-0.016 738 891 601 562 496 530 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 72(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 72(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 72| = 0.016 738 891 601 562 496 530 72


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 72.

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 72 × 2 = 0 + 0.033 477 783 203 124 993 061 44;
  • 2) 0.033 477 783 203 124 993 061 44 × 2 = 0 + 0.066 955 566 406 249 986 122 88;
  • 3) 0.066 955 566 406 249 986 122 88 × 2 = 0 + 0.133 911 132 812 499 972 245 76;
  • 4) 0.133 911 132 812 499 972 245 76 × 2 = 0 + 0.267 822 265 624 999 944 491 52;
  • 5) 0.267 822 265 624 999 944 491 52 × 2 = 0 + 0.535 644 531 249 999 888 983 04;
  • 6) 0.535 644 531 249 999 888 983 04 × 2 = 1 + 0.071 289 062 499 999 777 966 08;
  • 7) 0.071 289 062 499 999 777 966 08 × 2 = 0 + 0.142 578 124 999 999 555 932 16;
  • 8) 0.142 578 124 999 999 555 932 16 × 2 = 0 + 0.285 156 249 999 999 111 864 32;
  • 9) 0.285 156 249 999 999 111 864 32 × 2 = 0 + 0.570 312 499 999 998 223 728 64;
  • 10) 0.570 312 499 999 998 223 728 64 × 2 = 1 + 0.140 624 999 999 996 447 457 28;
  • 11) 0.140 624 999 999 996 447 457 28 × 2 = 0 + 0.281 249 999 999 992 894 914 56;
  • 12) 0.281 249 999 999 992 894 914 56 × 2 = 0 + 0.562 499 999 999 985 789 829 12;
  • 13) 0.562 499 999 999 985 789 829 12 × 2 = 1 + 0.124 999 999 999 971 579 658 24;
  • 14) 0.124 999 999 999 971 579 658 24 × 2 = 0 + 0.249 999 999 999 943 159 316 48;
  • 15) 0.249 999 999 999 943 159 316 48 × 2 = 0 + 0.499 999 999 999 886 318 632 96;
  • 16) 0.499 999 999 999 886 318 632 96 × 2 = 0 + 0.999 999 999 999 772 637 265 92;
  • 17) 0.999 999 999 999 772 637 265 92 × 2 = 1 + 0.999 999 999 999 545 274 531 84;
  • 18) 0.999 999 999 999 545 274 531 84 × 2 = 1 + 0.999 999 999 999 090 549 063 68;
  • 19) 0.999 999 999 999 090 549 063 68 × 2 = 1 + 0.999 999 999 998 181 098 127 36;
  • 20) 0.999 999 999 998 181 098 127 36 × 2 = 1 + 0.999 999 999 996 362 196 254 72;
  • 21) 0.999 999 999 996 362 196 254 72 × 2 = 1 + 0.999 999 999 992 724 392 509 44;
  • 22) 0.999 999 999 992 724 392 509 44 × 2 = 1 + 0.999 999 999 985 448 785 018 88;
  • 23) 0.999 999 999 985 448 785 018 88 × 2 = 1 + 0.999 999 999 970 897 570 037 76;
  • 24) 0.999 999 999 970 897 570 037 76 × 2 = 1 + 0.999 999 999 941 795 140 075 52;
  • 25) 0.999 999 999 941 795 140 075 52 × 2 = 1 + 0.999 999 999 883 590 280 151 04;
  • 26) 0.999 999 999 883 590 280 151 04 × 2 = 1 + 0.999 999 999 767 180 560 302 08;
  • 27) 0.999 999 999 767 180 560 302 08 × 2 = 1 + 0.999 999 999 534 361 120 604 16;
  • 28) 0.999 999 999 534 361 120 604 16 × 2 = 1 + 0.999 999 999 068 722 241 208 32;
  • 29) 0.999 999 999 068 722 241 208 32 × 2 = 1 + 0.999 999 998 137 444 482 416 64;
