-0.016 738 891 601 562 496 343 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 343(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
-0.016 738 891 601 562 496 343(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:

|-0.016 738 891 601 562 496 343| = 0.016 738 891 601 562 496 343


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: 0.016 738 891 601 562 496 343.

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) 0.016 738 891 601 562 496 343 × 2 = 0 + 0.033 477 783 203 124 992 686;
  • 2) 0.033 477 783 203 124 992 686 × 2 = 0 + 0.066 955 566 406 249 985 372;
  • 3) 0.066 955 566 406 249 985 372 × 2 = 0 + 0.133 911 132 812 499 970 744;
  • 4) 0.133 911 132 812 499 970 744 × 2 = 0 + 0.267 822 265 624 999 941 488;
  • 5) 0.267 822 265 624 999 941 488 × 2 = 0 + 0.535 644 531 249 999 882 976;
  • 6) 0.535 644 531 249 999 882 976 × 2 = 1 + 0.071 289 062 499 999 765 952;
  • 7) 0.071 289 062 499 999 765 952 × 2 = 0 + 0.142 578 124 999 999 531 904;
  • 8) 0.142 578 124 999 999 531 904 × 2 = 0 + 0.285 156 249 999 999 063 808;
  • 9) 0.285 156 249 999 999 063 808 × 2 = 0 + 0.570 312 499 999 998 127 616;
  • 10) 0.570 312 499 999 998 127 616 × 2 = 1 + 0.140 624 999 999 996 255 232;
  • 11) 0.140 624 999 999 996 255 232 × 2 = 0 + 0.281 249 999 999 992 510 464;
  • 12) 0.281 249 999 999 992 510 464 × 2 = 0 + 0.562 499 999 999 985 020 928;
  • 13) 0.562 499 999 999 985 020 928 × 2 = 1 + 0.124 999 999 999 970 041 856;
  • 14) 0.124 999 999 999 970 041 856 × 2 = 0 + 0.249 999 999 999 940 083 712;
  • 15) 0.249 999 999 999 940 083 712 × 2 = 0 + 0.499 999 999 999 880 167 424;
  • 16) 0.499 999 999 999 880 167 424 × 2 = 0 + 0.999 999 999 999 760 334 848;
  • 17) 0.999 999 999 999 760 334 848 × 2 = 1 + 0.999 999 999 999 520 669 696;
  • 18) 0.999 999 999 999 520 669 696 × 2 = 1 + 0.999 999 999 999 041 339 392;
  • 19) 0.999 999 999 999 041 339 392 × 2 = 1 + 0.999 999 999 998 082 678 784;
  • 20) 0.999 999 999 998 082 678 784 × 2 = 1 + 0.999 999 999 996 165 357 568;
  • 21) 0.999 999 999 996 165 357 568 × 2 = 1 + 0.999 999 999 992 330 715 136;
  • 22) 0.999 999 999 992 330 715 136 × 2 = 1 + 0.999 999 999 984 661 430 272;
  • 23) 0.999 999 999 984 661 430 272 × 2 = 1 + 0.999 999 999 969 322 860 544;
  • 24) 0.999 999 999 969 322 860 544 × 2 = 1 + 0.999 999 999 938 645 721 088;
  • 25) 0.999 999 999 938 645 721 088 × 2 = 1 + 0.999 999 999 877 291 442 176;
  • 26) 0.999 999 999 877 291 442 176 × 2 = 1 + 0.999 999 999 754 582 884 352;
  • 27) 0.999 999 999 754 582 884 352 × 2 = 1 + 0.999 999 999 509 165 768 704;
  • 28) 0.999 999 999 509 165 768 704 × 2 = 1 + 0.999 999 999 018 331 537 408;
  • 29) 0.999 999 999 018 331 537 408 × 2 = 1 + 0.999 999 998 036 663 074 816;
  • 30) 0.999 999 998 036 663 074 816 × 2 = 1 + 0.999 999 996 073 326 149 632;
  • 31) 0.999 999 996 073 326 149 632 × 2 = 1 + 0.999 999 992 146 652 299 264;
  • 32) 0.999 999 992 146 652 299 264 × 2 = 1 + 0.999 999 984 293 304 598 528;
  • 33) 0.999 999 984 293 304 598 528 × 2 = 1 + 0.999 999 968 586 609 197 056;
  • 34) 0.999 999 968 586 609 197 056 × 2 = 1 + 0.999 999 937 173 218 394 112;
  • 35) 0.999 999 937 173 218 394 112 × 2 = 1 + 0.999 999 874 346 436 788 224;
  • 36) 0.999 999 874 346 436 788 224 × 2 = 1 + 0.999 999 748 692 873 576 448;
  • 37) 0.999 999 748 692 873 576 448 × 2 = 1 + 0.999 999 497 385 747 152 896;
  • 38) 0.999 999 497 385 747 152 896 × 2 = 1 + 0.999 998 994 771 494 305 792;
  • 39) 0.999 998 994 771 494 305 792 × 2 = 1 + 0.999 997 989 542 988 611 584;
  • 40) 0.999 997 989 542 988 611 584 × 2 = 1 + 0.999 995 979 085 977 223 168;
  • 41) 0.999 995 979 085 977 223 168 × 2 = 1 + 0.999 991 958 171 954 446 336;
  • 42) 0.999 991 958 171 954 446 336 × 2 = 1 + 0.999 983 916 343 908 892 672;
  • 43) 0.999 983 916 343 908 892 672 × 2 = 1 + 0.999 967 832 687 817 785 344;
  • 44) 0.999 967 832 687 817 785 344 × 2 = 1 + 0.999 935 665 375 635 570 688;
  • 45) 0.999 935 665 375 635 570 688 × 2 = 1 + 0.999 871 330 751 271 141 376;
  • 46) 0.999 871 330 751 271 141 376 × 2 = 1 + 0.999 742 661 502 542 282 752;
  • 47) 0.999 742 661 502 542 282 752 × 2 = 1 + 0.999 485 323 005 084 565 504;
  • 48) 0.999 485 323 005 084 565 504 × 2 = 1 + 0.998 970 646 010 169 131 008;
  • 49) 0.998 970 646 010 169 131 008 × 2 = 1 + 0.997 941 292 020 338 262 016;
  • 50) 0.997 941 292 020 338 262 016 × 2 = 1 + 0.995 882 584 040 676 524 032;
  • 51) 0.995 882 584 040 676 524 032 × 2 = 1 + 0.991 765 168 081 353 048 064;
  • 52) 0.991 765 168 081 353 048 064 × 2 = 1 + 0.983 530 336 162 706 096 128;
  • 53) 0.983 530 336 162 706 096 128 × 2 = 1 + 0.967 060 672 325 412 192 256;
  • 54) 0.967 060 672 325 412 192 256 × 2 = 1 + 0.934 121 344 650 824 384 512;
  • 55) 0.934 121 344 650 824 384 512 × 2 = 1 + 0.868 242 689 301 648 769 024;
  • 56) 0.868 242 689 301 648 769 024 × 2 = 1 + 0.736 485 378 603 297 538 048;
  • 57) 0.736 485 378 603 297 538 048 × 2 = 1 + 0.472 970 757 206 595 076 096;
  • 58) 0.472 970 757 206 595 076 096 × 2 = 0 + 0.945 941 514 413 190 152 192;

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:


0.016 738 891 601 562 496 343(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 343(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:


0.016 738 891 601 562 496 343(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 -0.016 738 891 601 562 496 343 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