-0.016 738 891 601 562 496 832 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 832(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
-0.016 738 891 601 562 496 832(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:

|-0.016 738 891 601 562 496 832| = 0.016 738 891 601 562 496 832


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: 0.016 738 891 601 562 496 832.

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) 0.016 738 891 601 562 496 832 × 2 = 0 + 0.033 477 783 203 124 993 664;
  • 2) 0.033 477 783 203 124 993 664 × 2 = 0 + 0.066 955 566 406 249 987 328;
  • 3) 0.066 955 566 406 249 987 328 × 2 = 0 + 0.133 911 132 812 499 974 656;
  • 4) 0.133 911 132 812 499 974 656 × 2 = 0 + 0.267 822 265 624 999 949 312;
  • 5) 0.267 822 265 624 999 949 312 × 2 = 0 + 0.535 644 531 249 999 898 624;
  • 6) 0.535 644 531 249 999 898 624 × 2 = 1 + 0.071 289 062 499 999 797 248;
  • 7) 0.071 289 062 499 999 797 248 × 2 = 0 + 0.142 578 124 999 999 594 496;
  • 8) 0.142 578 124 999 999 594 496 × 2 = 0 + 0.285 156 249 999 999 188 992;
  • 9) 0.285 156 249 999 999 188 992 × 2 = 0 + 0.570 312 499 999 998 377 984;
  • 10) 0.570 312 499 999 998 377 984 × 2 = 1 + 0.140 624 999 999 996 755 968;
  • 11) 0.140 624 999 999 996 755 968 × 2 = 0 + 0.281 249 999 999 993 511 936;
  • 12) 0.281 249 999 999 993 511 936 × 2 = 0 + 0.562 499 999 999 987 023 872;
  • 13) 0.562 499 999 999 987 023 872 × 2 = 1 + 0.124 999 999 999 974 047 744;
  • 14) 0.124 999 999 999 974 047 744 × 2 = 0 + 0.249 999 999 999 948 095 488;
  • 15) 0.249 999 999 999 948 095 488 × 2 = 0 + 0.499 999 999 999 896 190 976;
  • 16) 0.499 999 999 999 896 190 976 × 2 = 0 + 0.999 999 999 999 792 381 952;
  • 17) 0.999 999 999 999 792 381 952 × 2 = 1 + 0.999 999 999 999 584 763 904;
  • 18) 0.999 999 999 999 584 763 904 × 2 = 1 + 0.999 999 999 999 169 527 808;
  • 19) 0.999 999 999 999 169 527 808 × 2 = 1 + 0.999 999 999 998 339 055 616;
  • 20) 0.999 999 999 998 339 055 616 × 2 = 1 + 0.999 999 999 996 678 111 232;
  • 21) 0.999 999 999 996 678 111 232 × 2 = 1 + 0.999 999 999 993 356 222 464;
  • 22) 0.999 999 999 993 356 222 464 × 2 = 1 + 0.999 999 999 986 712 444 928;
  • 23) 0.999 999 999 986 712 444 928 × 2 = 1 + 0.999 999 999 973 424 889 856;
  • 24) 0.999 999 999 973 424 889 856 × 2 = 1 + 0.999 999 999 946 849 779 712;
  • 25) 0.999 999 999 946 849 779 712 × 2 = 1 + 0.999 999 999 893 699 559 424;
  • 26) 0.999 999 999 893 699 559 424 × 2 = 1 + 0.999 999 999 787 399 118 848;
  • 27) 0.999 999 999 787 399 118 848 × 2 = 1 + 0.999 999 999 574 798 237 696;
  • 28) 0.999 999 999 574 798 237 696 × 2 = 1 + 0.999 999 999 149 596 475 392;
  • 29) 0.999 999 999 149 596 475 392 × 2 = 1 + 0.999 999 998 299 192 950 784;
  • 30) 0.999 999 998 299 192 950 784 × 2 = 1 + 0.999 999 996 598 385 901 568;
  • 31) 0.999 999 996 598 385 901 568 × 2 = 1 + 0.999 999 993 196 771 803 136;
  • 32) 0.999 999 993 196 771 803 136 × 2 = 1 + 0.999 999 986 393 543 606 272;
  • 33) 0.999 999 986 393 543 606 272 × 2 = 1 + 0.999 999 972 787 087 212 544;
  • 34) 0.999 999 972 787 087 212 544 × 2 = 1 + 0.999 999 945 574 174 425 088;
  • 35) 0.999 999 945 574 174 425 088 × 2 = 1 + 0.999 999 891 148 348 850 176;
  • 36) 0.999 999 891 148 348 850 176 × 2 = 1 + 0.999 999 782 296 697 700 352;
  • 37) 0.999 999 782 296 697 700 352 × 2 = 1 + 0.999 999 564 593 395 400 704;
  • 38) 0.999 999 564 593 395 400 704 × 2 = 1 + 0.999 999 129 186 790 801 408;
  • 39) 0.999 999 129 186 790 801 408 × 2 = 1 + 0.999 998 258 373 581 602 816;
  • 40) 0.999 998 258 373 581 602 816 × 2 = 1 + 0.999 996 516 747 163 205 632;
  • 41) 0.999 996 516 747 163 205 632 × 2 = 1 + 0.999 993 033 494 326 411 264;
  • 42) 0.999 993 033 494 326 411 264 × 2 = 1 + 0.999 986 066 988 652 822 528;
  • 43) 0.999 986 066 988 652 822 528 × 2 = 1 + 0.999 972 133 977 305 645 056;
  • 44) 0.999 972 133 977 305 645 056 × 2 = 1 + 0.999 944 267 954 611 290 112;
  • 45) 0.999 944 267 954 611 290 112 × 2 = 1 + 0.999 888 535 909 222 580 224;
  • 46) 0.999 888 535 909 222 580 224 × 2 = 1 + 0.999 777 071 818 445 160 448;
  • 47) 0.999 777 071 818 445 160 448 × 2 = 1 + 0.999 554 143 636 890 320 896;
  • 48) 0.999 554 143 636 890 320 896 × 2 = 1 + 0.999 108 287 273 780 641 792;
  • 49) 0.999 108 287 273 780 641 792 × 2 = 1 + 0.998 216 574 547 561 283 584;
  • 50) 0.998 216 574 547 561 283 584 × 2 = 1 + 0.996 433 149 095 122 567 168;
  • 51) 0.996 433 149 095 122 567 168 × 2 = 1 + 0.992 866 298 190 245 134 336;
  • 52) 0.992 866 298 190 245 134 336 × 2 = 1 + 0.985 732 596 380 490 268 672;
  • 53) 0.985 732 596 380 490 268 672 × 2 = 1 + 0.971 465 192 760 980 537 344;
  • 54) 0.971 465 192 760 980 537 344 × 2 = 1 + 0.942 930 385 521 961 074 688;
  • 55) 0.942 930 385 521 961 074 688 × 2 = 1 + 0.885 860 771 043 922 149 376;
  • 56) 0.885 860 771 043 922 149 376 × 2 = 1 + 0.771 721 542 087 844 298 752;
  • 57) 0.771 721 542 087 844 298 752 × 2 = 1 + 0.543 443 084 175 688 597 504;
  • 58) 0.543 443 084 175 688 597 504 × 2 = 1 + 0.086 886 168 351 377 195 008;

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:


0.016 738 891 601 562 496 832(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 832(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

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


0.016 738 891 601 562 496 832(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 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) =

1017(10) =

011 1111 1001(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 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 832 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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