-0.016 738 891 601 562 496 540 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 540 4(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
-0.016 738 891 601 562 496 540 4(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:

|-0.016 738 891 601 562 496 540 4| = 0.016 738 891 601 562 496 540 4


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: 0.016 738 891 601 562 496 540 4.

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) 0.016 738 891 601 562 496 540 4 × 2 = 0 + 0.033 477 783 203 124 993 080 8;
  • 2) 0.033 477 783 203 124 993 080 8 × 2 = 0 + 0.066 955 566 406 249 986 161 6;
  • 3) 0.066 955 566 406 249 986 161 6 × 2 = 0 + 0.133 911 132 812 499 972 323 2;
  • 4) 0.133 911 132 812 499 972 323 2 × 2 = 0 + 0.267 822 265 624 999 944 646 4;
  • 5) 0.267 822 265 624 999 944 646 4 × 2 = 0 + 0.535 644 531 249 999 889 292 8;
  • 6) 0.535 644 531 249 999 889 292 8 × 2 = 1 + 0.071 289 062 499 999 778 585 6;
  • 7) 0.071 289 062 499 999 778 585 6 × 2 = 0 + 0.142 578 124 999 999 557 171 2;
  • 8) 0.142 578 124 999 999 557 171 2 × 2 = 0 + 0.285 156 249 999 999 114 342 4;
  • 9) 0.285 156 249 999 999 114 342 4 × 2 = 0 + 0.570 312 499 999 998 228 684 8;
  • 10) 0.570 312 499 999 998 228 684 8 × 2 = 1 + 0.140 624 999 999 996 457 369 6;
  • 11) 0.140 624 999 999 996 457 369 6 × 2 = 0 + 0.281 249 999 999 992 914 739 2;
  • 12) 0.281 249 999 999 992 914 739 2 × 2 = 0 + 0.562 499 999 999 985 829 478 4;
  • 13) 0.562 499 999 999 985 829 478 4 × 2 = 1 + 0.124 999 999 999 971 658 956 8;
  • 14) 0.124 999 999 999 971 658 956 8 × 2 = 0 + 0.249 999 999 999 943 317 913 6;
  • 15) 0.249 999 999 999 943 317 913 6 × 2 = 0 + 0.499 999 999 999 886 635 827 2;
  • 16) 0.499 999 999 999 886 635 827 2 × 2 = 0 + 0.999 999 999 999 773 271 654 4;
  • 17) 0.999 999 999 999 773 271 654 4 × 2 = 1 + 0.999 999 999 999 546 543 308 8;
  • 18) 0.999 999 999 999 546 543 308 8 × 2 = 1 + 0.999 999 999 999 093 086 617 6;
  • 19) 0.999 999 999 999 093 086 617 6 × 2 = 1 + 0.999 999 999 998 186 173 235 2;
  • 20) 0.999 999 999 998 186 173 235 2 × 2 = 1 + 0.999 999 999 996 372 346 470 4;
  • 21) 0.999 999 999 996 372 346 470 4 × 2 = 1 + 0.999 999 999 992 744 692 940 8;
  • 22) 0.999 999 999 992 744 692 940 8 × 2 = 1 + 0.999 999 999 985 489 385 881 6;
  • 23) 0.999 999 999 985 489 385 881 6 × 2 = 1 + 0.999 999 999 970 978 771 763 2;
  • 24) 0.999 999 999 970 978 771 763 2 × 2 = 1 + 0.999 999 999 941 957 543 526 4;
  • 25) 0.999 999 999 941 957 543 526 4 × 2 = 1 + 0.999 999 999 883 915 087 052 8;
  • 26) 0.999 999 999 883 915 087 052 8 × 2 = 1 + 0.999 999 999 767 830 174 105 6;
  • 27) 0.999 999 999 767 830 174 105 6 × 2 = 1 + 0.999 999 999 535 660 348 211 2;
  • 28) 0.999 999 999 535 660 348 211 2 × 2 = 1 + 0.999 999 999 071 320 696 422 4;
  • 29) 0.999 999 999 071 320 696 422 4 × 2 = 1 + 0.999 999 998 142 641 392 844 8;
