-0.016 738 891 601 562 496 604 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 604(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
-0.016 738 891 601 562 496 604(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:

|-0.016 738 891 601 562 496 604| = 0.016 738 891 601 562 496 604


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: 0.016 738 891 601 562 496 604.

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) 0.016 738 891 601 562 496 604 × 2 = 0 + 0.033 477 783 203 124 993 208;
  • 2) 0.033 477 783 203 124 993 208 × 2 = 0 + 0.066 955 566 406 249 986 416;
  • 3) 0.066 955 566 406 249 986 416 × 2 = 0 + 0.133 911 132 812 499 972 832;
  • 4) 0.133 911 132 812 499 972 832 × 2 = 0 + 0.267 822 265 624 999 945 664;
  • 5) 0.267 822 265 624 999 945 664 × 2 = 0 + 0.535 644 531 249 999 891 328;
  • 6) 0.535 644 531 249 999 891 328 × 2 = 1 + 0.071 289 062 499 999 782 656;
  • 7) 0.071 289 062 499 999 782 656 × 2 = 0 + 0.142 578 124 999 999 565 312;
  • 8) 0.142 578 124 999 999 565 312 × 2 = 0 + 0.285 156 249 999 999 130 624;
  • 9) 0.285 156 249 999 999 130 624 × 2 = 0 + 0.570 312 499 999 998 261 248;
  • 10) 0.570 312 499 999 998 261 248 × 2 = 1 + 0.140 624 999 999 996 522 496;
  • 11) 0.140 624 999 999 996 522 496 × 2 = 0 + 0.281 249 999 999 993 044 992;
  • 12) 0.281 249 999 999 993 044 992 × 2 = 0 + 0.562 499 999 999 986 089 984;
  • 13) 0.562 499 999 999 986 089 984 × 2 = 1 + 0.124 999 999 999 972 179 968;
  • 14) 0.124 999 999 999 972 179 968 × 2 = 0 + 0.249 999 999 999 944 359 936;
  • 15) 0.249 999 999 999 944 359 936 × 2 = 0 + 0.499 999 999 999 888 719 872;
  • 16) 0.499 999 999 999 888 719 872 × 2 = 0 + 0.999 999 999 999 777 439 744;
  • 17) 0.999 999 999 999 777 439 744 × 2 = 1 + 0.999 999 999 999 554 879 488;
  • 18) 0.999 999 999 999 554 879 488 × 2 = 1 + 0.999 999 999 999 109 758 976;
  • 19) 0.999 999 999 999 109 758 976 × 2 = 1 + 0.999 999 999 998 219 517 952;
  • 20) 0.999 999 999 998 219 517 952 × 2 = 1 + 0.999 999 999 996 439 035 904;
  • 21) 0.999 999 999 996 439 035 904 × 2 = 1 + 0.999 999 999 992 878 071 808;
  • 22) 0.999 999 999 992 878 071 808 × 2 = 1 + 0.999 999 999 985 756 143 616;
  • 23) 0.999 999 999 985 756 143 616 × 2 = 1 + 0.999 999 999 971 512 287 232;
  • 24) 0.999 999 999 971 512 287 232 × 2 = 1 + 0.999 999 999 943 024 574 464;
  • 25) 0.999 999 999 943 024 574 464 × 2 = 1 + 0.999 999 999 886 049 148 928;
  • 26) 0.999 999 999 886 049 148 928 × 2 = 1 + 0.999 999 999 772 098 297 856;
  • 27) 0.999 999 999 772 098 297 856 × 2 = 1 + 0.999 999 999 544 196 595 712;
  • 28) 0.999 999 999 544 196 595 712 × 2 = 1 + 0.999 999 999 088 393 191 424;
  • 29) 0.999 999 999 088 393 191 424 × 2 = 1 + 0.999 999 998 176 786 382 848;
  • 30) 0.999 999 998 176 786 382 848 × 2 = 1 + 0.999 999 996 353 572 765 696;
  • 31) 0.999 999 996 353 572 765 696 × 2 = 1 + 0.999 999 992 707 145 531 392;
  • 32) 0.999 999 992 707 145 531 392 × 2 = 1 + 0.999 999 985 414 291 062 784;
  • 33) 0.999 999 985 414 291 062 784 × 2 = 1 + 0.999 999 970 828 582 125 568;
  • 34) 0.999 999 970 828 582 125 568 × 2 = 1 + 0.999 999 941 657 164 251 136;
  • 35) 0.999 999 941 657 164 251 136 × 2 = 1 + 0.999 999 883 314 328 502 272;
  • 36) 0.999 999 883 314 328 502 272 × 2 = 1 + 0.999 999 766 628 657 004 544;
  • 37) 0.999 999 766 628 657 004 544 × 2 = 1 + 0.999 999 533 257 314 009 088;
  • 38) 0.999 999 533 257 314 009 088 × 2 = 1 + 0.999 999 066 514 628 018 176;
  • 39) 0.999 999 066 514 628 018 176 × 2 = 1 + 0.999 998 133 029 256 036 352;
  • 40) 0.999 998 133 029 256 036 352 × 2 = 1 + 0.999 996 266 058 512 072 704;
  • 41) 0.999 996 266 058 512 072 704 × 2 = 1 + 0.999 992 532 117 024 145 408;
  • 42) 0.999 992 532 117 024 145 408 × 2 = 1 + 0.999 985 064 234 048 290 816;
  • 43) 0.999 985 064 234 048 290 816 × 2 = 1 + 0.999 970 128 468 096 581 632;
  • 44) 0.999 970 128 468 096 581 632 × 2 = 1 + 0.999 940 256 936 193 163 264;
  • 45) 0.999 940 256 936 193 163 264 × 2 = 1 + 0.999 880 513 872 386 326 528;
  • 46) 0.999 880 513 872 386 326 528 × 2 = 1 + 0.999 761 027 744 772 653 056;
  • 47) 0.999 761 027 744 772 653 056 × 2 = 1 + 0.999 522 055 489 545 306 112;
  • 48) 0.999 522 055 489 545 306 112 × 2 = 1 + 0.999 044 110 979 090 612 224;
  • 49) 0.999 044 110 979 090 612 224 × 2 = 1 + 0.998 088 221 958 181 224 448;
  • 50) 0.998 088 221 958 181 224 448 × 2 = 1 + 0.996 176 443 916 362 448 896;
  • 51) 0.996 176 443 916 362 448 896 × 2 = 1 + 0.992 352 887 832 724 897 792;
  • 52) 0.992 352 887 832 724 897 792 × 2 = 1 + 0.984 705 775 665 449 795 584;
  • 53) 0.984 705 775 665 449 795 584 × 2 = 1 + 0.969 411 551 330 899 591 168;
  • 54) 0.969 411 551 330 899 591 168 × 2 = 1 + 0.938 823 102 661 799 182 336;
  • 55) 0.938 823 102 661 799 182 336 × 2 = 1 + 0.877 646 205 323 598 364 672;
  • 56) 0.877 646 205 323 598 364 672 × 2 = 1 + 0.755 292 410 647 196 729 344;
  • 57) 0.755 292 410 647 196 729 344 × 2 = 1 + 0.510 584 821 294 393 458 688;
  • 58) 0.510 584 821 294 393 458 688 × 2 = 1 + 0.021 169 642 588 786 917 376;

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:


0.016 738 891 601 562 496 604(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 604(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:


0.016 738 891 601 562 496 604(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 -0.016 738 891 601 562 496 604 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