-0.016 738 891 601 562 496 839 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 839(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
-0.016 738 891 601 562 496 839(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:

|-0.016 738 891 601 562 496 839| = 0.016 738 891 601 562 496 839


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: 0.016 738 891 601 562 496 839.

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) 0.016 738 891 601 562 496 839 × 2 = 0 + 0.033 477 783 203 124 993 678;
  • 2) 0.033 477 783 203 124 993 678 × 2 = 0 + 0.066 955 566 406 249 987 356;
  • 3) 0.066 955 566 406 249 987 356 × 2 = 0 + 0.133 911 132 812 499 974 712;
  • 4) 0.133 911 132 812 499 974 712 × 2 = 0 + 0.267 822 265 624 999 949 424;
  • 5) 0.267 822 265 624 999 949 424 × 2 = 0 + 0.535 644 531 249 999 898 848;
  • 6) 0.535 644 531 249 999 898 848 × 2 = 1 + 0.071 289 062 499 999 797 696;
  • 7) 0.071 289 062 499 999 797 696 × 2 = 0 + 0.142 578 124 999 999 595 392;
  • 8) 0.142 578 124 999 999 595 392 × 2 = 0 + 0.285 156 249 999 999 190 784;
  • 9) 0.285 156 249 999 999 190 784 × 2 = 0 + 0.570 312 499 999 998 381 568;
  • 10) 0.570 312 499 999 998 381 568 × 2 = 1 + 0.140 624 999 999 996 763 136;
  • 11) 0.140 624 999 999 996 763 136 × 2 = 0 + 0.281 249 999 999 993 526 272;
  • 12) 0.281 249 999 999 993 526 272 × 2 = 0 + 0.562 499 999 999 987 052 544;
  • 13) 0.562 499 999 999 987 052 544 × 2 = 1 + 0.124 999 999 999 974 105 088;
  • 14) 0.124 999 999 999 974 105 088 × 2 = 0 + 0.249 999 999 999 948 210 176;
  • 15) 0.249 999 999 999 948 210 176 × 2 = 0 + 0.499 999 999 999 896 420 352;
  • 16) 0.499 999 999 999 896 420 352 × 2 = 0 + 0.999 999 999 999 792 840 704;
  • 17) 0.999 999 999 999 792 840 704 × 2 = 1 + 0.999 999 999 999 585 681 408;
  • 18) 0.999 999 999 999 585 681 408 × 2 = 1 + 0.999 999 999 999 171 362 816;
  • 19) 0.999 999 999 999 171 362 816 × 2 = 1 + 0.999 999 999 998 342 725 632;
  • 20) 0.999 999 999 998 342 725 632 × 2 = 1 + 0.999 999 999 996 685 451 264;
  • 21) 0.999 999 999 996 685 451 264 × 2 = 1 + 0.999 999 999 993 370 902 528;
  • 22) 0.999 999 999 993 370 902 528 × 2 = 1 + 0.999 999 999 986 741 805 056;
  • 23) 0.999 999 999 986 741 805 056 × 2 = 1 + 0.999 999 999 973 483 610 112;
  • 24) 0.999 999 999 973 483 610 112 × 2 = 1 + 0.999 999 999 946 967 220 224;
  • 25) 0.999 999 999 946 967 220 224 × 2 = 1 + 0.999 999 999 893 934 440 448;
  • 26) 0.999 999 999 893 934 440 448 × 2 = 1 + 0.999 999 999 787 868 880 896;
  • 27) 0.999 999 999 787 868 880 896 × 2 = 1 + 0.999 999 999 575 737 761 792;
  • 28) 0.999 999 999 575 737 761 792 × 2 = 1 + 0.999 999 999 151 475 523 584;
  • 29) 0.999 999 999 151 475 523 584 × 2 = 1 + 0.999 999 998 302 951 047 168;
  • 30) 0.999 999 998 302 951 047 168 × 2 = 1 + 0.999 999 996 605 902 094 336;
  • 31) 0.999 999 996 605 902 094 336 × 2 = 1 + 0.999 999 993 211 804 188 672;
  • 32) 0.999 999 993 211 804 188 672 × 2 = 1 + 0.999 999 986 423 608 377 344;
  • 33) 0.999 999 986 423 608 377 344 × 2 = 1 + 0.999 999 972 847 216 754 688;
  • 34) 0.999 999 972 847 216 754 688 × 2 = 1 + 0.999 999 945 694 433 509 376;
  • 35) 0.999 999 945 694 433 509 376 × 2 = 1 + 0.999 999 891 388 867 018 752;
  • 36) 0.999 999 891 388 867 018 752 × 2 = 1 + 0.999 999 782 777 734 037 504;
  • 37) 0.999 999 782 777 734 037 504 × 2 = 1 + 0.999 999 565 555 468 075 008;
  • 38) 0.999 999 565 555 468 075 008 × 2 = 1 + 0.999 999 131 110 936 150 016;
  • 39) 0.999 999 131 110 936 150 016 × 2 = 1 + 0.999 998 262 221 872 300 032;
  • 40) 0.999 998 262 221 872 300 032 × 2 = 1 + 0.999 996 524 443 744 600 064;
  • 41) 0.999 996 524 443 744 600 064 × 2 = 1 + 0.999 993 048 887 489 200 128;
  • 42) 0.999 993 048 887 489 200 128 × 2 = 1 + 0.999 986 097 774 978 400 256;
  • 43) 0.999 986 097 774 978 400 256 × 2 = 1 + 0.999 972 195 549 956 800 512;
  • 44) 0.999 972 195 549 956 800 512 × 2 = 1 + 0.999 944 391 099 913 601 024;
  • 45) 0.999 944 391 099 913 601 024 × 2 = 1 + 0.999 888 782 199 827 202 048;
  • 46) 0.999 888 782 199 827 202 048 × 2 = 1 + 0.999 777 564 399 654 404 096;
  • 47) 0.999 777 564 399 654 404 096 × 2 = 1 + 0.999 555 128 799 308 808 192;
  • 48) 0.999 555 128 799 308 808 192 × 2 = 1 + 0.999 110 257 598 617 616 384;
  • 49) 0.999 110 257 598 617 616 384 × 2 = 1 + 0.998 220 515 197 235 232 768;
  • 50) 0.998 220 515 197 235 232 768 × 2 = 1 + 0.996 441 030 394 470 465 536;
  • 51) 0.996 441 030 394 470 465 536 × 2 = 1 + 0.992 882 060 788 940 931 072;
  • 52) 0.992 882 060 788 940 931 072 × 2 = 1 + 0.985 764 121 577 881 862 144;
  • 53) 0.985 764 121 577 881 862 144 × 2 = 1 + 0.971 528 243 155 763 724 288;
  • 54) 0.971 528 243 155 763 724 288 × 2 = 1 + 0.943 056 486 311 527 448 576;
  • 55) 0.943 056 486 311 527 448 576 × 2 = 1 + 0.886 112 972 623 054 897 152;
  • 56) 0.886 112 972 623 054 897 152 × 2 = 1 + 0.772 225 945 246 109 794 304;
  • 57) 0.772 225 945 246 109 794 304 × 2 = 1 + 0.544 451 890 492 219 588 608;
  • 58) 0.544 451 890 492 219 588 608 × 2 = 1 + 0.088 903 780 984 439 177 216;

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:


0.016 738 891 601 562 496 839(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 839(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:


0.016 738 891 601 562 496 839(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 -0.016 738 891 601 562 496 839 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