-0.016 738 891 601 562 798 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 798(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
-0.016 738 891 601 562 798(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:

|-0.016 738 891 601 562 798| = 0.016 738 891 601 562 798


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: 0.016 738 891 601 562 798.

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) 0.016 738 891 601 562 798 × 2 = 0 + 0.033 477 783 203 125 596;
  • 2) 0.033 477 783 203 125 596 × 2 = 0 + 0.066 955 566 406 251 192;
  • 3) 0.066 955 566 406 251 192 × 2 = 0 + 0.133 911 132 812 502 384;
  • 4) 0.133 911 132 812 502 384 × 2 = 0 + 0.267 822 265 625 004 768;
  • 5) 0.267 822 265 625 004 768 × 2 = 0 + 0.535 644 531 250 009 536;
  • 6) 0.535 644 531 250 009 536 × 2 = 1 + 0.071 289 062 500 019 072;
  • 7) 0.071 289 062 500 019 072 × 2 = 0 + 0.142 578 125 000 038 144;
  • 8) 0.142 578 125 000 038 144 × 2 = 0 + 0.285 156 250 000 076 288;
  • 9) 0.285 156 250 000 076 288 × 2 = 0 + 0.570 312 500 000 152 576;
  • 10) 0.570 312 500 000 152 576 × 2 = 1 + 0.140 625 000 000 305 152;
  • 11) 0.140 625 000 000 305 152 × 2 = 0 + 0.281 250 000 000 610 304;
  • 12) 0.281 250 000 000 610 304 × 2 = 0 + 0.562 500 000 001 220 608;
  • 13) 0.562 500 000 001 220 608 × 2 = 1 + 0.125 000 000 002 441 216;
  • 14) 0.125 000 000 002 441 216 × 2 = 0 + 0.250 000 000 004 882 432;
  • 15) 0.250 000 000 004 882 432 × 2 = 0 + 0.500 000 000 009 764 864;
  • 16) 0.500 000 000 009 764 864 × 2 = 1 + 0.000 000 000 019 529 728;
  • 17) 0.000 000 000 019 529 728 × 2 = 0 + 0.000 000 000 039 059 456;
  • 18) 0.000 000 000 039 059 456 × 2 = 0 + 0.000 000 000 078 118 912;
  • 19) 0.000 000 000 078 118 912 × 2 = 0 + 0.000 000 000 156 237 824;
  • 20) 0.000 000 000 156 237 824 × 2 = 0 + 0.000 000 000 312 475 648;
  • 21) 0.000 000 000 312 475 648 × 2 = 0 + 0.000 000 000 624 951 296;
  • 22) 0.000 000 000 624 951 296 × 2 = 0 + 0.000 000 001 249 902 592;
  • 23) 0.000 000 001 249 902 592 × 2 = 0 + 0.000 000 002 499 805 184;
  • 24) 0.000 000 002 499 805 184 × 2 = 0 + 0.000 000 004 999 610 368;
  • 25) 0.000 000 004 999 610 368 × 2 = 0 + 0.000 000 009 999 220 736;
  • 26) 0.000 000 009 999 220 736 × 2 = 0 + 0.000 000 019 998 441 472;
  • 27) 0.000 000 019 998 441 472 × 2 = 0 + 0.000 000 039 996 882 944;
  • 28) 0.000 000 039 996 882 944 × 2 = 0 + 0.000 000 079 993 765 888;
  • 29) 0.000 000 079 993 765 888 × 2 = 0 + 0.000 000 159 987 531 776;
  • 30) 0.000 000 159 987 531 776 × 2 = 0 + 0.000 000 319 975 063 552;
  • 31) 0.000 000 319 975 063 552 × 2 = 0 + 0.000 000 639 950 127 104;
  • 32) 0.000 000 639 950 127 104 × 2 = 0 + 0.000 001 279 900 254 208;
  • 33) 0.000 001 279 900 254 208 × 2 = 0 + 0.000 002 559 800 508 416;
  • 34) 0.000 002 559 800 508 416 × 2 = 0 + 0.000 005 119 601 016 832;
  • 35) 0.000 005 119 601 016 832 × 2 = 0 + 0.000 010 239 202 033 664;
  • 36) 0.000 010 239 202 033 664 × 2 = 0 + 0.000 020 478 404 067 328;
  • 37) 0.000 020 478 404 067 328 × 2 = 0 + 0.000 040 956 808 134 656;
  • 38) 0.000 040 956 808 134 656 × 2 = 0 + 0.000 081 913 616 269 312;
  • 39) 0.000 081 913 616 269 312 × 2 = 0 + 0.000 163 827 232 538 624;
  • 40) 0.000 163 827 232 538 624 × 2 = 0 + 0.000 327 654 465 077 248;
  • 41) 0.000 327 654 465 077 248 × 2 = 0 + 0.000 655 308 930 154 496;
  • 42) 0.000 655 308 930 154 496 × 2 = 0 + 0.001 310 617 860 308 992;
  • 43) 0.001 310 617 860 308 992 × 2 = 0 + 0.002 621 235 720 617 984;
  • 44) 0.002 621 235 720 617 984 × 2 = 0 + 0.005 242 471 441 235 968;
  • 45) 0.005 242 471 441 235 968 × 2 = 0 + 0.010 484 942 882 471 936;
  • 46) 0.010 484 942 882 471 936 × 2 = 0 + 0.020 969 885 764 943 872;
  • 47) 0.020 969 885 764 943 872 × 2 = 0 + 0.041 939 771 529 887 744;
  • 48) 0.041 939 771 529 887 744 × 2 = 0 + 0.083 879 543 059 775 488;
  • 49) 0.083 879 543 059 775 488 × 2 = 0 + 0.167 759 086 119 550 976;
  • 50) 0.167 759 086 119 550 976 × 2 = 0 + 0.335 518 172 239 101 952;
  • 51) 0.335 518 172 239 101 952 × 2 = 0 + 0.671 036 344 478 203 904;
  • 52) 0.671 036 344 478 203 904 × 2 = 1 + 0.342 072 688 956 407 808;
  • 53) 0.342 072 688 956 407 808 × 2 = 0 + 0.684 145 377 912 815 616;
  • 54) 0.684 145 377 912 815 616 × 2 = 1 + 0.368 290 755 825 631 232;
  • 55) 0.368 290 755 825 631 232 × 2 = 0 + 0.736 581 511 651 262 464;
  • 56) 0.736 581 511 651 262 464 × 2 = 1 + 0.473 163 023 302 524 928;
  • 57) 0.473 163 023 302 524 928 × 2 = 0 + 0.946 326 046 605 049 856;
  • 58) 0.946 326 046 605 049 856 × 2 = 1 + 0.892 652 093 210 099 712;

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:


0.016 738 891 601 562 798(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0101 01(2)

6. Positive number before normalization:

0.016 738 891 601 562 798(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0101 01(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:


0.016 738 891 601 562 798(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0101 01(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0001 0101 01(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101(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 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101


Decimal number -0.016 738 891 601 562 798 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0101 0101

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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