-0.016 738 891 601 562 496 530 668 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 668(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
-0.016 738 891 601 562 496 530 668(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:

|-0.016 738 891 601 562 496 530 668| = 0.016 738 891 601 562 496 530 668


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: 0.016 738 891 601 562 496 530 668.

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) 0.016 738 891 601 562 496 530 668 × 2 = 0 + 0.033 477 783 203 124 993 061 336;
  • 2) 0.033 477 783 203 124 993 061 336 × 2 = 0 + 0.066 955 566 406 249 986 122 672;
  • 3) 0.066 955 566 406 249 986 122 672 × 2 = 0 + 0.133 911 132 812 499 972 245 344;
  • 4) 0.133 911 132 812 499 972 245 344 × 2 = 0 + 0.267 822 265 624 999 944 490 688;
  • 5) 0.267 822 265 624 999 944 490 688 × 2 = 0 + 0.535 644 531 249 999 888 981 376;
  • 6) 0.535 644 531 249 999 888 981 376 × 2 = 1 + 0.071 289 062 499 999 777 962 752;
  • 7) 0.071 289 062 499 999 777 962 752 × 2 = 0 + 0.142 578 124 999 999 555 925 504;
  • 8) 0.142 578 124 999 999 555 925 504 × 2 = 0 + 0.285 156 249 999 999 111 851 008;
  • 9) 0.285 156 249 999 999 111 851 008 × 2 = 0 + 0.570 312 499 999 998 223 702 016;
  • 10) 0.570 312 499 999 998 223 702 016 × 2 = 1 + 0.140 624 999 999 996 447 404 032;
  • 11) 0.140 624 999 999 996 447 404 032 × 2 = 0 + 0.281 249 999 999 992 894 808 064;
  • 12) 0.281 249 999 999 992 894 808 064 × 2 = 0 + 0.562 499 999 999 985 789 616 128;
  • 13) 0.562 499 999 999 985 789 616 128 × 2 = 1 + 0.124 999 999 999 971 579 232 256;
  • 14) 0.124 999 999 999 971 579 232 256 × 2 = 0 + 0.249 999 999 999 943 158 464 512;
  • 15) 0.249 999 999 999 943 158 464 512 × 2 = 0 + 0.499 999 999 999 886 316 929 024;
  • 16) 0.499 999 999 999 886 316 929 024 × 2 = 0 + 0.999 999 999 999 772 633 858 048;
  • 17) 0.999 999 999 999 772 633 858 048 × 2 = 1 + 0.999 999 999 999 545 267 716 096;
  • 18) 0.999 999 999 999 545 267 716 096 × 2 = 1 + 0.999 999 999 999 090 535 432 192;
  • 19) 0.999 999 999 999 090 535 432 192 × 2 = 1 + 0.999 999 999 998 181 070 864 384;
  • 20) 0.999 999 999 998 181 070 864 384 × 2 = 1 + 0.999 999 999 996 362 141 728 768;
  • 21) 0.999 999 999 996 362 141 728 768 × 2 = 1 + 0.999 999 999 992 724 283 457 536;
  • 22) 0.999 999 999 992 724 283 457 536 × 2 = 1 + 0.999 999 999 985 448 566 915 072;
  • 23) 0.999 999 999 985 448 566 915 072 × 2 = 1 + 0.999 999 999 970 897 133 830 144;
  • 24) 0.999 999 999 970 897 133 830 144 × 2 = 1 + 0.999 999 999 941 794 267 660 288;
  • 25) 0.999 999 999 941 794 267 660 288 × 2 = 1 + 0.999 999 999 883 588 535 320 576;
  • 26) 0.999 999 999 883 588 535 320 576 × 2 = 1 + 0.999 999 999 767 177 070 641 152;
  • 27) 0.999 999 999 767 177 070 641 152 × 2 = 1 + 0.999 999 999 534 354 141 282 304;
  • 28) 0.999 999 999 534 354 141 282 304 × 2 = 1 + 0.999 999 999 068 708 282 564 608;
  • 29) 0.999 999 999 068 708 282 564 608 × 2 = 1 + 0.999 999 998 137 416 565 129 216;
