-0.016 738 891 601 562 496 583 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 583(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
-0.016 738 891 601 562 496 583(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:

|-0.016 738 891 601 562 496 583| = 0.016 738 891 601 562 496 583


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: 0.016 738 891 601 562 496 583.

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) 0.016 738 891 601 562 496 583 × 2 = 0 + 0.033 477 783 203 124 993 166;
  • 2) 0.033 477 783 203 124 993 166 × 2 = 0 + 0.066 955 566 406 249 986 332;
  • 3) 0.066 955 566 406 249 986 332 × 2 = 0 + 0.133 911 132 812 499 972 664;
  • 4) 0.133 911 132 812 499 972 664 × 2 = 0 + 0.267 822 265 624 999 945 328;
  • 5) 0.267 822 265 624 999 945 328 × 2 = 0 + 0.535 644 531 249 999 890 656;
  • 6) 0.535 644 531 249 999 890 656 × 2 = 1 + 0.071 289 062 499 999 781 312;
  • 7) 0.071 289 062 499 999 781 312 × 2 = 0 + 0.142 578 124 999 999 562 624;
  • 8) 0.142 578 124 999 999 562 624 × 2 = 0 + 0.285 156 249 999 999 125 248;
  • 9) 0.285 156 249 999 999 125 248 × 2 = 0 + 0.570 312 499 999 998 250 496;
  • 10) 0.570 312 499 999 998 250 496 × 2 = 1 + 0.140 624 999 999 996 500 992;
  • 11) 0.140 624 999 999 996 500 992 × 2 = 0 + 0.281 249 999 999 993 001 984;
  • 12) 0.281 249 999 999 993 001 984 × 2 = 0 + 0.562 499 999 999 986 003 968;
  • 13) 0.562 499 999 999 986 003 968 × 2 = 1 + 0.124 999 999 999 972 007 936;
  • 14) 0.124 999 999 999 972 007 936 × 2 = 0 + 0.249 999 999 999 944 015 872;
  • 15) 0.249 999 999 999 944 015 872 × 2 = 0 + 0.499 999 999 999 888 031 744;
  • 16) 0.499 999 999 999 888 031 744 × 2 = 0 + 0.999 999 999 999 776 063 488;
  • 17) 0.999 999 999 999 776 063 488 × 2 = 1 + 0.999 999 999 999 552 126 976;
  • 18) 0.999 999 999 999 552 126 976 × 2 = 1 + 0.999 999 999 999 104 253 952;
  • 19) 0.999 999 999 999 104 253 952 × 2 = 1 + 0.999 999 999 998 208 507 904;
  • 20) 0.999 999 999 998 208 507 904 × 2 = 1 + 0.999 999 999 996 417 015 808;
  • 21) 0.999 999 999 996 417 015 808 × 2 = 1 + 0.999 999 999 992 834 031 616;
  • 22) 0.999 999 999 992 834 031 616 × 2 = 1 + 0.999 999 999 985 668 063 232;
  • 23) 0.999 999 999 985 668 063 232 × 2 = 1 + 0.999 999 999 971 336 126 464;
  • 24) 0.999 999 999 971 336 126 464 × 2 = 1 + 0.999 999 999 942 672 252 928;
  • 25) 0.999 999 999 942 672 252 928 × 2 = 1 + 0.999 999 999 885 344 505 856;
  • 26) 0.999 999 999 885 344 505 856 × 2 = 1 + 0.999 999 999 770 689 011 712;
  • 27) 0.999 999 999 770 689 011 712 × 2 = 1 + 0.999 999 999 541 378 023 424;
  • 28) 0.999 999 999 541 378 023 424 × 2 = 1 + 0.999 999 999 082 756 046 848;
  • 29) 0.999 999 999 082 756 046 848 × 2 = 1 + 0.999 999 998 165 512 093 696;
  • 30) 0.999 999 998 165 512 093 696 × 2 = 1 + 0.999 999 996 331 024 187 392;
  • 31) 0.999 999 996 331 024 187 392 × 2 = 1 + 0.999 999 992 662 048 374 784;
  • 32) 0.999 999 992 662 048 374 784 × 2 = 1 + 0.999 999 985 324 096 749 568;
  • 33) 0.999 999 985 324 096 749 568 × 2 = 1 + 0.999 999 970 648 193 499 136;
  • 34) 0.999 999 970 648 193 499 136 × 2 = 1 + 0.999 999 941 296 386 998 272;
  • 35) 0.999 999 941 296 386 998 272 × 2 = 1 + 0.999 999 882 592 773 996 544;
  • 36) 0.999 999 882 592 773 996 544 × 2 = 1 + 0.999 999 765 185 547 993 088;
  • 37) 0.999 999 765 185 547 993 088 × 2 = 1 + 0.999 999 530 371 095 986 176;
  • 38) 0.999 999 530 371 095 986 176 × 2 = 1 + 0.999 999 060 742 191 972 352;
  • 39) 0.999 999 060 742 191 972 352 × 2 = 1 + 0.999 998 121 484 383 944 704;
  • 40) 0.999 998 121 484 383 944 704 × 2 = 1 + 0.999 996 242 968 767 889 408;
  • 41) 0.999 996 242 968 767 889 408 × 2 = 1 + 0.999 992 485 937 535 778 816;
  • 42) 0.999 992 485 937 535 778 816 × 2 = 1 + 0.999 984 971 875 071 557 632;
  • 43) 0.999 984 971 875 071 557 632 × 2 = 1 + 0.999 969 943 750 143 115 264;
  • 44) 0.999 969 943 750 143 115 264 × 2 = 1 + 0.999 939 887 500 286 230 528;
  • 45) 0.999 939 887 500 286 230 528 × 2 = 1 + 0.999 879 775 000 572 461 056;
  • 46) 0.999 879 775 000 572 461 056 × 2 = 1 + 0.999 759 550 001 144 922 112;
  • 47) 0.999 759 550 001 144 922 112 × 2 = 1 + 0.999 519 100 002 289 844 224;
  • 48) 0.999 519 100 002 289 844 224 × 2 = 1 + 0.999 038 200 004 579 688 448;
  • 49) 0.999 038 200 004 579 688 448 × 2 = 1 + 0.998 076 400 009 159 376 896;
  • 50) 0.998 076 400 009 159 376 896 × 2 = 1 + 0.996 152 800 018 318 753 792;
  • 51) 0.996 152 800 018 318 753 792 × 2 = 1 + 0.992 305 600 036 637 507 584;
  • 52) 0.992 305 600 036 637 507 584 × 2 = 1 + 0.984 611 200 073 275 015 168;
  • 53) 0.984 611 200 073 275 015 168 × 2 = 1 + 0.969 222 400 146 550 030 336;
  • 54) 0.969 222 400 146 550 030 336 × 2 = 1 + 0.938 444 800 293 100 060 672;
  • 55) 0.938 444 800 293 100 060 672 × 2 = 1 + 0.876 889 600 586 200 121 344;
  • 56) 0.876 889 600 586 200 121 344 × 2 = 1 + 0.753 779 201 172 400 242 688;
  • 57) 0.753 779 201 172 400 242 688 × 2 = 1 + 0.507 558 402 344 800 485 376;
  • 58) 0.507 558 402 344 800 485 376 × 2 = 1 + 0.015 116 804 689 600 970 752;

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:


0.016 738 891 601 562 496 583(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 583(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:


0.016 738 891 601 562 496 583(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 -0.016 738 891 601 562 496 583 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