-0.016 738 891 601 562 496 530 795 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 795(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
-0.016 738 891 601 562 496 530 795(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:

|-0.016 738 891 601 562 496 530 795| = 0.016 738 891 601 562 496 530 795


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: 0.016 738 891 601 562 496 530 795.

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) 0.016 738 891 601 562 496 530 795 × 2 = 0 + 0.033 477 783 203 124 993 061 59;
  • 2) 0.033 477 783 203 124 993 061 59 × 2 = 0 + 0.066 955 566 406 249 986 123 18;
  • 3) 0.066 955 566 406 249 986 123 18 × 2 = 0 + 0.133 911 132 812 499 972 246 36;
  • 4) 0.133 911 132 812 499 972 246 36 × 2 = 0 + 0.267 822 265 624 999 944 492 72;
  • 5) 0.267 822 265 624 999 944 492 72 × 2 = 0 + 0.535 644 531 249 999 888 985 44;
  • 6) 0.535 644 531 249 999 888 985 44 × 2 = 1 + 0.071 289 062 499 999 777 970 88;
  • 7) 0.071 289 062 499 999 777 970 88 × 2 = 0 + 0.142 578 124 999 999 555 941 76;
  • 8) 0.142 578 124 999 999 555 941 76 × 2 = 0 + 0.285 156 249 999 999 111 883 52;
  • 9) 0.285 156 249 999 999 111 883 52 × 2 = 0 + 0.570 312 499 999 998 223 767 04;
  • 10) 0.570 312 499 999 998 223 767 04 × 2 = 1 + 0.140 624 999 999 996 447 534 08;
  • 11) 0.140 624 999 999 996 447 534 08 × 2 = 0 + 0.281 249 999 999 992 895 068 16;
  • 12) 0.281 249 999 999 992 895 068 16 × 2 = 0 + 0.562 499 999 999 985 790 136 32;
  • 13) 0.562 499 999 999 985 790 136 32 × 2 = 1 + 0.124 999 999 999 971 580 272 64;
  • 14) 0.124 999 999 999 971 580 272 64 × 2 = 0 + 0.249 999 999 999 943 160 545 28;
  • 15) 0.249 999 999 999 943 160 545 28 × 2 = 0 + 0.499 999 999 999 886 321 090 56;
  • 16) 0.499 999 999 999 886 321 090 56 × 2 = 0 + 0.999 999 999 999 772 642 181 12;
  • 17) 0.999 999 999 999 772 642 181 12 × 2 = 1 + 0.999 999 999 999 545 284 362 24;
  • 18) 0.999 999 999 999 545 284 362 24 × 2 = 1 + 0.999 999 999 999 090 568 724 48;
  • 19) 0.999 999 999 999 090 568 724 48 × 2 = 1 + 0.999 999 999 998 181 137 448 96;
  • 20) 0.999 999 999 998 181 137 448 96 × 2 = 1 + 0.999 999 999 996 362 274 897 92;
  • 21) 0.999 999 999 996 362 274 897 92 × 2 = 1 + 0.999 999 999 992 724 549 795 84;
  • 22) 0.999 999 999 992 724 549 795 84 × 2 = 1 + 0.999 999 999 985 449 099 591 68;
  • 23) 0.999 999 999 985 449 099 591 68 × 2 = 1 + 0.999 999 999 970 898 199 183 36;
  • 24) 0.999 999 999 970 898 199 183 36 × 2 = 1 + 0.999 999 999 941 796 398 366 72;
  • 25) 0.999 999 999 941 796 398 366 72 × 2 = 1 + 0.999 999 999 883 592 796 733 44;
  • 26) 0.999 999 999 883 592 796 733 44 × 2 = 1 + 0.999 999 999 767 185 593 466 88;
  • 27) 0.999 999 999 767 185 593 466 88 × 2 = 1 + 0.999 999 999 534 371 186 933 76;
  • 28) 0.999 999 999 534 371 186 933 76 × 2 = 1 + 0.999 999 999 068 742 373 867 52;
  • 29) 0.999 999 999 068 742 373 867 52 × 2 = 1 + 0.999 999 998 137 484 747 735 04;
