-0.016 738 891 601 562 496 521 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 521 3(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
-0.016 738 891 601 562 496 521 3(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:

|-0.016 738 891 601 562 496 521 3| = 0.016 738 891 601 562 496 521 3


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: 0.016 738 891 601 562 496 521 3.

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) 0.016 738 891 601 562 496 521 3 × 2 = 0 + 0.033 477 783 203 124 993 042 6;
  • 2) 0.033 477 783 203 124 993 042 6 × 2 = 0 + 0.066 955 566 406 249 986 085 2;
  • 3) 0.066 955 566 406 249 986 085 2 × 2 = 0 + 0.133 911 132 812 499 972 170 4;
  • 4) 0.133 911 132 812 499 972 170 4 × 2 = 0 + 0.267 822 265 624 999 944 340 8;
  • 5) 0.267 822 265 624 999 944 340 8 × 2 = 0 + 0.535 644 531 249 999 888 681 6;
  • 6) 0.535 644 531 249 999 888 681 6 × 2 = 1 + 0.071 289 062 499 999 777 363 2;
  • 7) 0.071 289 062 499 999 777 363 2 × 2 = 0 + 0.142 578 124 999 999 554 726 4;
  • 8) 0.142 578 124 999 999 554 726 4 × 2 = 0 + 0.285 156 249 999 999 109 452 8;
  • 9) 0.285 156 249 999 999 109 452 8 × 2 = 0 + 0.570 312 499 999 998 218 905 6;
  • 10) 0.570 312 499 999 998 218 905 6 × 2 = 1 + 0.140 624 999 999 996 437 811 2;
  • 11) 0.140 624 999 999 996 437 811 2 × 2 = 0 + 0.281 249 999 999 992 875 622 4;
  • 12) 0.281 249 999 999 992 875 622 4 × 2 = 0 + 0.562 499 999 999 985 751 244 8;
  • 13) 0.562 499 999 999 985 751 244 8 × 2 = 1 + 0.124 999 999 999 971 502 489 6;
  • 14) 0.124 999 999 999 971 502 489 6 × 2 = 0 + 0.249 999 999 999 943 004 979 2;
  • 15) 0.249 999 999 999 943 004 979 2 × 2 = 0 + 0.499 999 999 999 886 009 958 4;
  • 16) 0.499 999 999 999 886 009 958 4 × 2 = 0 + 0.999 999 999 999 772 019 916 8;
  • 17) 0.999 999 999 999 772 019 916 8 × 2 = 1 + 0.999 999 999 999 544 039 833 6;
  • 18) 0.999 999 999 999 544 039 833 6 × 2 = 1 + 0.999 999 999 999 088 079 667 2;
  • 19) 0.999 999 999 999 088 079 667 2 × 2 = 1 + 0.999 999 999 998 176 159 334 4;
  • 20) 0.999 999 999 998 176 159 334 4 × 2 = 1 + 0.999 999 999 996 352 318 668 8;
  • 21) 0.999 999 999 996 352 318 668 8 × 2 = 1 + 0.999 999 999 992 704 637 337 6;
  • 22) 0.999 999 999 992 704 637 337 6 × 2 = 1 + 0.999 999 999 985 409 274 675 2;
  • 23) 0.999 999 999 985 409 274 675 2 × 2 = 1 + 0.999 999 999 970 818 549 350 4;
  • 24) 0.999 999 999 970 818 549 350 4 × 2 = 1 + 0.999 999 999 941 637 098 700 8;
  • 25) 0.999 999 999 941 637 098 700 8 × 2 = 1 + 0.999 999 999 883 274 197 401 6;
  • 26) 0.999 999 999 883 274 197 401 6 × 2 = 1 + 0.999 999 999 766 548 394 803 2;
  • 27) 0.999 999 999 766 548 394 803 2 × 2 = 1 + 0.999 999 999 533 096 789 606 4;
  • 28) 0.999 999 999 533 096 789 606 4 × 2 = 1 + 0.999 999 999 066 193 579 212 8;
  • 29) 0.999 999 999 066 193 579 212 8 × 2 = 1 + 0.999 999 998 132 387 158 425 6;
