-0.016 738 891 601 562 496 535 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 535 9(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
-0.016 738 891 601 562 496 535 9(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:

|-0.016 738 891 601 562 496 535 9| = 0.016 738 891 601 562 496 535 9


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: 0.016 738 891 601 562 496 535 9.

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) 0.016 738 891 601 562 496 535 9 × 2 = 0 + 0.033 477 783 203 124 993 071 8;
  • 2) 0.033 477 783 203 124 993 071 8 × 2 = 0 + 0.066 955 566 406 249 986 143 6;
  • 3) 0.066 955 566 406 249 986 143 6 × 2 = 0 + 0.133 911 132 812 499 972 287 2;
  • 4) 0.133 911 132 812 499 972 287 2 × 2 = 0 + 0.267 822 265 624 999 944 574 4;
  • 5) 0.267 822 265 624 999 944 574 4 × 2 = 0 + 0.535 644 531 249 999 889 148 8;
  • 6) 0.535 644 531 249 999 889 148 8 × 2 = 1 + 0.071 289 062 499 999 778 297 6;
  • 7) 0.071 289 062 499 999 778 297 6 × 2 = 0 + 0.142 578 124 999 999 556 595 2;
  • 8) 0.142 578 124 999 999 556 595 2 × 2 = 0 + 0.285 156 249 999 999 113 190 4;
  • 9) 0.285 156 249 999 999 113 190 4 × 2 = 0 + 0.570 312 499 999 998 226 380 8;
  • 10) 0.570 312 499 999 998 226 380 8 × 2 = 1 + 0.140 624 999 999 996 452 761 6;
  • 11) 0.140 624 999 999 996 452 761 6 × 2 = 0 + 0.281 249 999 999 992 905 523 2;
  • 12) 0.281 249 999 999 992 905 523 2 × 2 = 0 + 0.562 499 999 999 985 811 046 4;
  • 13) 0.562 499 999 999 985 811 046 4 × 2 = 1 + 0.124 999 999 999 971 622 092 8;
  • 14) 0.124 999 999 999 971 622 092 8 × 2 = 0 + 0.249 999 999 999 943 244 185 6;
  • 15) 0.249 999 999 999 943 244 185 6 × 2 = 0 + 0.499 999 999 999 886 488 371 2;
  • 16) 0.499 999 999 999 886 488 371 2 × 2 = 0 + 0.999 999 999 999 772 976 742 4;
  • 17) 0.999 999 999 999 772 976 742 4 × 2 = 1 + 0.999 999 999 999 545 953 484 8;
  • 18) 0.999 999 999 999 545 953 484 8 × 2 = 1 + 0.999 999 999 999 091 906 969 6;
  • 19) 0.999 999 999 999 091 906 969 6 × 2 = 1 + 0.999 999 999 998 183 813 939 2;
  • 20) 0.999 999 999 998 183 813 939 2 × 2 = 1 + 0.999 999 999 996 367 627 878 4;
  • 21) 0.999 999 999 996 367 627 878 4 × 2 = 1 + 0.999 999 999 992 735 255 756 8;
  • 22) 0.999 999 999 992 735 255 756 8 × 2 = 1 + 0.999 999 999 985 470 511 513 6;
  • 23) 0.999 999 999 985 470 511 513 6 × 2 = 1 + 0.999 999 999 970 941 023 027 2;
  • 24) 0.999 999 999 970 941 023 027 2 × 2 = 1 + 0.999 999 999 941 882 046 054 4;
  • 25) 0.999 999 999 941 882 046 054 4 × 2 = 1 + 0.999 999 999 883 764 092 108 8;
  • 26) 0.999 999 999 883 764 092 108 8 × 2 = 1 + 0.999 999 999 767 528 184 217 6;
  • 27) 0.999 999 999 767 528 184 217 6 × 2 = 1 + 0.999 999 999 535 056 368 435 2;
  • 28) 0.999 999 999 535 056 368 435 2 × 2 = 1 + 0.999 999 999 070 112 736 870 4;
  • 29) 0.999 999 999 070 112 736 870 4 × 2 = 1 + 0.999 999 998 140 225 473 740 8;
