-0.016 738 891 601 562 496 548 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 548(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
-0.016 738 891 601 562 496 548(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:

|-0.016 738 891 601 562 496 548| = 0.016 738 891 601 562 496 548


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: 0.016 738 891 601 562 496 548.

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) 0.016 738 891 601 562 496 548 × 2 = 0 + 0.033 477 783 203 124 993 096;
  • 2) 0.033 477 783 203 124 993 096 × 2 = 0 + 0.066 955 566 406 249 986 192;
  • 3) 0.066 955 566 406 249 986 192 × 2 = 0 + 0.133 911 132 812 499 972 384;
  • 4) 0.133 911 132 812 499 972 384 × 2 = 0 + 0.267 822 265 624 999 944 768;
  • 5) 0.267 822 265 624 999 944 768 × 2 = 0 + 0.535 644 531 249 999 889 536;
  • 6) 0.535 644 531 249 999 889 536 × 2 = 1 + 0.071 289 062 499 999 779 072;
  • 7) 0.071 289 062 499 999 779 072 × 2 = 0 + 0.142 578 124 999 999 558 144;
  • 8) 0.142 578 124 999 999 558 144 × 2 = 0 + 0.285 156 249 999 999 116 288;
  • 9) 0.285 156 249 999 999 116 288 × 2 = 0 + 0.570 312 499 999 998 232 576;
  • 10) 0.570 312 499 999 998 232 576 × 2 = 1 + 0.140 624 999 999 996 465 152;
  • 11) 0.140 624 999 999 996 465 152 × 2 = 0 + 0.281 249 999 999 992 930 304;
  • 12) 0.281 249 999 999 992 930 304 × 2 = 0 + 0.562 499 999 999 985 860 608;
  • 13) 0.562 499 999 999 985 860 608 × 2 = 1 + 0.124 999 999 999 971 721 216;
  • 14) 0.124 999 999 999 971 721 216 × 2 = 0 + 0.249 999 999 999 943 442 432;
  • 15) 0.249 999 999 999 943 442 432 × 2 = 0 + 0.499 999 999 999 886 884 864;
  • 16) 0.499 999 999 999 886 884 864 × 2 = 0 + 0.999 999 999 999 773 769 728;
  • 17) 0.999 999 999 999 773 769 728 × 2 = 1 + 0.999 999 999 999 547 539 456;
  • 18) 0.999 999 999 999 547 539 456 × 2 = 1 + 0.999 999 999 999 095 078 912;
  • 19) 0.999 999 999 999 095 078 912 × 2 = 1 + 0.999 999 999 998 190 157 824;
  • 20) 0.999 999 999 998 190 157 824 × 2 = 1 + 0.999 999 999 996 380 315 648;
  • 21) 0.999 999 999 996 380 315 648 × 2 = 1 + 0.999 999 999 992 760 631 296;
  • 22) 0.999 999 999 992 760 631 296 × 2 = 1 + 0.999 999 999 985 521 262 592;
  • 23) 0.999 999 999 985 521 262 592 × 2 = 1 + 0.999 999 999 971 042 525 184;
  • 24) 0.999 999 999 971 042 525 184 × 2 = 1 + 0.999 999 999 942 085 050 368;
  • 25) 0.999 999 999 942 085 050 368 × 2 = 1 + 0.999 999 999 884 170 100 736;
  • 26) 0.999 999 999 884 170 100 736 × 2 = 1 + 0.999 999 999 768 340 201 472;
  • 27) 0.999 999 999 768 340 201 472 × 2 = 1 + 0.999 999 999 536 680 402 944;
  • 28) 0.999 999 999 536 680 402 944 × 2 = 1 + 0.999 999 999 073 360 805 888;
  • 29) 0.999 999 999 073 360 805 888 × 2 = 1 + 0.999 999 998 146 721 611 776;
  • 30) 0.999 999 998 146 721 611 776 × 2 = 1 + 0.999 999 996 293 443 223 552;
  • 31) 0.999 999 996 293 443 223 552 × 2 = 1 + 0.999 999 992 586 886 447 104;
  • 32) 0.999 999 992 586 886 447 104 × 2 = 1 + 0.999 999 985 173 772 894 208;
  • 33) 0.999 999 985 173 772 894 208 × 2 = 1 + 0.999 999 970 347 545 788 416;
  • 34) 0.999 999 970 347 545 788 416 × 2 = 1 + 0.999 999 940 695 091 576 832;
  • 35) 0.999 999 940 695 091 576 832 × 2 = 1 + 0.999 999 881 390 183 153 664;
  • 36) 0.999 999 881 390 183 153 664 × 2 = 1 + 0.999 999 762 780 366 307 328;
  • 37) 0.999 999 762 780 366 307 328 × 2 = 1 + 0.999 999 525 560 732 614 656;
  • 38) 0.999 999 525 560 732 614 656 × 2 = 1 + 0.999 999 051 121 465 229 312;
  • 39) 0.999 999 051 121 465 229 312 × 2 = 1 + 0.999 998 102 242 930 458 624;
  • 40) 0.999 998 102 242 930 458 624 × 2 = 1 + 0.999 996 204 485 860 917 248;
  • 41) 0.999 996 204 485 860 917 248 × 2 = 1 + 0.999 992 408 971 721 834 496;
  • 42) 0.999 992 408 971 721 834 496 × 2 = 1 + 0.999 984 817 943 443 668 992;
  • 43) 0.999 984 817 943 443 668 992 × 2 = 1 + 0.999 969 635 886 887 337 984;
  • 44) 0.999 969 635 886 887 337 984 × 2 = 1 + 0.999 939 271 773 774 675 968;
  • 45) 0.999 939 271 773 774 675 968 × 2 = 1 + 0.999 878 543 547 549 351 936;
  • 46) 0.999 878 543 547 549 351 936 × 2 = 1 + 0.999 757 087 095 098 703 872;
  • 47) 0.999 757 087 095 098 703 872 × 2 = 1 + 0.999 514 174 190 197 407 744;
  • 48) 0.999 514 174 190 197 407 744 × 2 = 1 + 0.999 028 348 380 394 815 488;
  • 49) 0.999 028 348 380 394 815 488 × 2 = 1 + 0.998 056 696 760 789 630 976;
  • 50) 0.998 056 696 760 789 630 976 × 2 = 1 + 0.996 113 393 521 579 261 952;
  • 51) 0.996 113 393 521 579 261 952 × 2 = 1 + 0.992 226 787 043 158 523 904;
  • 52) 0.992 226 787 043 158 523 904 × 2 = 1 + 0.984 453 574 086 317 047 808;
  • 53) 0.984 453 574 086 317 047 808 × 2 = 1 + 0.968 907 148 172 634 095 616;
  • 54) 0.968 907 148 172 634 095 616 × 2 = 1 + 0.937 814 296 345 268 191 232;
  • 55) 0.937 814 296 345 268 191 232 × 2 = 1 + 0.875 628 592 690 536 382 464;
  • 56) 0.875 628 592 690 536 382 464 × 2 = 1 + 0.751 257 185 381 072 764 928;
  • 57) 0.751 257 185 381 072 764 928 × 2 = 1 + 0.502 514 370 762 145 529 856;
  • 58) 0.502 514 370 762 145 529 856 × 2 = 1 + 0.005 028 741 524 291 059 712;

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:


0.016 738 891 601 562 496 548(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 548(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:


0.016 738 891 601 562 496 548(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 -0.016 738 891 601 562 496 548 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