  • 30) 0.999 999 998 137 484 747 735 04 × 2 = 1 + 0.999 999 996 274 969 495 470 08;
  • 31) 0.999 999 996 274 969 495 470 08 × 2 = 1 + 0.999 999 992 549 938 990 940 16;
  • 32) 0.999 999 992 549 938 990 940 16 × 2 = 1 + 0.999 999 985 099 877 981 880 32;
  • 33) 0.999 999 985 099 877 981 880 32 × 2 = 1 + 0.999 999 970 199 755 963 760 64;
  • 34) 0.999 999 970 199 755 963 760 64 × 2 = 1 + 0.999 999 940 399 511 927 521 28;
  • 35) 0.999 999 940 399 511 927 521 28 × 2 = 1 + 0.999 999 880 799 023 855 042 56;
  • 36) 0.999 999 880 799 023 855 042 56 × 2 = 1 + 0.999 999 761 598 047 710 085 12;
  • 37) 0.999 999 761 598 047 710 085 12 × 2 = 1 + 0.999 999 523 196 095 420 170 24;
  • 38) 0.999 999 523 196 095 420 170 24 × 2 = 1 + 0.999 999 046 392 190 840 340 48;
  • 39) 0.999 999 046 392 190 840 340 48 × 2 = 1 + 0.999 998 092 784 381 680 680 96;
  • 40) 0.999 998 092 784 381 680 680 96 × 2 = 1 + 0.999 996 185 568 763 361 361 92;
  • 41) 0.999 996 185 568 763 361 361 92 × 2 = 1 + 0.999 992 371 137 526 722 723 84;
  • 42) 0.999 992 371 137 526 722 723 84 × 2 = 1 + 0.999 984 742 275 053 445 447 68;
  • 43) 0.999 984 742 275 053 445 447 68 × 2 = 1 + 0.999 969 484 550 106 890 895 36;
  • 44) 0.999 969 484 550 106 890 895 36 × 2 = 1 + 0.999 938 969 100 213 781 790 72;
  • 45) 0.999 938 969 100 213 781 790 72 × 2 = 1 + 0.999 877 938 200 427 563 581 44;
  • 46) 0.999 877 938 200 427 563 581 44 × 2 = 1 + 0.999 755 876 400 855 127 162 88;
  • 47) 0.999 755 876 400 855 127 162 88 × 2 = 1 + 0.999 511 752 801 710 254 325 76;
  • 48) 0.999 511 752 801 710 254 325 76 × 2 = 1 + 0.999 023 505 603 420 508 651 52;
  • 49) 0.999 023 505 603 420 508 651 52 × 2 = 1 + 0.998 047 011 206 841 017 303 04;
  • 50) 0.998 047 011 206 841 017 303 04 × 2 = 1 + 0.996 094 022 413 682 034 606 08;
  • 51) 0.996 094 022 413 682 034 606 08 × 2 = 1 + 0.992 188 044 827 364 069 212 16;
  • 52) 0.992 188 044 827 364 069 212 16 × 2 = 1 + 0.984 376 089 654 728 138 424 32;
  • 53) 0.984 376 089 654 728 138 424 32 × 2 = 1 + 0.968 752 179 309 456 276 848 64;
  • 54) 0.968 752 179 309 456 276 848 64 × 2 = 1 + 0.937 504 358 618 912 553 697 28;
  • 55) 0.937 504 358 618 912 553 697 28 × 2 = 1 + 0.875 008 717 237 825 107 394 56;
  • 56) 0.875 008 717 237 825 107 394 56 × 2 = 1 + 0.750 017 434 475 650 214 789 12;
  • 57) 0.750 017 434 475 650 214 789 12 × 2 = 1 + 0.500 034 868 951 300 429 578 24;
  • 58) 0.500 034 868 951 300 429 578 24 × 2 = 1 + 0.000 069 737 902 600 859 156 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:


0.016 738 891 601 562 496 530 795(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 795(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:


0.016 738 891 601 562 496 530 795(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 -0.016 738 891 601 562 496 530 795 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