-0.016 738 891 601 562 496 530 549 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 549 6(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
-0.016 738 891 601 562 496 530 549 6(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:

|-0.016 738 891 601 562 496 530 549 6| = 0.016 738 891 601 562 496 530 549 6


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: 0.016 738 891 601 562 496 530 549 6.

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) 0.016 738 891 601 562 496 530 549 6 × 2 = 0 + 0.033 477 783 203 124 993 061 099 2;
  • 2) 0.033 477 783 203 124 993 061 099 2 × 2 = 0 + 0.066 955 566 406 249 986 122 198 4;
  • 3) 0.066 955 566 406 249 986 122 198 4 × 2 = 0 + 0.133 911 132 812 499 972 244 396 8;
  • 4) 0.133 911 132 812 499 972 244 396 8 × 2 = 0 + 0.267 822 265 624 999 944 488 793 6;
  • 5) 0.267 822 265 624 999 944 488 793 6 × 2 = 0 + 0.535 644 531 249 999 888 977 587 2;
  • 6) 0.535 644 531 249 999 888 977 587 2 × 2 = 1 + 0.071 289 062 499 999 777 955 174 4;
  • 7) 0.071 289 062 499 999 777 955 174 4 × 2 = 0 + 0.142 578 124 999 999 555 910 348 8;
  • 8) 0.142 578 124 999 999 555 910 348 8 × 2 = 0 + 0.285 156 249 999 999 111 820 697 6;
  • 9) 0.285 156 249 999 999 111 820 697 6 × 2 = 0 + 0.570 312 499 999 998 223 641 395 2;
  • 10) 0.570 312 499 999 998 223 641 395 2 × 2 = 1 + 0.140 624 999 999 996 447 282 790 4;
  • 11) 0.140 624 999 999 996 447 282 790 4 × 2 = 0 + 0.281 249 999 999 992 894 565 580 8;
  • 12) 0.281 249 999 999 992 894 565 580 8 × 2 = 0 + 0.562 499 999 999 985 789 131 161 6;
  • 13) 0.562 499 999 999 985 789 131 161 6 × 2 = 1 + 0.124 999 999 999 971 578 262 323 2;
  • 14) 0.124 999 999 999 971 578 262 323 2 × 2 = 0 + 0.249 999 999 999 943 156 524 646 4;
  • 15) 0.249 999 999 999 943 156 524 646 4 × 2 = 0 + 0.499 999 999 999 886 313 049 292 8;
  • 16) 0.499 999 999 999 886 313 049 292 8 × 2 = 0 + 0.999 999 999 999 772 626 098 585 6;
  • 17) 0.999 999 999 999 772 626 098 585 6 × 2 = 1 + 0.999 999 999 999 545 252 197 171 2;
  • 18) 0.999 999 999 999 545 252 197 171 2 × 2 = 1 + 0.999 999 999 999 090 504 394 342 4;
  • 19) 0.999 999 999 999 090 504 394 342 4 × 2 = 1 + 0.999 999 999 998 181 008 788 684 8;
  • 20) 0.999 999 999 998 181 008 788 684 8 × 2 = 1 + 0.999 999 999 996 362 017 577 369 6;
  • 21) 0.999 999 999 996 362 017 577 369 6 × 2 = 1 + 0.999 999 999 992 724 035 154 739 2;
  • 22) 0.999 999 999 992 724 035 154 739 2 × 2 = 1 + 0.999 999 999 985 448 070 309 478 4;
  • 23) 0.999 999 999 985 448 070 309 478 4 × 2 = 1 + 0.999 999 999 970 896 140 618 956 8;
  • 24) 0.999 999 999 970 896 140 618 956 8 × 2 = 1 + 0.999 999 999 941 792 281 237 913 6;
  • 25) 0.999 999 999 941 792 281 237 913 6 × 2 = 1 + 0.999 999 999 883 584 562 475 827 2;
  • 26) 0.999 999 999 883 584 562 475 827 2 × 2 = 1 + 0.999 999 999 767 169 124 951 654 4;
  • 27) 0.999 999 999 767 169 124 951 654 4 × 2 = 1 + 0.999 999 999 534 338 249 903 308 8;
  • 28) 0.999 999 999 534 338 249 903 308 8 × 2 = 1 + 0.999 999 999 068 676 499 806 617 6;
  • 29) 0.999 999 999 068 676 499 806 617 6 × 2 = 1 + 0.999 999 998 137 352 999 613 235 2;
