 0.319 999 999 999 708 961 696 128;
  • 8) 0.319 999 999 999 708 961 696 128 × 2 = 0 + 0.639 999 999 999 417 923 392 256;
  • 9) 0.639 999 999 999 417 923 392 256 × 2 = 1 + 0.279 999 999 998 835 846 784 512;
  • 10) 0.279 999 999 998 835 846 784 512 × 2 = 0 + 0.559 999 999 997 671 693 569 024;
  • 11) 0.559 999 999 997 671 693 569 024 × 2 = 1 + 0.119 999 999 995 343 387 138 048;
  • 12) 0.119 999 999 995 343 387 138 048 × 2 = 0 + 0.239 999 999 990 686 774 276 096;
  • 13) 0.239 999 999 990 686 774 276 096 × 2 = 0 + 0.479 999 999 981 373 548 552 192;
  • 14) 0.479 999 999 981 373 548 552 192 × 2 = 0 + 0.959 999 999 962 747 097 104 384;
  • 15) 0.959 999 999 962 747 097 104 384 × 2 = 1 + 0.919 999 999 925 494 194 208 768;
  • 16) 0.919 999 999 925 494 194 208 768 × 2 = 1 + 0.839 999 999 850 988 388 417 536;
  • 17) 0.839 999 999 850 988 388 417 536 × 2 = 1 + 0.679 999 999 701 976 776 835 072;
  • 18) 0.679 999 999 701 976 776 835 072 × 2 = 1 + 0.359 999 999 403 953 553 670 144;
  • 19) 0.359 999 999 403 953 553 670 144 × 2 = 0 + 0.719 999 998 807 907 107 340 288;
  • 20) 0.719 999 998 807 907 107 340 288 × 2 = 1 + 0.439 999 997 615 814 214 680 576;
  • 21) 0.439 999 997 615 814 214 680 576 × 2 = 0 + 0.879 999 995 231 628 429 361 152;
  • 22) 0.879 999 995 231 628 429 361 152 × 2 = 1 + 0.759 999 990 463 256 858 722 304;
  • 23) 0.759 999 990 463 256 858 722 304 × 2 = 1 + 0.519 999 980 926 513 717 444 608;
  • 24) 0.519 999 980 926 513 717 444 608 × 2 = 1 + 0.039 999 961 853 027 434 889 216;
  • 25) 0.039 999 961 853 027 434 889 216 × 2 = 0 + 0.079 999 923 706 054 869 778 432;
  • 26) 0.079 999 923 706 054 869 778 432 × 2 = 0 + 0.159 999 847 412 109 739 556 864;
  • 27) 0.159 999 847 412 109 739 556 864 × 2 = 0 + 0.319 999 694 824 219 479 113 728;
  • 28) 0.319 999 694 824 219 479 113 728 × 2 = 0 + 0.639 999 389 648 438 958 227 456;
  • 29) 0.639 999 389 648 438 958 227 456 × 2 = 1 + 0.279 998 779 296 877 916 454 912;
  • 30) 0.279 998 779 296 877 916 454 912 × 2 = 0 + 0.559 997 558 593 755 832 909 824;
  • 31) 0.559 997 558 593 755 832 909 824 × 2 = 1 + 0.119 995 117 187 511 665 819 648;
  • 32) 0.119 995 117 187 511 665 819 648 × 2 = 0 + 0.239 990 234 375 023 331 639 296;
  • 33) 0.239 990 234 375 023 331 639 296 × 2 = 0 + 0.479 980 468 750 046 663 278 592;
  • 34) 0.479 980 468 750 046 663 278 592 × 2 = 0 + 0.959 960 937 500 093 326 557 184;
  • 35) 0.959 960 937 500 093 326 557 184 × 2 = 1 + 0.919 921 875 000 186 653 114 368;
  • 36) 0.919 921 875 000 186 653 114 368 × 2 = 1 + 0.839 843 750 000 373 306 228 736;
  • 37) 0.839 843 750 000 373 306 228 736 × 2 = 1 + 0.679 687 500 000 746 612 457 472;
  • 38) 0.679 687 500 000 746 612 457 472 × 2 = 1 + 0.359 375 000 001 493 224 914 944;
  • 39) 0.359 375 000 001 493 224 914 944 × 2 = 0 + 0.718 750 000 002 986 449 829 888;
  • 40) 0.718 750 000 002 986 449 829 888 × 2 = 1 + 0.437 500 000 005 972 899 659 776;
  • 41) 0.437 500 000 005 972 899 659 776 × 2 = 0 + 0.875 000 000 011 945 799 319 552;
  • 42) 0.875 000 000 011 945 799 319 552 × 2 = 1 + 0.750 000 000 023 891 598 639 104;
  • 43) 0.750 000 000 023 891 598 639 104 × 2 = 1 + 0.500 000 000 047 783 197 278 208;
  • 44) 0.500 000 000 047 783 197 278 208 × 2 = 1 + 0.000 000 000 095 566 394 556 416;
  • 45) 0.000 000 000 095 566 394 556 416 × 2 = 0 + 0.000 000 000 191 132 789 112 832;
  • 46) 0.000 000 000 191 132 789 112 832 × 2 = 0 + 0.000 000 000 382 265 578 225 664;
  • 47) 0.000 000 000 382 265 578 225 664 × 2 = 0 + 0.000 000 000 764 531 156 451 328;
  • 48) 0.000 000 000 764 531 156 451 328 × 2 = 0 + 0.000 000 001 529 062 312 902 656;
  • 49) 0.000 000 001 529 062 312 902 656 × 2 = 0 + 0.000 000 003 058 124 625 805 312;
  • 50) 0.000 000 003 058 124 625 805 312 × 2 = 0 + 0.000 000 006 116 249 251 610 624;
  • 51) 0.000 000 006 116 249 251 610 624 × 2 = 0 + 0.000 000 012 232 498 503 221 248;
  • 52) 0.000 000 012 232 498 503 221 248 × 2 = 0 + 0.000 000 024 464 997 006 442 496;
  • 53) 0.000 000 024 464 997 006 442 496 × 2 = 0 + 0.000 000 048 929 994 012 884 992;

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:


0.814 999 999 999 997 726 263 251(10) =

0.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0(2)

6. Positive number before normalization:

69.814 999 999 999 997 726 263 251(10) =

100 0101.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


69.814 999 999 999 997 726 263 251(10) =

100 0101.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0(2) =

100 0101.1101 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0000 0000 0(2) × 20 =

1.0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0000 000(2) × 26


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): 6

Mantissa (not normalized):
1.0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0000 000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

6 + 2(11-1) - 1 =

(6 + 1 023)(10) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1029(10) =

100 0000 0101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 000 0000 =

0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100


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) =
100 0000 0101

Mantissa (52 bits) =
0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100


Decimal number -69.814 999 999 999 997 726 263 251 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0101 - 0001 0111 0100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100

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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