-0.016 738 891 601 562 496 530 628 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 628(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
-0.016 738 891 601 562 496 530 628(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:

|-0.016 738 891 601 562 496 530 628| = 0.016 738 891 601 562 496 530 628


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: 0.016 738 891 601 562 496 530 628.

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) 0.016 738 891 601 562 496 530 628 × 2 = 0 + 0.033 477 783 203 124 993 061 256;
  • 2) 0.033 477 783 203 124 993 061 256 × 2 = 0 + 0.066 955 566 406 249 986 122 512;
  • 3) 0.066 955 566 406 249 986 122 512 × 2 = 0 + 0.133 911 132 812 499 972 245 024;
  • 4) 0.133 911 132 812 499 972 245 024 × 2 = 0 + 0.267 822 265 624 999 944 490 048;
  • 5) 0.267 822 265 624 999 944 490 048 × 2 = 0 + 0.535 644 531 249 999 888 980 096;
  • 6) 0.535 644 531 249 999 888 980 096 × 2 = 1 + 0.071 289 062 499 999 777 960 192;
  • 7) 0.071 289 062 499 999 777 960 192 × 2 = 0 + 0.142 578 124 999 999 555 920 384;
  • 8) 0.142 578 124 999 999 555 920 384 × 2 = 0 + 0.285 156 249 999 999 111 840 768;
  • 9) 0.285 156 249 999 999 111 840 768 × 2 = 0 + 0.570 312 499 999 998 223 681 536;
  • 10) 0.570 312 499 999 998 223 681 536 × 2 = 1 + 0.140 624 999 999 996 447 363 072;
  • 11) 0.140 624 999 999 996 447 363 072 × 2 = 0 + 0.281 249 999 999 992 894 726 144;
  • 12) 0.281 249 999 999 992 894 726 144 × 2 = 0 + 0.562 499 999 999 985 789 452 288;
  • 13) 0.562 499 999 999 985 789 452 288 × 2 = 1 + 0.124 999 999 999 971 578 904 576;
  • 14) 0.124 999 999 999 971 578 904 576 × 2 = 0 + 0.249 999 999 999 943 157 809 152;
  • 15) 0.249 999 999 999 943 157 809 152 × 2 = 0 + 0.499 999 999 999 886 315 618 304;
  • 16) 0.499 999 999 999 886 315 618 304 × 2 = 0 + 0.999 999 999 999 772 631 236 608;
  • 17) 0.999 999 999 999 772 631 236 608 × 2 = 1 + 0.999 999 999 999 545 262 473 216;
  • 18) 0.999 999 999 999 545 262 473 216 × 2 = 1 + 0.999 999 999 999 090 524 946 432;
  • 19) 0.999 999 999 999 090 524 946 432 × 2 = 1 + 0.999 999 999 998 181 049 892 864;
  • 20) 0.999 999 999 998 181 049 892 864 × 2 = 1 + 0.999 999 999 996 362 099 785 728;
  • 21) 0.999 999 999 996 362 099 785 728 × 2 = 1 + 0.999 999 999 992 724 199 571 456;
  • 22) 0.999 999 999 992 724 199 571 456 × 2 = 1 + 0.999 999 999 985 448 399 142 912;
  • 23) 0.999 999 999 985 448 399 142 912 × 2 = 1 + 0.999 999 999 970 896 798 285 824;
  • 24) 0.999 999 999 970 896 798 285 824 × 2 = 1 + 0.999 999 999 941 793 596 571 648;
  • 25) 0.999 999 999 941 793 596 571 648 × 2 = 1 + 0.999 999 999 883 587 193 143 296;
  • 26) 0.999 999 999 883 587 193 143 296 × 2 = 1 + 0.999 999 999 767 174 386 286 592;
  • 27) 0.999 999 999 767 174 386 286 592 × 2 = 1 + 0.999 999 999 534 348 772 573 184;
  • 28) 0.999 999 999 534 348 772 573 184 × 2 = 1 + 0.999 999 999 068 697 545 146 368;
  • 29) 0.999 999 999 068 697 545 146 368 × 2 = 1 + 0.999 999 998 137 395 090 292 736;