  • 30) 0.999 999 998 137 352 999 613 235 2 × 2 = 1 + 0.999 999 996 274 705 999 226 470 4;
  • 31) 0.999 999 996 274 705 999 226 470 4 × 2 = 1 + 0.999 999 992 549 411 998 452 940 8;
  • 32) 0.999 999 992 549 411 998 452 940 8 × 2 = 1 + 0.999 999 985 098 823 996 905 881 6;
  • 33) 0.999 999 985 098 823 996 905 881 6 × 2 = 1 + 0.999 999 970 197 647 993 811 763 2;
  • 34) 0.999 999 970 197 647 993 811 763 2 × 2 = 1 + 0.999 999 940 395 295 987 623 526 4;
  • 35) 0.999 999 940 395 295 987 623 526 4 × 2 = 1 + 0.999 999 880 790 591 975 247 052 8;
  • 36) 0.999 999 880 790 591 975 247 052 8 × 2 = 1 + 0.999 999 761 581 183 950 494 105 6;
  • 37) 0.999 999 761 581 183 950 494 105 6 × 2 = 1 + 0.999 999 523 162 367 900 988 211 2;
  • 38) 0.999 999 523 162 367 900 988 211 2 × 2 = 1 + 0.999 999 046 324 735 801 976 422 4;
  • 39) 0.999 999 046 324 735 801 976 422 4 × 2 = 1 + 0.999 998 092 649 471 603 952 844 8;
  • 40) 0.999 998 092 649 471 603 952 844 8 × 2 = 1 + 0.999 996 185 298 943 207 905 689 6;
  • 41) 0.999 996 185 298 943 207 905 689 6 × 2 = 1 + 0.999 992 370 597 886 415 811 379 2;
  • 42) 0.999 992 370 597 886 415 811 379 2 × 2 = 1 + 0.999 984 741 195 772 831 622 758 4;
  • 43) 0.999 984 741 195 772 831 622 758 4 × 2 = 1 + 0.999 969 482 391 545 663 245 516 8;
  • 44) 0.999 969 482 391 545 663 245 516 8 × 2 = 1 + 0.999 938 964 783 091 326 491 033 6;
  • 45) 0.999 938 964 783 091 326 491 033 6 × 2 = 1 + 0.999 877 929 566 182 652 982 067 2;
  • 46) 0.999 877 929 566 182 652 982 067 2 × 2 = 1 + 0.999 755 859 132 365 305 964 134 4;
  • 47) 0.999 755 859 132 365 305 964 134 4 × 2 = 1 + 0.999 511 718 264 730 611 928 268 8;
  • 48) 0.999 511 718 264 730 611 928 268 8 × 2 = 1 + 0.999 023 436 529 461 223 856 537 6;
  • 49) 0.999 023 436 529 461 223 856 537 6 × 2 = 1 + 0.998 046 873 058 922 447 713 075 2;
  • 50) 0.998 046 873 058 922 447 713 075 2 × 2 = 1 + 0.996 093 746 117 844 895 426 150 4;
  • 51) 0.996 093 746 117 844 895 426 150 4 × 2 = 1 + 0.992 187 492 235 689 790 852 300 8;
  • 52) 0.992 187 492 235 689 790 852 300 8 × 2 = 1 + 0.984 374 984 471 379 581 704 601 6;
  • 53) 0.984 374 984 471 379 581 704 601 6 × 2 = 1 + 0.968 749 968 942 759 163 409 203 2;
  • 54) 0.968 749 968 942 759 163 409 203 2 × 2 = 1 + 0.937 499 937 885 518 326 818 406 4;
  • 55) 0.937 499 937 885 518 326 818 406 4 × 2 = 1 + 0.874 999 875 771 036 653 636 812 8;
  • 56) 0.874 999 875 771 036 653 636 812 8 × 2 = 1 + 0.749 999 751 542 073 307 273 625 6;
  • 57) 0.749 999 751 542 073 307 273 625 6 × 2 = 1 + 0.499 999 503 084 146 614 547 251 2;
  • 58) 0.499 999 503 084 146 614 547 251 2 × 2 = 0 + 0.999 999 006 168 293 229 094 502 4;

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:


0.016 738 891 601 562 496 530 549 6(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 549 6(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:


0.016 738 891 601 562 496 530 549 6(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 -0.016 738 891 601 562 496 530 549 6 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