-0.016 738 891 601 562 496 532 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 532 92(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
-0.016 738 891 601 562 496 532 92(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:

|-0.016 738 891 601 562 496 532 92| = 0.016 738 891 601 562 496 532 92


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: 0.016 738 891 601 562 496 532 92.

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) 0.016 738 891 601 562 496 532 92 × 2 = 0 + 0.033 477 783 203 124 993 065 84;
  • 2) 0.033 477 783 203 124 993 065 84 × 2 = 0 + 0.066 955 566 406 249 986 131 68;
  • 3) 0.066 955 566 406 249 986 131 68 × 2 = 0 + 0.133 911 132 812 499 972 263 36;
  • 4) 0.133 911 132 812 499 972 263 36 × 2 = 0 + 0.267 822 265 624 999 944 526 72;
  • 5) 0.267 822 265 624 999 944 526 72 × 2 = 0 + 0.535 644 531 249 999 889 053 44;
  • 6) 0.535 644 531 249 999 889 053 44 × 2 = 1 + 0.071 289 062 499 999 778 106 88;
  • 7) 0.071 289 062 499 999 778 106 88 × 2 = 0 + 0.142 578 124 999 999 556 213 76;
  • 8) 0.142 578 124 999 999 556 213 76 × 2 = 0 + 0.285 156 249 999 999 112 427 52;
  • 9) 0.285 156 249 999 999 112 427 52 × 2 = 0 + 0.570 312 499 999 998 224 855 04;
  • 10) 0.570 312 499 999 998 224 855 04 × 2 = 1 + 0.140 624 999 999 996 449 710 08;
  • 11) 0.140 624 999 999 996 449 710 08 × 2 = 0 + 0.281 249 999 999 992 899 420 16;
  • 12) 0.281 249 999 999 992 899 420 16 × 2 = 0 + 0.562 499 999 999 985 798 840 32;
  • 13) 0.562 499 999 999 985 798 840 32 × 2 = 1 + 0.124 999 999 999 971 597 680 64;
  • 14) 0.124 999 999 999 971 597 680 64 × 2 = 0 + 0.249 999 999 999 943 195 361 28;
  • 15) 0.249 999 999 999 943 195 361 28 × 2 = 0 + 0.499 999 999 999 886 390 722 56;
  • 16) 0.499 999 999 999 886 390 722 56 × 2 = 0 + 0.999 999 999 999 772 781 445 12;
  • 17) 0.999 999 999 999 772 781 445 12 × 2 = 1 + 0.999 999 999 999 545 562 890 24;
  • 18) 0.999 999 999 999 545 562 890 24 × 2 = 1 + 0.999 999 999 999 091 125 780 48;
  • 19) 0.999 999 999 999 091 125 780 48 × 2 = 1 + 0.999 999 999 998 182 251 560 96;
  • 20) 0.999 999 999 998 182 251 560 96 × 2 = 1 + 0.999 999 999 996 364 503 121 92;
  • 21) 0.999 999 999 996 364 503 121 92 × 2 = 1 + 0.999 999 999 992 729 006 243 84;
  • 22) 0.999 999 999 992 729 006 243 84 × 2 = 1 + 0.999 999 999 985 458 012 487 68;
  • 23) 0.999 999 999 985 458 012 487 68 × 2 = 1 + 0.999 999 999 970 916 024 975 36;
  • 24) 0.999 999 999 970 916 024 975 36 × 2 = 1 + 0.999 999 999 941 832 049 950 72;
  • 25) 0.999 999 999 941 832 049 950 72 × 2 = 1 + 0.999 999 999 883 664 099 901 44;
  • 26) 0.999 999 999 883 664 099 901 44 × 2 = 1 + 0.999 999 999 767 328 199 802 88;
  • 27) 0.999 999 999 767 328 199 802 88 × 2 = 1 + 0.999 999 999 534 656 399 605 76;
  • 28) 0.999 999 999 534 656 399 605 76 × 2 = 1 + 0.999 999 999 069 312 799 211 52;
  • 29) 0.999 999 999 069 312 799 211 52 × 2 = 1 + 0.999 999 998 138 625 598 423 04;