  • 30) 0.999 999 998 137 395 090 292 736 × 2 = 1 + 0.999 999 996 274 790 180 585 472;
  • 31) 0.999 999 996 274 790 180 585 472 × 2 = 1 + 0.999 999 992 549 580 361 170 944;
  • 32) 0.999 999 992 549 580 361 170 944 × 2 = 1 + 0.999 999 985 099 160 722 341 888;
  • 33) 0.999 999 985 099 160 722 341 888 × 2 = 1 + 0.999 999 970 198 321 444 683 776;
  • 34) 0.999 999 970 198 321 444 683 776 × 2 = 1 + 0.999 999 940 396 642 889 367 552;
  • 35) 0.999 999 940 396 642 889 367 552 × 2 = 1 + 0.999 999 880 793 285 778 735 104;
  • 36) 0.999 999 880 793 285 778 735 104 × 2 = 1 + 0.999 999 761 586 571 557 470 208;
  • 37) 0.999 999 761 586 571 557 470 208 × 2 = 1 + 0.999 999 523 173 143 114 940 416;
  • 38) 0.999 999 523 173 143 114 940 416 × 2 = 1 + 0.999 999 046 346 286 229 880 832;
  • 39) 0.999 999 046 346 286 229 880 832 × 2 = 1 + 0.999 998 092 692 572 459 761 664;
  • 40) 0.999 998 092 692 572 459 761 664 × 2 = 1 + 0.999 996 185 385 144 919 523 328;
  • 41) 0.999 996 185 385 144 919 523 328 × 2 = 1 + 0.999 992 370 770 289 839 046 656;
  • 42) 0.999 992 370 770 289 839 046 656 × 2 = 1 + 0.999 984 741 540 579 678 093 312;
  • 43) 0.999 984 741 540 579 678 093 312 × 2 = 1 + 0.999 969 483 081 159 356 186 624;
  • 44) 0.999 969 483 081 159 356 186 624 × 2 = 1 + 0.999 938 966 162 318 712 373 248;
  • 45) 0.999 938 966 162 318 712 373 248 × 2 = 1 + 0.999 877 932 324 637 424 746 496;
  • 46) 0.999 877 932 324 637 424 746 496 × 2 = 1 + 0.999 755 864 649 274 849 492 992;
  • 47) 0.999 755 864 649 274 849 492 992 × 2 = 1 + 0.999 511 729 298 549 698 985 984;
  • 48) 0.999 511 729 298 549 698 985 984 × 2 = 1 + 0.999 023 458 597 099 397 971 968;
  • 49) 0.999 023 458 597 099 397 971 968 × 2 = 1 + 0.998 046 917 194 198 795 943 936;
  • 50) 0.998 046 917 194 198 795 943 936 × 2 = 1 + 0.996 093 834 388 397 591 887 872;
  • 51) 0.996 093 834 388 397 591 887 872 × 2 = 1 + 0.992 187 668 776 795 183 775 744;
  • 52) 0.992 187 668 776 795 183 775 744 × 2 = 1 + 0.984 375 337 553 590 367 551 488;
  • 53) 0.984 375 337 553 590 367 551 488 × 2 = 1 + 0.968 750 675 107 180 735 102 976;
  • 54) 0.968 750 675 107 180 735 102 976 × 2 = 1 + 0.937 501 350 214 361 470 205 952;
  • 55) 0.937 501 350 214 361 470 205 952 × 2 = 1 + 0.875 002 700 428 722 940 411 904;
  • 56) 0.875 002 700 428 722 940 411 904 × 2 = 1 + 0.750 005 400 857 445 880 823 808;
  • 57) 0.750 005 400 857 445 880 823 808 × 2 = 1 + 0.500 010 801 714 891 761 647 616;
  • 58) 0.500 010 801 714 891 761 647 616 × 2 = 1 + 0.000 021 603 429 783 523 295 232;

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:


0.016 738 891 601 562 496 530 628(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 628(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:


0.016 738 891 601 562 496 530 628(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 -0.016 738 891 601 562 496 530 628 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