-0.016 738 891 601 562 496 681 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 681(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
-0.016 738 891 601 562 496 681(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:

|-0.016 738 891 601 562 496 681| = 0.016 738 891 601 562 496 681


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: 0.016 738 891 601 562 496 681.

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) 0.016 738 891 601 562 496 681 × 2 = 0 + 0.033 477 783 203 124 993 362;
  • 2) 0.033 477 783 203 124 993 362 × 2 = 0 + 0.066 955 566 406 249 986 724;
  • 3) 0.066 955 566 406 249 986 724 × 2 = 0 + 0.133 911 132 812 499 973 448;
  • 4) 0.133 911 132 812 499 973 448 × 2 = 0 + 0.267 822 265 624 999 946 896;
  • 5) 0.267 822 265 624 999 946 896 × 2 = 0 + 0.535 644 531 249 999 893 792;
  • 6) 0.535 644 531 249 999 893 792 × 2 = 1 + 0.071 289 062 499 999 787 584;
  • 7) 0.071 289 062 499 999 787 584 × 2 = 0 + 0.142 578 124 999 999 575 168;
  • 8) 0.142 578 124 999 999 575 168 × 2 = 0 + 0.285 156 249 999 999 150 336;
  • 9) 0.285 156 249 999 999 150 336 × 2 = 0 + 0.570 312 499 999 998 300 672;
  • 10) 0.570 312 499 999 998 300 672 × 2 = 1 + 0.140 624 999 999 996 601 344;
  • 11) 0.140 624 999 999 996 601 344 × 2 = 0 + 0.281 249 999 999 993 202 688;
  • 12) 0.281 249 999 999 993 202 688 × 2 = 0 + 0.562 499 999 999 986 405 376;
  • 13) 0.562 499 999 999 986 405 376 × 2 = 1 + 0.124 999 999 999 972 810 752;
  • 14) 0.124 999 999 999 972 810 752 × 2 = 0 + 0.249 999 999 999 945 621 504;
  • 15) 0.249 999 999 999 945 621 504 × 2 = 0 + 0.499 999 999 999 891 243 008;
  • 16) 0.499 999 999 999 891 243 008 × 2 = 0 + 0.999 999 999 999 782 486 016;
  • 17) 0.999 999 999 999 782 486 016 × 2 = 1 + 0.999 999 999 999 564 972 032;
  • 18) 0.999 999 999 999 564 972 032 × 2 = 1 + 0.999 999 999 999 129 944 064;
  • 19) 0.999 999 999 999 129 944 064 × 2 = 1 + 0.999 999 999 998 259 888 128;
  • 20) 0.999 999 999 998 259 888 128 × 2 = 1 + 0.999 999 999 996 519 776 256;
  • 21) 0.999 999 999 996 519 776 256 × 2 = 1 + 0.999 999 999 993 039 552 512;
  • 22) 0.999 999 999 993 039 552 512 × 2 = 1 + 0.999 999 999 986 079 105 024;
  • 23) 0.999 999 999 986 079 105 024 × 2 = 1 + 0.999 999 999 972 158 210 048;
  • 24) 0.999 999 999 972 158 210 048 × 2 = 1 + 0.999 999 999 944 316 420 096;
  • 25) 0.999 999 999 944 316 420 096 × 2 = 1 + 0.999 999 999 888 632 840 192;
  • 26) 0.999 999 999 888 632 840 192 × 2 = 1 + 0.999 999 999 777 265 680 384;
  • 27) 0.999 999 999 777 265 680 384 × 2 = 1 + 0.999 999 999 554 531 360 768;
  • 28) 0.999 999 999 554 531 360 768 × 2 = 1 + 0.999 999 999 109 062 721 536;
  • 29) 0.999 999 999 109 062 721 536 × 2 = 1 + 0.999 999 998 218 125 443 072;
  • 30) 0.999 999 998 218 125 443 072 × 2 = 1 + 0.999 999 996 436 250 886 144;
  • 31) 0.999 999 996 436 250 886 144 × 2 = 1 + 0.999 999 992 872 501 772 288;
  • 32) 0.999 999 992 872 501 772 288 × 2 = 1 + 0.999 999 985 745 003 544 576;
  • 33) 0.999 999 985 745 003 544 576 × 2 = 1 + 0.999 999 971 490 007 089 152;
  • 34) 0.999 999 971 490 007 089 152 × 2 = 1 + 0.999 999 942 980 014 178 304;
  • 35) 0.999 999 942 980 014 178 304 × 2 = 1 + 0.999 999 885 960 028 356 608;
  • 36) 0.999 999 885 960 028 356 608 × 2 = 1 + 0.999 999 771 920 056 713 216;
  • 37) 0.999 999 771 920 056 713 216 × 2 = 1 + 0.999 999 543 840 113 426 432;
  • 38) 0.999 999 543 840 113 426 432 × 2 = 1 + 0.999 999 087 680 226 852 864;
  • 39) 0.999 999 087 680 226 852 864 × 2 = 1 + 0.999 998 175 360 453 705 728;
  • 40) 0.999 998 175 360 453 705 728 × 2 = 1 + 0.999 996 350 720 907 411 456;
  • 41) 0.999 996 350 720 907 411 456 × 2 = 1 + 0.999 992 701 441 814 822 912;
  • 42) 0.999 992 701 441 814 822 912 × 2 = 1 + 0.999 985 402 883 629 645 824;
  • 43) 0.999 985 402 883 629 645 824 × 2 = 1 + 0.999 970 805 767 259 291 648;
  • 44) 0.999 970 805 767 259 291 648 × 2 = 1 + 0.999 941 611 534 518 583 296;
  • 45) 0.999 941 611 534 518 583 296 × 2 = 1 + 0.999 883 223 069 037 166 592;
  • 46) 0.999 883 223 069 037 166 592 × 2 = 1 + 0.999 766 446 138 074 333 184;
  • 47) 0.999 766 446 138 074 333 184 × 2 = 1 + 0.999 532 892 276 148 666 368;
  • 48) 0.999 532 892 276 148 666 368 × 2 = 1 + 0.999 065 784 552 297 332 736;
  • 49) 0.999 065 784 552 297 332 736 × 2 = 1 + 0.998 131 569 104 594 665 472;
  • 50) 0.998 131 569 104 594 665 472 × 2 = 1 + 0.996 263 138 209 189 330 944;
  • 51) 0.996 263 138 209 189 330 944 × 2 = 1 + 0.992 526 276 418 378 661 888;
  • 52) 0.992 526 276 418 378 661 888 × 2 = 1 + 0.985 052 552 836 757 323 776;
  • 53) 0.985 052 552 836 757 323 776 × 2 = 1 + 0.970 105 105 673 514 647 552;
  • 54) 0.970 105 105 673 514 647 552 × 2 = 1 + 0.940 210 211 347 029 295 104;
  • 55) 0.940 210 211 347 029 295 104 × 2 = 1 + 0.880 420 422 694 058 590 208;
  • 56) 0.880 420 422 694 058 590 208 × 2 = 1 + 0.760 840 845 388 117 180 416;
  • 57) 0.760 840 845 388 117 180 416 × 2 = 1 + 0.521 681 690 776 234 360 832;
  • 58) 0.521 681 690 776 234 360 832 × 2 = 1 + 0.043 363 381 552 468 721 664;

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:


0.016 738 891 601 562 496 681(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 681(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:


0.016 738 891 601 562 496 681(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 -0.016 738 891 601 562 496 681 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