-0.016 738 891 601 562 496 851 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 851(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
-0.016 738 891 601 562 496 851(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:

|-0.016 738 891 601 562 496 851| = 0.016 738 891 601 562 496 851


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: 0.016 738 891 601 562 496 851.

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) 0.016 738 891 601 562 496 851 × 2 = 0 + 0.033 477 783 203 124 993 702;
  • 2) 0.033 477 783 203 124 993 702 × 2 = 0 + 0.066 955 566 406 249 987 404;
  • 3) 0.066 955 566 406 249 987 404 × 2 = 0 + 0.133 911 132 812 499 974 808;
  • 4) 0.133 911 132 812 499 974 808 × 2 = 0 + 0.267 822 265 624 999 949 616;
  • 5) 0.267 822 265 624 999 949 616 × 2 = 0 + 0.535 644 531 249 999 899 232;
  • 6) 0.535 644 531 249 999 899 232 × 2 = 1 + 0.071 289 062 499 999 798 464;
  • 7) 0.071 289 062 499 999 798 464 × 2 = 0 + 0.142 578 124 999 999 596 928;
  • 8) 0.142 578 124 999 999 596 928 × 2 = 0 + 0.285 156 249 999 999 193 856;
  • 9) 0.285 156 249 999 999 193 856 × 2 = 0 + 0.570 312 499 999 998 387 712;
  • 10) 0.570 312 499 999 998 387 712 × 2 = 1 + 0.140 624 999 999 996 775 424;
  • 11) 0.140 624 999 999 996 775 424 × 2 = 0 + 0.281 249 999 999 993 550 848;
  • 12) 0.281 249 999 999 993 550 848 × 2 = 0 + 0.562 499 999 999 987 101 696;
  • 13) 0.562 499 999 999 987 101 696 × 2 = 1 + 0.124 999 999 999 974 203 392;
  • 14) 0.124 999 999 999 974 203 392 × 2 = 0 + 0.249 999 999 999 948 406 784;
  • 15) 0.249 999 999 999 948 406 784 × 2 = 0 + 0.499 999 999 999 896 813 568;
  • 16) 0.499 999 999 999 896 813 568 × 2 = 0 + 0.999 999 999 999 793 627 136;
  • 17) 0.999 999 999 999 793 627 136 × 2 = 1 + 0.999 999 999 999 587 254 272;
  • 18) 0.999 999 999 999 587 254 272 × 2 = 1 + 0.999 999 999 999 174 508 544;
  • 19) 0.999 999 999 999 174 508 544 × 2 = 1 + 0.999 999 999 998 349 017 088;
  • 20) 0.999 999 999 998 349 017 088 × 2 = 1 + 0.999 999 999 996 698 034 176;
  • 21) 0.999 999 999 996 698 034 176 × 2 = 1 + 0.999 999 999 993 396 068 352;
  • 22) 0.999 999 999 993 396 068 352 × 2 = 1 + 0.999 999 999 986 792 136 704;
  • 23) 0.999 999 999 986 792 136 704 × 2 = 1 + 0.999 999 999 973 584 273 408;
  • 24) 0.999 999 999 973 584 273 408 × 2 = 1 + 0.999 999 999 947 168 546 816;
  • 25) 0.999 999 999 947 168 546 816 × 2 = 1 + 0.999 999 999 894 337 093 632;
  • 26) 0.999 999 999 894 337 093 632 × 2 = 1 + 0.999 999 999 788 674 187 264;
  • 27) 0.999 999 999 788 674 187 264 × 2 = 1 + 0.999 999 999 577 348 374 528;
  • 28) 0.999 999 999 577 348 374 528 × 2 = 1 + 0.999 999 999 154 696 749 056;
  • 29) 0.999 999 999 154 696 749 056 × 2 = 1 + 0.999 999 998 309 393 498 112;
  • 30) 0.999 999 998 309 393 498 112 × 2 = 1 + 0.999 999 996 618 786 996 224;
  • 31) 0.999 999 996 618 786 996 224 × 2 = 1 + 0.999 999 993 237 573 992 448;
  • 32) 0.999 999 993 237 573 992 448 × 2 = 1 + 0.999 999 986 475 147 984 896;
  • 33) 0.999 999 986 475 147 984 896 × 2 = 1 + 0.999 999 972 950 295 969 792;
  • 34) 0.999 999 972 950 295 969 792 × 2 = 1 + 0.999 999 945 900 591 939 584;
  • 35) 0.999 999 945 900 591 939 584 × 2 = 1 + 0.999 999 891 801 183 879 168;
  • 36) 0.999 999 891 801 183 879 168 × 2 = 1 + 0.999 999 783 602 367 758 336;
  • 37) 0.999 999 783 602 367 758 336 × 2 = 1 + 0.999 999 567 204 735 516 672;
  • 38) 0.999 999 567 204 735 516 672 × 2 = 1 + 0.999 999 134 409 471 033 344;
  • 39) 0.999 999 134 409 471 033 344 × 2 = 1 + 0.999 998 268 818 942 066 688;
  • 40) 0.999 998 268 818 942 066 688 × 2 = 1 + 0.999 996 537 637 884 133 376;
  • 41) 0.999 996 537 637 884 133 376 × 2 = 1 + 0.999 993 075 275 768 266 752;
  • 42) 0.999 993 075 275 768 266 752 × 2 = 1 + 0.999 986 150 551 536 533 504;
  • 43) 0.999 986 150 551 536 533 504 × 2 = 1 + 0.999 972 301 103 073 067 008;
  • 44) 0.999 972 301 103 073 067 008 × 2 = 1 + 0.999 944 602 206 146 134 016;
  • 45) 0.999 944 602 206 146 134 016 × 2 = 1 + 0.999 889 204 412 292 268 032;
  • 46) 0.999 889 204 412 292 268 032 × 2 = 1 + 0.999 778 408 824 584 536 064;
  • 47) 0.999 778 408 824 584 536 064 × 2 = 1 + 0.999 556 817 649 169 072 128;
  • 48) 0.999 556 817 649 169 072 128 × 2 = 1 + 0.999 113 635 298 338 144 256;
  • 49) 0.999 113 635 298 338 144 256 × 2 = 1 + 0.998 227 270 596 676 288 512;
  • 50) 0.998 227 270 596 676 288 512 × 2 = 1 + 0.996 454 541 193 352 577 024;
  • 51) 0.996 454 541 193 352 577 024 × 2 = 1 + 0.992 909 082 386 705 154 048;
  • 52) 0.992 909 082 386 705 154 048 × 2 = 1 + 0.985 818 164 773 410 308 096;
  • 53) 0.985 818 164 773 410 308 096 × 2 = 1 + 0.971 636 329 546 820 616 192;
  • 54) 0.971 636 329 546 820 616 192 × 2 = 1 + 0.943 272 659 093 641 232 384;
  • 55) 0.943 272 659 093 641 232 384 × 2 = 1 + 0.886 545 318 187 282 464 768;
  • 56) 0.886 545 318 187 282 464 768 × 2 = 1 + 0.773 090 636 374 564 929 536;
  • 57) 0.773 090 636 374 564 929 536 × 2 = 1 + 0.546 181 272 749 129 859 072;
  • 58) 0.546 181 272 749 129 859 072 × 2 = 1 + 0.092 362 545 498 259 718 144;

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:


0.016 738 891 601 562 496 851(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 851(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:


0.016 738 891 601 562 496 851(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 -0.016 738 891 601 562 496 851 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