  • 30) 0.999 999 998 142 641 392 844 8 × 2 = 1 + 0.999 999 996 285 282 785 689 6;
  • 31) 0.999 999 996 285 282 785 689 6 × 2 = 1 + 0.999 999 992 570 565 571 379 2;
  • 32) 0.999 999 992 570 565 571 379 2 × 2 = 1 + 0.999 999 985 141 131 142 758 4;
  • 33) 0.999 999 985 141 131 142 758 4 × 2 = 1 + 0.999 999 970 282 262 285 516 8;
  • 34) 0.999 999 970 282 262 285 516 8 × 2 = 1 + 0.999 999 940 564 524 571 033 6;
  • 35) 0.999 999 940 564 524 571 033 6 × 2 = 1 + 0.999 999 881 129 049 142 067 2;
  • 36) 0.999 999 881 129 049 142 067 2 × 2 = 1 + 0.999 999 762 258 098 284 134 4;
  • 37) 0.999 999 762 258 098 284 134 4 × 2 = 1 + 0.999 999 524 516 196 568 268 8;
  • 38) 0.999 999 524 516 196 568 268 8 × 2 = 1 + 0.999 999 049 032 393 136 537 6;
  • 39) 0.999 999 049 032 393 136 537 6 × 2 = 1 + 0.999 998 098 064 786 273 075 2;
  • 40) 0.999 998 098 064 786 273 075 2 × 2 = 1 + 0.999 996 196 129 572 546 150 4;
  • 41) 0.999 996 196 129 572 546 150 4 × 2 = 1 + 0.999 992 392 259 145 092 300 8;
  • 42) 0.999 992 392 259 145 092 300 8 × 2 = 1 + 0.999 984 784 518 290 184 601 6;
  • 43) 0.999 984 784 518 290 184 601 6 × 2 = 1 + 0.999 969 569 036 580 369 203 2;
  • 44) 0.999 969 569 036 580 369 203 2 × 2 = 1 + 0.999 939 138 073 160 738 406 4;
  • 45) 0.999 939 138 073 160 738 406 4 × 2 = 1 + 0.999 878 276 146 321 476 812 8;
  • 46) 0.999 878 276 146 321 476 812 8 × 2 = 1 + 0.999 756 552 292 642 953 625 6;
  • 47) 0.999 756 552 292 642 953 625 6 × 2 = 1 + 0.999 513 104 585 285 907 251 2;
  • 48) 0.999 513 104 585 285 907 251 2 × 2 = 1 + 0.999 026 209 170 571 814 502 4;
  • 49) 0.999 026 209 170 571 814 502 4 × 2 = 1 + 0.998 052 418 341 143 629 004 8;
  • 50) 0.998 052 418 341 143 629 004 8 × 2 = 1 + 0.996 104 836 682 287 258 009 6;
  • 51) 0.996 104 836 682 287 258 009 6 × 2 = 1 + 0.992 209 673 364 574 516 019 2;
  • 52) 0.992 209 673 364 574 516 019 2 × 2 = 1 + 0.984 419 346 729 149 032 038 4;
  • 53) 0.984 419 346 729 149 032 038 4 × 2 = 1 + 0.968 838 693 458 298 064 076 8;
  • 54) 0.968 838 693 458 298 064 076 8 × 2 = 1 + 0.937 677 386 916 596 128 153 6;
  • 55) 0.937 677 386 916 596 128 153 6 × 2 = 1 + 0.875 354 773 833 192 256 307 2;
  • 56) 0.875 354 773 833 192 256 307 2 × 2 = 1 + 0.750 709 547 666 384 512 614 4;
  • 57) 0.750 709 547 666 384 512 614 4 × 2 = 1 + 0.501 419 095 332 769 025 228 8;
  • 58) 0.501 419 095 332 769 025 228 8 × 2 = 1 + 0.002 838 190 665 538 050 457 6;

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:


0.016 738 891 601 562 496 540 4(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 540 4(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:


0.016 738 891 601 562 496 540 4(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 -0.016 738 891 601 562 496 540 4 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