  • 30) 0.999 999 998 140 225 473 740 8 × 2 = 1 + 0.999 999 996 280 450 947 481 6;
  • 31) 0.999 999 996 280 450 947 481 6 × 2 = 1 + 0.999 999 992 560 901 894 963 2;
  • 32) 0.999 999 992 560 901 894 963 2 × 2 = 1 + 0.999 999 985 121 803 789 926 4;
  • 33) 0.999 999 985 121 803 789 926 4 × 2 = 1 + 0.999 999 970 243 607 579 852 8;
  • 34) 0.999 999 970 243 607 579 852 8 × 2 = 1 + 0.999 999 940 487 215 159 705 6;
  • 35) 0.999 999 940 487 215 159 705 6 × 2 = 1 + 0.999 999 880 974 430 319 411 2;
  • 36) 0.999 999 880 974 430 319 411 2 × 2 = 1 + 0.999 999 761 948 860 638 822 4;
  • 37) 0.999 999 761 948 860 638 822 4 × 2 = 1 + 0.999 999 523 897 721 277 644 8;
  • 38) 0.999 999 523 897 721 277 644 8 × 2 = 1 + 0.999 999 047 795 442 555 289 6;
  • 39) 0.999 999 047 795 442 555 289 6 × 2 = 1 + 0.999 998 095 590 885 110 579 2;
  • 40) 0.999 998 095 590 885 110 579 2 × 2 = 1 + 0.999 996 191 181 770 221 158 4;
  • 41) 0.999 996 191 181 770 221 158 4 × 2 = 1 + 0.999 992 382 363 540 442 316 8;
  • 42) 0.999 992 382 363 540 442 316 8 × 2 = 1 + 0.999 984 764 727 080 884 633 6;
  • 43) 0.999 984 764 727 080 884 633 6 × 2 = 1 + 0.999 969 529 454 161 769 267 2;
  • 44) 0.999 969 529 454 161 769 267 2 × 2 = 1 + 0.999 939 058 908 323 538 534 4;
  • 45) 0.999 939 058 908 323 538 534 4 × 2 = 1 + 0.999 878 117 816 647 077 068 8;
  • 46) 0.999 878 117 816 647 077 068 8 × 2 = 1 + 0.999 756 235 633 294 154 137 6;
  • 47) 0.999 756 235 633 294 154 137 6 × 2 = 1 + 0.999 512 471 266 588 308 275 2;
  • 48) 0.999 512 471 266 588 308 275 2 × 2 = 1 + 0.999 024 942 533 176 616 550 4;
  • 49) 0.999 024 942 533 176 616 550 4 × 2 = 1 + 0.998 049 885 066 353 233 100 8;
  • 50) 0.998 049 885 066 353 233 100 8 × 2 = 1 + 0.996 099 770 132 706 466 201 6;
  • 51) 0.996 099 770 132 706 466 201 6 × 2 = 1 + 0.992 199 540 265 412 932 403 2;
  • 52) 0.992 199 540 265 412 932 403 2 × 2 = 1 + 0.984 399 080 530 825 864 806 4;
  • 53) 0.984 399 080 530 825 864 806 4 × 2 = 1 + 0.968 798 161 061 651 729 612 8;
  • 54) 0.968 798 161 061 651 729 612 8 × 2 = 1 + 0.937 596 322 123 303 459 225 6;
  • 55) 0.937 596 322 123 303 459 225 6 × 2 = 1 + 0.875 192 644 246 606 918 451 2;
  • 56) 0.875 192 644 246 606 918 451 2 × 2 = 1 + 0.750 385 288 493 213 836 902 4;
  • 57) 0.750 385 288 493 213 836 902 4 × 2 = 1 + 0.500 770 576 986 427 673 804 8;
  • 58) 0.500 770 576 986 427 673 804 8 × 2 = 1 + 0.001 541 153 972 855 347 609 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:


0.016 738 891 601 562 496 535 9(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 535 9(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:


0.016 738 891 601 562 496 535 9(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 -0.016 738 891 601 562 496 535 9 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