-0.016 738 891 601 562 496 530 553 177 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 177(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
-0.016 738 891 601 562 496 530 553 177(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:

|-0.016 738 891 601 562 496 530 553 177| = 0.016 738 891 601 562 496 530 553 177


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: 0.016 738 891 601 562 496 530 553 177.

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) 0.016 738 891 601 562 496 530 553 177 × 2 = 0 + 0.033 477 783 203 124 993 061 106 354;
  • 2) 0.033 477 783 203 124 993 061 106 354 × 2 = 0 + 0.066 955 566 406 249 986 122 212 708;
  • 3) 0.066 955 566 406 249 986 122 212 708 × 2 = 0 + 0.133 911 132 812 499 972 244 425 416;
  • 4) 0.133 911 132 812 499 972 244 425 416 × 2 = 0 + 0.267 822 265 624 999 944 488 850 832;
  • 5) 0.267 822 265 624 999 944 488 850 832 × 2 = 0 + 0.535 644 531 249 999 888 977 701 664;
  • 6) 0.535 644 531 249 999 888 977 701 664 × 2 = 1 + 0.071 289 062 499 999 777 955 403 328;
  • 7) 0.071 289 062 499 999 777 955 403 328 × 2 = 0 + 0.142 578 124 999 999 555 910 806 656;
  • 8) 0.142 578 124 999 999 555 910 806 656 × 2 = 0 + 0.285 156 249 999 999 111 821 613 312;
  • 9) 0.285 156 249 999 999 111 821 613 312 × 2 = 0 + 0.570 312 499 999 998 223 643 226 624;
  • 10) 0.570 312 499 999 998 223 643 226 624 × 2 = 1 + 0.140 624 999 999 996 447 286 453 248;
  • 11) 0.140 624 999 999 996 447 286 453 248 × 2 = 0 + 0.281 249 999 999 992 894 572 906 496;
  • 12) 0.281 249 999 999 992 894 572 906 496 × 2 = 0 + 0.562 499 999 999 985 789 145 812 992;
  • 13) 0.562 499 999 999 985 789 145 812 992 × 2 = 1 + 0.124 999 999 999 971 578 291 625 984;
  • 14) 0.124 999 999 999 971 578 291 625 984 × 2 = 0 + 0.249 999 999 999 943 156 583 251 968;
  • 15) 0.249 999 999 999 943 156 583 251 968 × 2 = 0 + 0.499 999 999 999 886 313 166 503 936;
  • 16) 0.499 999 999 999 886 313 166 503 936 × 2 = 0 + 0.999 999 999 999 772 626 333 007 872;
  • 17) 0.999 999 999 999 772 626 333 007 872 × 2 = 1 + 0.999 999 999 999 545 252 666 015 744;
  • 18) 0.999 999 999 999 545 252 666 015 744 × 2 = 1 + 0.999 999 999 999 090 505 332 031 488;
  • 19) 0.999 999 999 999 090 505 332 031 488 × 2 = 1 + 0.999 999 999 998 181 010 664 062 976;
  • 20) 0.999 999 999 998 181 010 664 062 976 × 2 = 1 + 0.999 999 999 996 362 021 328 125 952;
  • 21) 0.999 999 999 996 362 021 328 125 952 × 2 = 1 + 0.999 999 999 992 724 042 656 251 904;
  • 22) 0.999 999 999 992 724 042 656 251 904 × 2 = 1 + 0.999 999 999 985 448 085 312 503 808;
  • 23) 0.999 999 999 985 448 085 312 503 808 × 2 = 1 + 0.999 999 999 970 896 170 625 007 616;
  • 24) 0.999 999 999 970 896 170 625 007 616 × 2 = 1 + 0.999 999 999 941 792 341 250 015 232;
  • 25) 0.999 999 999 941 792 341 250 015 232 × 2 = 1 + 0.999 999 999 883 584 682 500 030 464;
  • 26) 0.999 999 999 883 584 682 500 030 464 × 2 = 1 + 0.999 999 999 767 169 365 000 060 928;
  • 27) 0.999 999 999 767 169 365 000 060 928 × 2 = 1 + 0.999 999 999 534 338 730 000 121 856;
  • 28) 0.999 999 999 534 338 730 000 121 856 × 2 = 1 + 0.999 999 999 068 677 460 000 243 712;
  • 29) 0.999 999 999 068 677 460 000 243 712 × 2 = 1 + 0.999 999 998 137 354 920 000 487 424;