  • 30) 0.999 999 998 132 387 158 425 6 × 2 = 1 + 0.999 999 996 264 774 316 851 2;
  • 31) 0.999 999 996 264 774 316 851 2 × 2 = 1 + 0.999 999 992 529 548 633 702 4;
  • 32) 0.999 999 992 529 548 633 702 4 × 2 = 1 + 0.999 999 985 059 097 267 404 8;
  • 33) 0.999 999 985 059 097 267 404 8 × 2 = 1 + 0.999 999 970 118 194 534 809 6;
  • 34) 0.999 999 970 118 194 534 809 6 × 2 = 1 + 0.999 999 940 236 389 069 619 2;
  • 35) 0.999 999 940 236 389 069 619 2 × 2 = 1 + 0.999 999 880 472 778 139 238 4;
  • 36) 0.999 999 880 472 778 139 238 4 × 2 = 1 + 0.999 999 760 945 556 278 476 8;
  • 37) 0.999 999 760 945 556 278 476 8 × 2 = 1 + 0.999 999 521 891 112 556 953 6;
  • 38) 0.999 999 521 891 112 556 953 6 × 2 = 1 + 0.999 999 043 782 225 113 907 2;
  • 39) 0.999 999 043 782 225 113 907 2 × 2 = 1 + 0.999 998 087 564 450 227 814 4;
  • 40) 0.999 998 087 564 450 227 814 4 × 2 = 1 + 0.999 996 175 128 900 455 628 8;
  • 41) 0.999 996 175 128 900 455 628 8 × 2 = 1 + 0.999 992 350 257 800 911 257 6;
  • 42) 0.999 992 350 257 800 911 257 6 × 2 = 1 + 0.999 984 700 515 601 822 515 2;
  • 43) 0.999 984 700 515 601 822 515 2 × 2 = 1 + 0.999 969 401 031 203 645 030 4;
  • 44) 0.999 969 401 031 203 645 030 4 × 2 = 1 + 0.999 938 802 062 407 290 060 8;
  • 45) 0.999 938 802 062 407 290 060 8 × 2 = 1 + 0.999 877 604 124 814 580 121 6;
  • 46) 0.999 877 604 124 814 580 121 6 × 2 = 1 + 0.999 755 208 249 629 160 243 2;
  • 47) 0.999 755 208 249 629 160 243 2 × 2 = 1 + 0.999 510 416 499 258 320 486 4;
  • 48) 0.999 510 416 499 258 320 486 4 × 2 = 1 + 0.999 020 832 998 516 640 972 8;
  • 49) 0.999 020 832 998 516 640 972 8 × 2 = 1 + 0.998 041 665 997 033 281 945 6;
  • 50) 0.998 041 665 997 033 281 945 6 × 2 = 1 + 0.996 083 331 994 066 563 891 2;
  • 51) 0.996 083 331 994 066 563 891 2 × 2 = 1 + 0.992 166 663 988 133 127 782 4;
  • 52) 0.992 166 663 988 133 127 782 4 × 2 = 1 + 0.984 333 327 976 266 255 564 8;
  • 53) 0.984 333 327 976 266 255 564 8 × 2 = 1 + 0.968 666 655 952 532 511 129 6;
  • 54) 0.968 666 655 952 532 511 129 6 × 2 = 1 + 0.937 333 311 905 065 022 259 2;
  • 55) 0.937 333 311 905 065 022 259 2 × 2 = 1 + 0.874 666 623 810 130 044 518 4;
  • 56) 0.874 666 623 810 130 044 518 4 × 2 = 1 + 0.749 333 247 620 260 089 036 8;
  • 57) 0.749 333 247 620 260 089 036 8 × 2 = 1 + 0.498 666 495 240 520 178 073 6;
  • 58) 0.498 666 495 240 520 178 073 6 × 2 = 0 + 0.997 332 990 481 040 356 147 2;

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:


0.016 738 891 601 562 496 521 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 521 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 521 3(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 521 3 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 1110

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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