  • 30) 0.999 999 998 137 444 482 416 64 × 2 = 1 + 0.999 999 996 274 888 964 833 28;
  • 31) 0.999 999 996 274 888 964 833 28 × 2 = 1 + 0.999 999 992 549 777 929 666 56;
  • 32) 0.999 999 992 549 777 929 666 56 × 2 = 1 + 0.999 999 985 099 555 859 333 12;
  • 33) 0.999 999 985 099 555 859 333 12 × 2 = 1 + 0.999 999 970 199 111 718 666 24;
  • 34) 0.999 999 970 199 111 718 666 24 × 2 = 1 + 0.999 999 940 398 223 437 332 48;
  • 35) 0.999 999 940 398 223 437 332 48 × 2 = 1 + 0.999 999 880 796 446 874 664 96;
  • 36) 0.999 999 880 796 446 874 664 96 × 2 = 1 + 0.999 999 761 592 893 749 329 92;
  • 37) 0.999 999 761 592 893 749 329 92 × 2 = 1 + 0.999 999 523 185 787 498 659 84;
  • 38) 0.999 999 523 185 787 498 659 84 × 2 = 1 + 0.999 999 046 371 574 997 319 68;
  • 39) 0.999 999 046 371 574 997 319 68 × 2 = 1 + 0.999 998 092 743 149 994 639 36;
  • 40) 0.999 998 092 743 149 994 639 36 × 2 = 1 + 0.999 996 185 486 299 989 278 72;
  • 41) 0.999 996 185 486 299 989 278 72 × 2 = 1 + 0.999 992 370 972 599 978 557 44;
  • 42) 0.999 992 370 972 599 978 557 44 × 2 = 1 + 0.999 984 741 945 199 957 114 88;
  • 43) 0.999 984 741 945 199 957 114 88 × 2 = 1 + 0.999 969 483 890 399 914 229 76;
  • 44) 0.999 969 483 890 399 914 229 76 × 2 = 1 + 0.999 938 967 780 799 828 459 52;
  • 45) 0.999 938 967 780 799 828 459 52 × 2 = 1 + 0.999 877 935 561 599 656 919 04;
  • 46) 0.999 877 935 561 599 656 919 04 × 2 = 1 + 0.999 755 871 123 199 313 838 08;
  • 47) 0.999 755 871 123 199 313 838 08 × 2 = 1 + 0.999 511 742 246 398 627 676 16;
  • 48) 0.999 511 742 246 398 627 676 16 × 2 = 1 + 0.999 023 484 492 797 255 352 32;
  • 49) 0.999 023 484 492 797 255 352 32 × 2 = 1 + 0.998 046 968 985 594 510 704 64;
  • 50) 0.998 046 968 985 594 510 704 64 × 2 = 1 + 0.996 093 937 971 189 021 409 28;
  • 51) 0.996 093 937 971 189 021 409 28 × 2 = 1 + 0.992 187 875 942 378 042 818 56;
  • 52) 0.992 187 875 942 378 042 818 56 × 2 = 1 + 0.984 375 751 884 756 085 637 12;
  • 53) 0.984 375 751 884 756 085 637 12 × 2 = 1 + 0.968 751 503 769 512 171 274 24;
  • 54) 0.968 751 503 769 512 171 274 24 × 2 = 1 + 0.937 503 007 539 024 342 548 48;
  • 55) 0.937 503 007 539 024 342 548 48 × 2 = 1 + 0.875 006 015 078 048 685 096 96;
  • 56) 0.875 006 015 078 048 685 096 96 × 2 = 1 + 0.750 012 030 156 097 370 193 92;
  • 57) 0.750 012 030 156 097 370 193 92 × 2 = 1 + 0.500 024 060 312 194 740 387 84;
  • 58) 0.500 024 060 312 194 740 387 84 × 2 = 1 + 0.000 048 120 624 389 480 775 68;

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 72(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 72(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 72(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 72 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