-0.016 738 891 601 562 496 530 560 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 560 9(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
-0.016 738 891 601 562 496 530 560 9(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:

|-0.016 738 891 601 562 496 530 560 9| = 0.016 738 891 601 562 496 530 560 9


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: 0.016 738 891 601 562 496 530 560 9.

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) 0.016 738 891 601 562 496 530 560 9 × 2 = 0 + 0.033 477 783 203 124 993 061 121 8;
  • 2) 0.033 477 783 203 124 993 061 121 8 × 2 = 0 + 0.066 955 566 406 249 986 122 243 6;
  • 3) 0.066 955 566 406 249 986 122 243 6 × 2 = 0 + 0.133 911 132 812 499 972 244 487 2;
  • 4) 0.133 911 132 812 499 972 244 487 2 × 2 = 0 + 0.267 822 265 624 999 944 488 974 4;
  • 5) 0.267 822 265 624 999 944 488 974 4 × 2 = 0 + 0.535 644 531 249 999 888 977 948 8;
  • 6) 0.535 644 531 249 999 888 977 948 8 × 2 = 1 + 0.071 289 062 499 999 777 955 897 6;
  • 7) 0.071 289 062 499 999 777 955 897 6 × 2 = 0 + 0.142 578 124 999 999 555 911 795 2;
  • 8) 0.142 578 124 999 999 555 911 795 2 × 2 = 0 + 0.285 156 249 999 999 111 823 590 4;
  • 9) 0.285 156 249 999 999 111 823 590 4 × 2 = 0 + 0.570 312 499 999 998 223 647 180 8;
  • 10) 0.570 312 499 999 998 223 647 180 8 × 2 = 1 + 0.140 624 999 999 996 447 294 361 6;
  • 11) 0.140 624 999 999 996 447 294 361 6 × 2 = 0 + 0.281 249 999 999 992 894 588 723 2;
  • 12) 0.281 249 999 999 992 894 588 723 2 × 2 = 0 + 0.562 499 999 999 985 789 177 446 4;
  • 13) 0.562 499 999 999 985 789 177 446 4 × 2 = 1 + 0.124 999 999 999 971 578 354 892 8;
  • 14) 0.124 999 999 999 971 578 354 892 8 × 2 = 0 + 0.249 999 999 999 943 156 709 785 6;
  • 15) 0.249 999 999 999 943 156 709 785 6 × 2 = 0 + 0.499 999 999 999 886 313 419 571 2;
  • 16) 0.499 999 999 999 886 313 419 571 2 × 2 = 0 + 0.999 999 999 999 772 626 839 142 4;
  • 17) 0.999 999 999 999 772 626 839 142 4 × 2 = 1 + 0.999 999 999 999 545 253 678 284 8;
  • 18) 0.999 999 999 999 545 253 678 284 8 × 2 = 1 + 0.999 999 999 999 090 507 356 569 6;
  • 19) 0.999 999 999 999 090 507 356 569 6 × 2 = 1 + 0.999 999 999 998 181 014 713 139 2;
  • 20) 0.999 999 999 998 181 014 713 139 2 × 2 = 1 + 0.999 999 999 996 362 029 426 278 4;
  • 21) 0.999 999 999 996 362 029 426 278 4 × 2 = 1 + 0.999 999 999 992 724 058 852 556 8;
  • 22) 0.999 999 999 992 724 058 852 556 8 × 2 = 1 + 0.999 999 999 985 448 117 705 113 6;
  • 23) 0.999 999 999 985 448 117 705 113 6 × 2 = 1 + 0.999 999 999 970 896 235 410 227 2;
  • 24) 0.999 999 999 970 896 235 410 227 2 × 2 = 1 + 0.999 999 999 941 792 470 820 454 4;
  • 25) 0.999 999 999 941 792 470 820 454 4 × 2 = 1 + 0.999 999 999 883 584 941 640 908 8;
  • 26) 0.999 999 999 883 584 941 640 908 8 × 2 = 1 + 0.999 999 999 767 169 883 281 817 6;
  • 27) 0.999 999 999 767 169 883 281 817 6 × 2 = 1 + 0.999 999 999 534 339 766 563 635 2;
  • 28) 0.999 999 999 534 339 766 563 635 2 × 2 = 1 + 0.999 999 999 068 679 533 127 270 4;
  • 29) 0.999 999 999 068 679 533 127 270 4 × 2 = 1 + 0.999 999 998 137 359 066 254 540 8;
