-0.016 738 891 601 562 495 01 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 495 01(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
-0.016 738 891 601 562 495 01(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:

|-0.016 738 891 601 562 495 01| = 0.016 738 891 601 562 495 01


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: 0.016 738 891 601 562 495 01.

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) 0.016 738 891 601 562 495 01 × 2 = 0 + 0.033 477 783 203 124 990 02;
  • 2) 0.033 477 783 203 124 990 02 × 2 = 0 + 0.066 955 566 406 249 980 04;
  • 3) 0.066 955 566 406 249 980 04 × 2 = 0 + 0.133 911 132 812 499 960 08;
  • 4) 0.133 911 132 812 499 960 08 × 2 = 0 + 0.267 822 265 624 999 920 16;
  • 5) 0.267 822 265 624 999 920 16 × 2 = 0 + 0.535 644 531 249 999 840 32;
  • 6) 0.535 644 531 249 999 840 32 × 2 = 1 + 0.071 289 062 499 999 680 64;
  • 7) 0.071 289 062 499 999 680 64 × 2 = 0 + 0.142 578 124 999 999 361 28;
  • 8) 0.142 578 124 999 999 361 28 × 2 = 0 + 0.285 156 249 999 998 722 56;
  • 9) 0.285 156 249 999 998 722 56 × 2 = 0 + 0.570 312 499 999 997 445 12;
  • 10) 0.570 312 499 999 997 445 12 × 2 = 1 + 0.140 624 999 999 994 890 24;
  • 11) 0.140 624 999 999 994 890 24 × 2 = 0 + 0.281 249 999 999 989 780 48;
  • 12) 0.281 249 999 999 989 780 48 × 2 = 0 + 0.562 499 999 999 979 560 96;
  • 13) 0.562 499 999 999 979 560 96 × 2 = 1 + 0.124 999 999 999 959 121 92;
  • 14) 0.124 999 999 999 959 121 92 × 2 = 0 + 0.249 999 999 999 918 243 84;
  • 15) 0.249 999 999 999 918 243 84 × 2 = 0 + 0.499 999 999 999 836 487 68;
  • 16) 0.499 999 999 999 836 487 68 × 2 = 0 + 0.999 999 999 999 672 975 36;
  • 17) 0.999 999 999 999 672 975 36 × 2 = 1 + 0.999 999 999 999 345 950 72;
  • 18) 0.999 999 999 999 345 950 72 × 2 = 1 + 0.999 999 999 998 691 901 44;
  • 19) 0.999 999 999 998 691 901 44 × 2 = 1 + 0.999 999 999 997 383 802 88;
  • 20) 0.999 999 999 997 383 802 88 × 2 = 1 + 0.999 999 999 994 767 605 76;
  • 21) 0.999 999 999 994 767 605 76 × 2 = 1 + 0.999 999 999 989 535 211 52;
  • 22) 0.999 999 999 989 535 211 52 × 2 = 1 + 0.999 999 999 979 070 423 04;
  • 23) 0.999 999 999 979 070 423 04 × 2 = 1 + 0.999 999 999 958 140 846 08;
  • 24) 0.999 999 999 958 140 846 08 × 2 = 1 + 0.999 999 999 916 281 692 16;
  • 25) 0.999 999 999 916 281 692 16 × 2 = 1 + 0.999 999 999 832 563 384 32;
  • 26) 0.999 999 999 832 563 384 32 × 2 = 1 + 0.999 999 999 665 126 768 64;
  • 27) 0.999 999 999 665 126 768 64 × 2 = 1 + 0.999 999 999 330 253 537 28;
  • 28) 0.999 999 999 330 253 537 28 × 2 = 1 + 0.999 999 998 660 507 074 56;
  • 29) 0.999 999 998 660 507 074 56 × 2 = 1 + 0.999 999 997 321 014 149 12;
  • 30) 0.999 999 997 321 014 149 12 × 2 = 1 + 0.999 999 994 642 028 298 24;
  • 31) 0.999 999 994 642 028 298 24 × 2 = 1 + 0.999 999 989 284 056 596 48;
  • 32) 0.999 999 989 284 056 596 48 × 2 = 1 + 0.999 999 978 568 113 192 96;
  • 33) 0.999 999 978 568 113 192 96 × 2 = 1 + 0.999 999 957 136 226 385 92;
  • 34) 0.999 999 957 136 226 385 92 × 2 = 1 + 0.999 999 914 272 452 771 84;
  • 35) 0.999 999 914 272 452 771 84 × 2 = 1 + 0.999 999 828 544 905 543 68;
  • 36) 0.999 999 828 544 905 543 68 × 2 = 1 + 0.999 999 657 089 811 087 36;
  • 37) 0.999 999 657 089 811 087 36 × 2 = 1 + 0.999 999 314 179 622 174 72;
  • 38) 0.999 999 314 179 622 174 72 × 2 = 1 + 0.999 998 628 359 244 349 44;
  • 39) 0.999 998 628 359 244 349 44 × 2 = 1 + 0.999 997 256 718 488 698 88;
  • 40) 0.999 997 256 718 488 698 88 × 2 = 1 + 0.999 994 513 436 977 397 76;
  • 41) 0.999 994 513 436 977 397 76 × 2 = 1 + 0.999 989 026 873 954 795 52;
  • 42) 0.999 989 026 873 954 795 52 × 2 = 1 + 0.999 978 053 747 909 591 04;
  • 43) 0.999 978 053 747 909 591 04 × 2 = 1 + 0.999 956 107 495 819 182 08;
  • 44) 0.999 956 107 495 819 182 08 × 2 = 1 + 0.999 912 214 991 638 364 16;
  • 45) 0.999 912 214 991 638 364 16 × 2 = 1 + 0.999 824 429 983 276 728 32;
  • 46) 0.999 824 429 983 276 728 32 × 2 = 1 + 0.999 648 859 966 553 456 64;
  • 47) 0.999 648 859 966 553 456 64 × 2 = 1 + 0.999 297 719 933 106 913 28;
  • 48) 0.999 297 719 933 106 913 28 × 2 = 1 + 0.998 595 439 866 213 826 56;
  • 49) 0.998 595 439 866 213 826 56 × 2 = 1 + 0.997 190 879 732 427 653 12;
  • 50) 0.997 190 879 732 427 653 12 × 2 = 1 + 0.994 381 759 464 855 306 24;
  • 51) 0.994 381 759 464 855 306 24 × 2 = 1 + 0.988 763 518 929 710 612 48;
  • 52) 0.988 763 518 929 710 612 48 × 2 = 1 + 0.977 527 037 859 421 224 96;
  • 53) 0.977 527 037 859 421 224 96 × 2 = 1 + 0.955 054 075 718 842 449 92;
  • 54) 0.955 054 075 718 842 449 92 × 2 = 1 + 0.910 108 151 437 684 899 84;
  • 55) 0.910 108 151 437 684 899 84 × 2 = 1 + 0.820 216 302 875 369 799 68;
  • 56) 0.820 216 302 875 369 799 68 × 2 = 1 + 0.640 432 605 750 739 599 36;
  • 57) 0.640 432 605 750 739 599 36 × 2 = 1 + 0.280 865 211 501 479 198 72;
  • 58) 0.280 865 211 501 479 198 72 × 2 = 0 + 0.561 730 423 002 958 397 44;

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:


0.016 738 891 601 562 495 01(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 495 01(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 495 01(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 495 01 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 1110

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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