-0.016 738 891 601 562 496 292 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 292(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
-0.016 738 891 601 562 496 292(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:

|-0.016 738 891 601 562 496 292| = 0.016 738 891 601 562 496 292


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: 0.016 738 891 601 562 496 292.

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) 0.016 738 891 601 562 496 292 × 2 = 0 + 0.033 477 783 203 124 992 584;
  • 2) 0.033 477 783 203 124 992 584 × 2 = 0 + 0.066 955 566 406 249 985 168;
  • 3) 0.066 955 566 406 249 985 168 × 2 = 0 + 0.133 911 132 812 499 970 336;
  • 4) 0.133 911 132 812 499 970 336 × 2 = 0 + 0.267 822 265 624 999 940 672;
  • 5) 0.267 822 265 624 999 940 672 × 2 = 0 + 0.535 644 531 249 999 881 344;
  • 6) 0.535 644 531 249 999 881 344 × 2 = 1 + 0.071 289 062 499 999 762 688;
  • 7) 0.071 289 062 499 999 762 688 × 2 = 0 + 0.142 578 124 999 999 525 376;
  • 8) 0.142 578 124 999 999 525 376 × 2 = 0 + 0.285 156 249 999 999 050 752;
  • 9) 0.285 156 249 999 999 050 752 × 2 = 0 + 0.570 312 499 999 998 101 504;
  • 10) 0.570 312 499 999 998 101 504 × 2 = 1 + 0.140 624 999 999 996 203 008;
  • 11) 0.140 624 999 999 996 203 008 × 2 = 0 + 0.281 249 999 999 992 406 016;
  • 12) 0.281 249 999 999 992 406 016 × 2 = 0 + 0.562 499 999 999 984 812 032;
  • 13) 0.562 499 999 999 984 812 032 × 2 = 1 + 0.124 999 999 999 969 624 064;
  • 14) 0.124 999 999 999 969 624 064 × 2 = 0 + 0.249 999 999 999 939 248 128;
  • 15) 0.249 999 999 999 939 248 128 × 2 = 0 + 0.499 999 999 999 878 496 256;
  • 16) 0.499 999 999 999 878 496 256 × 2 = 0 + 0.999 999 999 999 756 992 512;
  • 17) 0.999 999 999 999 756 992 512 × 2 = 1 + 0.999 999 999 999 513 985 024;
  • 18) 0.999 999 999 999 513 985 024 × 2 = 1 + 0.999 999 999 999 027 970 048;
  • 19) 0.999 999 999 999 027 970 048 × 2 = 1 + 0.999 999 999 998 055 940 096;
  • 20) 0.999 999 999 998 055 940 096 × 2 = 1 + 0.999 999 999 996 111 880 192;
  • 21) 0.999 999 999 996 111 880 192 × 2 = 1 + 0.999 999 999 992 223 760 384;
  • 22) 0.999 999 999 992 223 760 384 × 2 = 1 + 0.999 999 999 984 447 520 768;
  • 23) 0.999 999 999 984 447 520 768 × 2 = 1 + 0.999 999 999 968 895 041 536;
  • 24) 0.999 999 999 968 895 041 536 × 2 = 1 + 0.999 999 999 937 790 083 072;
  • 25) 0.999 999 999 937 790 083 072 × 2 = 1 + 0.999 999 999 875 580 166 144;
  • 26) 0.999 999 999 875 580 166 144 × 2 = 1 + 0.999 999 999 751 160 332 288;
  • 27) 0.999 999 999 751 160 332 288 × 2 = 1 + 0.999 999 999 502 320 664 576;
  • 28) 0.999 999 999 502 320 664 576 × 2 = 1 + 0.999 999 999 004 641 329 152;
  • 29) 0.999 999 999 004 641 329 152 × 2 = 1 + 0.999 999 998 009 282 658 304;
  • 30) 0.999 999 998 009 282 658 304 × 2 = 1 + 0.999 999 996 018 565 316 608;
  • 31) 0.999 999 996 018 565 316 608 × 2 = 1 + 0.999 999 992 037 130 633 216;
  • 32) 0.999 999 992 037 130 633 216 × 2 = 1 + 0.999 999 984 074 261 266 432;
  • 33) 0.999 999 984 074 261 266 432 × 2 = 1 + 0.999 999 968 148 522 532 864;
  • 34) 0.999 999 968 148 522 532 864 × 2 = 1 + 0.999 999 936 297 045 065 728;
  • 35) 0.999 999 936 297 045 065 728 × 2 = 1 + 0.999 999 872 594 090 131 456;
  • 36) 0.999 999 872 594 090 131 456 × 2 = 1 + 0.999 999 745 188 180 262 912;
  • 37) 0.999 999 745 188 180 262 912 × 2 = 1 + 0.999 999 490 376 360 525 824;
  • 38) 0.999 999 490 376 360 525 824 × 2 = 1 + 0.999 998 980 752 721 051 648;
  • 39) 0.999 998 980 752 721 051 648 × 2 = 1 + 0.999 997 961 505 442 103 296;
  • 40) 0.999 997 961 505 442 103 296 × 2 = 1 + 0.999 995 923 010 884 206 592;
  • 41) 0.999 995 923 010 884 206 592 × 2 = 1 + 0.999 991 846 021 768 413 184;
  • 42) 0.999 991 846 021 768 413 184 × 2 = 1 + 0.999 983 692 043 536 826 368;
  • 43) 0.999 983 692 043 536 826 368 × 2 = 1 + 0.999 967 384 087 073 652 736;
  • 44) 0.999 967 384 087 073 652 736 × 2 = 1 + 0.999 934 768 174 147 305 472;
  • 45) 0.999 934 768 174 147 305 472 × 2 = 1 + 0.999 869 536 348 294 610 944;
  • 46) 0.999 869 536 348 294 610 944 × 2 = 1 + 0.999 739 072 696 589 221 888;
  • 47) 0.999 739 072 696 589 221 888 × 2 = 1 + 0.999 478 145 393 178 443 776;
  • 48) 0.999 478 145 393 178 443 776 × 2 = 1 + 0.998 956 290 786 356 887 552;
  • 49) 0.998 956 290 786 356 887 552 × 2 = 1 + 0.997 912 581 572 713 775 104;
  • 50) 0.997 912 581 572 713 775 104 × 2 = 1 + 0.995 825 163 145 427 550 208;
  • 51) 0.995 825 163 145 427 550 208 × 2 = 1 + 0.991 650 326 290 855 100 416;
  • 52) 0.991 650 326 290 855 100 416 × 2 = 1 + 0.983 300 652 581 710 200 832;
  • 53) 0.983 300 652 581 710 200 832 × 2 = 1 + 0.966 601 305 163 420 401 664;
  • 54) 0.966 601 305 163 420 401 664 × 2 = 1 + 0.933 202 610 326 840 803 328;
  • 55) 0.933 202 610 326 840 803 328 × 2 = 1 + 0.866 405 220 653 681 606 656;
  • 56) 0.866 405 220 653 681 606 656 × 2 = 1 + 0.732 810 441 307 363 213 312;
  • 57) 0.732 810 441 307 363 213 312 × 2 = 1 + 0.465 620 882 614 726 426 624;
  • 58) 0.465 620 882 614 726 426 624 × 2 = 0 + 0.931 241 765 229 452 853 248;

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:


0.016 738 891 601 562 496 292(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 292(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 292(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 292 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 1110

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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