  • 30) 0.999 999 998 137 359 066 254 540 8 × 2 = 1 + 0.999 999 996 274 718 132 509 081 6;
  • 31) 0.999 999 996 274 718 132 509 081 6 × 2 = 1 + 0.999 999 992 549 436 265 018 163 2;
  • 32) 0.999 999 992 549 436 265 018 163 2 × 2 = 1 + 0.999 999 985 098 872 530 036 326 4;
  • 33) 0.999 999 985 098 872 530 036 326 4 × 2 = 1 + 0.999 999 970 197 745 060 072 652 8;
  • 34) 0.999 999 970 197 745 060 072 652 8 × 2 = 1 + 0.999 999 940 395 490 120 145 305 6;
  • 35) 0.999 999 940 395 490 120 145 305 6 × 2 = 1 + 0.999 999 880 790 980 240 290 611 2;
  • 36) 0.999 999 880 790 980 240 290 611 2 × 2 = 1 + 0.999 999 761 581 960 480 581 222 4;
  • 37) 0.999 999 761 581 960 480 581 222 4 × 2 = 1 + 0.999 999 523 163 920 961 162 444 8;
  • 38) 0.999 999 523 163 920 961 162 444 8 × 2 = 1 + 0.999 999 046 327 841 922 324 889 6;
  • 39) 0.999 999 046 327 841 922 324 889 6 × 2 = 1 + 0.999 998 092 655 683 844 649 779 2;
  • 40) 0.999 998 092 655 683 844 649 779 2 × 2 = 1 + 0.999 996 185 311 367 689 299 558 4;
  • 41) 0.999 996 185 311 367 689 299 558 4 × 2 = 1 + 0.999 992 370 622 735 378 599 116 8;
  • 42) 0.999 992 370 622 735 378 599 116 8 × 2 = 1 + 0.999 984 741 245 470 757 198 233 6;
  • 43) 0.999 984 741 245 470 757 198 233 6 × 2 = 1 + 0.999 969 482 490 941 514 396 467 2;
  • 44) 0.999 969 482 490 941 514 396 467 2 × 2 = 1 + 0.999 938 964 981 883 028 792 934 4;
  • 45) 0.999 938 964 981 883 028 792 934 4 × 2 = 1 + 0.999 877 929 963 766 057 585 868 8;
  • 46) 0.999 877 929 963 766 057 585 868 8 × 2 = 1 + 0.999 755 859 927 532 115 171 737 6;
  • 47) 0.999 755 859 927 532 115 171 737 6 × 2 = 1 + 0.999 511 719 855 064 230 343 475 2;
  • 48) 0.999 511 719 855 064 230 343 475 2 × 2 = 1 + 0.999 023 439 710 128 460 686 950 4;
  • 49) 0.999 023 439 710 128 460 686 950 4 × 2 = 1 + 0.998 046 879 420 256 921 373 900 8;
  • 50) 0.998 046 879 420 256 921 373 900 8 × 2 = 1 + 0.996 093 758 840 513 842 747 801 6;
  • 51) 0.996 093 758 840 513 842 747 801 6 × 2 = 1 + 0.992 187 517 681 027 685 495 603 2;
  • 52) 0.992 187 517 681 027 685 495 603 2 × 2 = 1 + 0.984 375 035 362 055 370 991 206 4;
  • 53) 0.984 375 035 362 055 370 991 206 4 × 2 = 1 + 0.968 750 070 724 110 741 982 412 8;
  • 54) 0.968 750 070 724 110 741 982 412 8 × 2 = 1 + 0.937 500 141 448 221 483 964 825 6;
  • 55) 0.937 500 141 448 221 483 964 825 6 × 2 = 1 + 0.875 000 282 896 442 967 929 651 2;
  • 56) 0.875 000 282 896 442 967 929 651 2 × 2 = 1 + 0.750 000 565 792 885 935 859 302 4;
  • 57) 0.750 000 565 792 885 935 859 302 4 × 2 = 1 + 0.500 001 131 585 771 871 718 604 8;
  • 58) 0.500 001 131 585 771 871 718 604 8 × 2 = 1 + 0.000 002 263 171 543 743 437 209 6;

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:


0.016 738 891 601 562 496 530 560 9(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 560 9(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:


0.016 738 891 601 562 496 530 560 9(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 -0.016 738 891 601 562 496 530 560 9 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