  • 30) 0.999 999 998 137 354 920 000 487 424 × 2 = 1 + 0.999 999 996 274 709 840 000 974 848;
  • 31) 0.999 999 996 274 709 840 000 974 848 × 2 = 1 + 0.999 999 992 549 419 680 001 949 696;
  • 32) 0.999 999 992 549 419 680 001 949 696 × 2 = 1 + 0.999 999 985 098 839 360 003 899 392;
  • 33) 0.999 999 985 098 839 360 003 899 392 × 2 = 1 + 0.999 999 970 197 678 720 007 798 784;
  • 34) 0.999 999 970 197 678 720 007 798 784 × 2 = 1 + 0.999 999 940 395 357 440 015 597 568;
  • 35) 0.999 999 940 395 357 440 015 597 568 × 2 = 1 + 0.999 999 880 790 714 880 031 195 136;
  • 36) 0.999 999 880 790 714 880 031 195 136 × 2 = 1 + 0.999 999 761 581 429 760 062 390 272;
  • 37) 0.999 999 761 581 429 760 062 390 272 × 2 = 1 + 0.999 999 523 162 859 520 124 780 544;
  • 38) 0.999 999 523 162 859 520 124 780 544 × 2 = 1 + 0.999 999 046 325 719 040 249 561 088;
  • 39) 0.999 999 046 325 719 040 249 561 088 × 2 = 1 + 0.999 998 092 651 438 080 499 122 176;
  • 40) 0.999 998 092 651 438 080 499 122 176 × 2 = 1 + 0.999 996 185 302 876 160 998 244 352;
  • 41) 0.999 996 185 302 876 160 998 244 352 × 2 = 1 + 0.999 992 370 605 752 321 996 488 704;
  • 42) 0.999 992 370 605 752 321 996 488 704 × 2 = 1 + 0.999 984 741 211 504 643 992 977 408;
  • 43) 0.999 984 741 211 504 643 992 977 408 × 2 = 1 + 0.999 969 482 423 009 287 985 954 816;
  • 44) 0.999 969 482 423 009 287 985 954 816 × 2 = 1 + 0.999 938 964 846 018 575 971 909 632;
  • 45) 0.999 938 964 846 018 575 971 909 632 × 2 = 1 + 0.999 877 929 692 037 151 943 819 264;
  • 46) 0.999 877 929 692 037 151 943 819 264 × 2 = 1 + 0.999 755 859 384 074 303 887 638 528;
  • 47) 0.999 755 859 384 074 303 887 638 528 × 2 = 1 + 0.999 511 718 768 148 607 775 277 056;
  • 48) 0.999 511 718 768 148 607 775 277 056 × 2 = 1 + 0.999 023 437 536 297 215 550 554 112;
  • 49) 0.999 023 437 536 297 215 550 554 112 × 2 = 1 + 0.998 046 875 072 594 431 101 108 224;
  • 50) 0.998 046 875 072 594 431 101 108 224 × 2 = 1 + 0.996 093 750 145 188 862 202 216 448;
  • 51) 0.996 093 750 145 188 862 202 216 448 × 2 = 1 + 0.992 187 500 290 377 724 404 432 896;
  • 52) 0.992 187 500 290 377 724 404 432 896 × 2 = 1 + 0.984 375 000 580 755 448 808 865 792;
  • 53) 0.984 375 000 580 755 448 808 865 792 × 2 = 1 + 0.968 750 001 161 510 897 617 731 584;
  • 54) 0.968 750 001 161 510 897 617 731 584 × 2 = 1 + 0.937 500 002 323 021 795 235 463 168;
  • 55) 0.937 500 002 323 021 795 235 463 168 × 2 = 1 + 0.875 000 004 646 043 590 470 926 336;
  • 56) 0.875 000 004 646 043 590 470 926 336 × 2 = 1 + 0.750 000 009 292 087 180 941 852 672;
  • 57) 0.750 000 009 292 087 180 941 852 672 × 2 = 1 + 0.500 000 018 584 174 361 883 705 344;
  • 58) 0.500 000 018 584 174 361 883 705 344 × 2 = 1 + 0.000 000 037 168 348 723 767 410 688;

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:


0.016 738 891 601 562 496 530 553 177(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 177(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:


0.016 738 891 601 562 496 530 553 177(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 -0.016 738 891 601 562 496 530 553 177 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