  • 30) 0.999 999 998 138 625 598 423 04 × 2 = 1 + 0.999 999 996 277 251 196 846 08;
  • 31) 0.999 999 996 277 251 196 846 08 × 2 = 1 + 0.999 999 992 554 502 393 692 16;
  • 32) 0.999 999 992 554 502 393 692 16 × 2 = 1 + 0.999 999 985 109 004 787 384 32;
  • 33) 0.999 999 985 109 004 787 384 32 × 2 = 1 + 0.999 999 970 218 009 574 768 64;
  • 34) 0.999 999 970 218 009 574 768 64 × 2 = 1 + 0.999 999 940 436 019 149 537 28;
  • 35) 0.999 999 940 436 019 149 537 28 × 2 = 1 + 0.999 999 880 872 038 299 074 56;
  • 36) 0.999 999 880 872 038 299 074 56 × 2 = 1 + 0.999 999 761 744 076 598 149 12;
  • 37) 0.999 999 761 744 076 598 149 12 × 2 = 1 + 0.999 999 523 488 153 196 298 24;
  • 38) 0.999 999 523 488 153 196 298 24 × 2 = 1 + 0.999 999 046 976 306 392 596 48;
  • 39) 0.999 999 046 976 306 392 596 48 × 2 = 1 + 0.999 998 093 952 612 785 192 96;
  • 40) 0.999 998 093 952 612 785 192 96 × 2 = 1 + 0.999 996 187 905 225 570 385 92;
  • 41) 0.999 996 187 905 225 570 385 92 × 2 = 1 + 0.999 992 375 810 451 140 771 84;
  • 42) 0.999 992 375 810 451 140 771 84 × 2 = 1 + 0.999 984 751 620 902 281 543 68;
  • 43) 0.999 984 751 620 902 281 543 68 × 2 = 1 + 0.999 969 503 241 804 563 087 36;
  • 44) 0.999 969 503 241 804 563 087 36 × 2 = 1 + 0.999 939 006 483 609 126 174 72;
  • 45) 0.999 939 006 483 609 126 174 72 × 2 = 1 + 0.999 878 012 967 218 252 349 44;
  • 46) 0.999 878 012 967 218 252 349 44 × 2 = 1 + 0.999 756 025 934 436 504 698 88;
  • 47) 0.999 756 025 934 436 504 698 88 × 2 = 1 + 0.999 512 051 868 873 009 397 76;
  • 48) 0.999 512 051 868 873 009 397 76 × 2 = 1 + 0.999 024 103 737 746 018 795 52;
  • 49) 0.999 024 103 737 746 018 795 52 × 2 = 1 + 0.998 048 207 475 492 037 591 04;
  • 50) 0.998 048 207 475 492 037 591 04 × 2 = 1 + 0.996 096 414 950 984 075 182 08;
  • 51) 0.996 096 414 950 984 075 182 08 × 2 = 1 + 0.992 192 829 901 968 150 364 16;
  • 52) 0.992 192 829 901 968 150 364 16 × 2 = 1 + 0.984 385 659 803 936 300 728 32;
  • 53) 0.984 385 659 803 936 300 728 32 × 2 = 1 + 0.968 771 319 607 872 601 456 64;
  • 54) 0.968 771 319 607 872 601 456 64 × 2 = 1 + 0.937 542 639 215 745 202 913 28;
  • 55) 0.937 542 639 215 745 202 913 28 × 2 = 1 + 0.875 085 278 431 490 405 826 56;
  • 56) 0.875 085 278 431 490 405 826 56 × 2 = 1 + 0.750 170 556 862 980 811 653 12;
  • 57) 0.750 170 556 862 980 811 653 12 × 2 = 1 + 0.500 341 113 725 961 623 306 24;
  • 58) 0.500 341 113 725 961 623 306 24 × 2 = 1 + 0.000 682 227 451 923 246 612 48;

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:


0.016 738 891 601 562 496 532 92(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 532 92(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:


0.016 738 891 601 562 496 532 92(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 -0.016 738 891 601 562 496 532 92 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