-0.016 738 891 601 563 041 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 563 041(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
-0.016 738 891 601 563 041(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:

|-0.016 738 891 601 563 041| = 0.016 738 891 601 563 041


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: 0.016 738 891 601 563 041.

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) 0.016 738 891 601 563 041 × 2 = 0 + 0.033 477 783 203 126 082;
  • 2) 0.033 477 783 203 126 082 × 2 = 0 + 0.066 955 566 406 252 164;
  • 3) 0.066 955 566 406 252 164 × 2 = 0 + 0.133 911 132 812 504 328;
  • 4) 0.133 911 132 812 504 328 × 2 = 0 + 0.267 822 265 625 008 656;
  • 5) 0.267 822 265 625 008 656 × 2 = 0 + 0.535 644 531 250 017 312;
  • 6) 0.535 644 531 250 017 312 × 2 = 1 + 0.071 289 062 500 034 624;
  • 7) 0.071 289 062 500 034 624 × 2 = 0 + 0.142 578 125 000 069 248;
  • 8) 0.142 578 125 000 069 248 × 2 = 0 + 0.285 156 250 000 138 496;
  • 9) 0.285 156 250 000 138 496 × 2 = 0 + 0.570 312 500 000 276 992;
  • 10) 0.570 312 500 000 276 992 × 2 = 1 + 0.140 625 000 000 553 984;
  • 11) 0.140 625 000 000 553 984 × 2 = 0 + 0.281 250 000 001 107 968;
  • 12) 0.281 250 000 001 107 968 × 2 = 0 + 0.562 500 000 002 215 936;
  • 13) 0.562 500 000 002 215 936 × 2 = 1 + 0.125 000 000 004 431 872;
  • 14) 0.125 000 000 004 431 872 × 2 = 0 + 0.250 000 000 008 863 744;
  • 15) 0.250 000 000 008 863 744 × 2 = 0 + 0.500 000 000 017 727 488;
  • 16) 0.500 000 000 017 727 488 × 2 = 1 + 0.000 000 000 035 454 976;
  • 17) 0.000 000 000 035 454 976 × 2 = 0 + 0.000 000 000 070 909 952;
  • 18) 0.000 000 000 070 909 952 × 2 = 0 + 0.000 000 000 141 819 904;
  • 19) 0.000 000 000 141 819 904 × 2 = 0 + 0.000 000 000 283 639 808;
  • 20) 0.000 000 000 283 639 808 × 2 = 0 + 0.000 000 000 567 279 616;
  • 21) 0.000 000 000 567 279 616 × 2 = 0 + 0.000 000 001 134 559 232;
  • 22) 0.000 000 001 134 559 232 × 2 = 0 + 0.000 000 002 269 118 464;
  • 23) 0.000 000 002 269 118 464 × 2 = 0 + 0.000 000 004 538 236 928;
  • 24) 0.000 000 004 538 236 928 × 2 = 0 + 0.000 000 009 076 473 856;
  • 25) 0.000 000 009 076 473 856 × 2 = 0 + 0.000 000 018 152 947 712;
  • 26) 0.000 000 018 152 947 712 × 2 = 0 + 0.000 000 036 305 895 424;
  • 27) 0.000 000 036 305 895 424 × 2 = 0 + 0.000 000 072 611 790 848;
  • 28) 0.000 000 072 611 790 848 × 2 = 0 + 0.000 000 145 223 581 696;
  • 29) 0.000 000 145 223 581 696 × 2 = 0 + 0.000 000 290 447 163 392;
  • 30) 0.000 000 290 447 163 392 × 2 = 0 + 0.000 000 580 894 326 784;
  • 31) 0.000 000 580 894 326 784 × 2 = 0 + 0.000 001 161 788 653 568;
  • 32) 0.000 001 161 788 653 568 × 2 = 0 + 0.000 002 323 577 307 136;
  • 33) 0.000 002 323 577 307 136 × 2 = 0 + 0.000 004 647 154 614 272;
  • 34) 0.000 004 647 154 614 272 × 2 = 0 + 0.000 009 294 309 228 544;
  • 35) 0.000 009 294 309 228 544 × 2 = 0 + 0.000 018 588 618 457 088;
  • 36) 0.000 018 588 618 457 088 × 2 = 0 + 0.000 037 177 236 914 176;
  • 37) 0.000 037 177 236 914 176 × 2 = 0 + 0.000 074 354 473 828 352;
  • 38) 0.000 074 354 473 828 352 × 2 = 0 + 0.000 148 708 947 656 704;
  • 39) 0.000 148 708 947 656 704 × 2 = 0 + 0.000 297 417 895 313 408;
  • 40) 0.000 297 417 895 313 408 × 2 = 0 + 0.000 594 835 790 626 816;
  • 41) 0.000 594 835 790 626 816 × 2 = 0 + 0.001 189 671 581 253 632;
  • 42) 0.001 189 671 581 253 632 × 2 = 0 + 0.002 379 343 162 507 264;
  • 43) 0.002 379 343 162 507 264 × 2 = 0 + 0.004 758 686 325 014 528;
  • 44) 0.004 758 686 325 014 528 × 2 = 0 + 0.009 517 372 650 029 056;
  • 45) 0.009 517 372 650 029 056 × 2 = 0 + 0.019 034 745 300 058 112;
  • 46) 0.019 034 745 300 058 112 × 2 = 0 + 0.038 069 490 600 116 224;
  • 47) 0.038 069 490 600 116 224 × 2 = 0 + 0.076 138 981 200 232 448;
  • 48) 0.076 138 981 200 232 448 × 2 = 0 + 0.152 277 962 400 464 896;
  • 49) 0.152 277 962 400 464 896 × 2 = 0 + 0.304 555 924 800 929 792;
  • 50) 0.304 555 924 800 929 792 × 2 = 0 + 0.609 111 849 601 859 584;
  • 51) 0.609 111 849 601 859 584 × 2 = 1 + 0.218 223 699 203 719 168;
  • 52) 0.218 223 699 203 719 168 × 2 = 0 + 0.436 447 398 407 438 336;
  • 53) 0.436 447 398 407 438 336 × 2 = 0 + 0.872 894 796 814 876 672;
  • 54) 0.872 894 796 814 876 672 × 2 = 1 + 0.745 789 593 629 753 344;
  • 55) 0.745 789 593 629 753 344 × 2 = 1 + 0.491 579 187 259 506 688;
  • 56) 0.491 579 187 259 506 688 × 2 = 0 + 0.983 158 374 519 013 376;
  • 57) 0.983 158 374 519 013 376 × 2 = 1 + 0.966 316 749 038 026 752;
  • 58) 0.966 316 749 038 026 752 × 2 = 1 + 0.932 633 498 076 053 504;

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:


0.016 738 891 601 563 041(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 11(2)

6. Positive number before normalization:

0.016 738 891 601 563 041(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 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:


0.016 738 891 601 563 041(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 11(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0010 0110 11(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 1001 1011(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 0100 0000 0000 0000 0000 0000 0000 0000 0000 1001 1011


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 1001 1011 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 1001 1011


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 0100 0000 0000 0000 0000 0000 0000 0000 0000 1001 1011


Decimal number -0.016 738 891 601 563 041 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 1001 1011

» Convert decimal -0.016 738 891 601 563 086 to 64 bit double precision IEEE 754 binary floating point representation standard

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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