  • 30) 0.999 999 998 137 416 565 129 216 × 2 = 1 + 0.999 999 996 274 833 130 258 432;
  • 31) 0.999 999 996 274 833 130 258 432 × 2 = 1 + 0.999 999 992 549 666 260 516 864;
  • 32) 0.999 999 992 549 666 260 516 864 × 2 = 1 + 0.999 999 985 099 332 521 033 728;
  • 33) 0.999 999 985 099 332 521 033 728 × 2 = 1 + 0.999 999 970 198 665 042 067 456;
  • 34) 0.999 999 970 198 665 042 067 456 × 2 = 1 + 0.999 999 940 397 330 084 134 912;
  • 35) 0.999 999 940 397 330 084 134 912 × 2 = 1 + 0.999 999 880 794 660 168 269 824;
  • 36) 0.999 999 880 794 660 168 269 824 × 2 = 1 + 0.999 999 761 589 320 336 539 648;
  • 37) 0.999 999 761 589 320 336 539 648 × 2 = 1 + 0.999 999 523 178 640 673 079 296;
  • 38) 0.999 999 523 178 640 673 079 296 × 2 = 1 + 0.999 999 046 357 281 346 158 592;
  • 39) 0.999 999 046 357 281 346 158 592 × 2 = 1 + 0.999 998 092 714 562 692 317 184;
  • 40) 0.999 998 092 714 562 692 317 184 × 2 = 1 + 0.999 996 185 429 125 384 634 368;
  • 41) 0.999 996 185 429 125 384 634 368 × 2 = 1 + 0.999 992 370 858 250 769 268 736;
  • 42) 0.999 992 370 858 250 769 268 736 × 2 = 1 + 0.999 984 741 716 501 538 537 472;
  • 43) 0.999 984 741 716 501 538 537 472 × 2 = 1 + 0.999 969 483 433 003 077 074 944;
  • 44) 0.999 969 483 433 003 077 074 944 × 2 = 1 + 0.999 938 966 866 006 154 149 888;
  • 45) 0.999 938 966 866 006 154 149 888 × 2 = 1 + 0.999 877 933 732 012 308 299 776;
  • 46) 0.999 877 933 732 012 308 299 776 × 2 = 1 + 0.999 755 867 464 024 616 599 552;
  • 47) 0.999 755 867 464 024 616 599 552 × 2 = 1 + 0.999 511 734 928 049 233 199 104;
  • 48) 0.999 511 734 928 049 233 199 104 × 2 = 1 + 0.999 023 469 856 098 466 398 208;
  • 49) 0.999 023 469 856 098 466 398 208 × 2 = 1 + 0.998 046 939 712 196 932 796 416;
  • 50) 0.998 046 939 712 196 932 796 416 × 2 = 1 + 0.996 093 879 424 393 865 592 832;
  • 51) 0.996 093 879 424 393 865 592 832 × 2 = 1 + 0.992 187 758 848 787 731 185 664;
  • 52) 0.992 187 758 848 787 731 185 664 × 2 = 1 + 0.984 375 517 697 575 462 371 328;
  • 53) 0.984 375 517 697 575 462 371 328 × 2 = 1 + 0.968 751 035 395 150 924 742 656;
  • 54) 0.968 751 035 395 150 924 742 656 × 2 = 1 + 0.937 502 070 790 301 849 485 312;
  • 55) 0.937 502 070 790 301 849 485 312 × 2 = 1 + 0.875 004 141 580 603 698 970 624;
  • 56) 0.875 004 141 580 603 698 970 624 × 2 = 1 + 0.750 008 283 161 207 397 941 248;
  • 57) 0.750 008 283 161 207 397 941 248 × 2 = 1 + 0.500 016 566 322 414 795 882 496;
  • 58) 0.500 016 566 322 414 795 882 496 × 2 = 1 + 0.000 033 132 644 829 591 764 992;

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:


0.016 738 891 601 562 496 530 668(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 668(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:


0.016 738 891 601 562 496 530 668(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 -0.016 738 891 601 562 496 530 668 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