-0.016 738 891 601 562 496 530 467 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 467(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
-0.016 738 891 601 562 496 530 467(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:

|-0.016 738 891 601 562 496 530 467| = 0.016 738 891 601 562 496 530 467


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: 0.016 738 891 601 562 496 530 467.

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) 0.016 738 891 601 562 496 530 467 × 2 = 0 + 0.033 477 783 203 124 993 060 934;
  • 2) 0.033 477 783 203 124 993 060 934 × 2 = 0 + 0.066 955 566 406 249 986 121 868;
  • 3) 0.066 955 566 406 249 986 121 868 × 2 = 0 + 0.133 911 132 812 499 972 243 736;
  • 4) 0.133 911 132 812 499 972 243 736 × 2 = 0 + 0.267 822 265 624 999 944 487 472;
  • 5) 0.267 822 265 624 999 944 487 472 × 2 = 0 + 0.535 644 531 249 999 888 974 944;
  • 6) 0.535 644 531 249 999 888 974 944 × 2 = 1 + 0.071 289 062 499 999 777 949 888;
  • 7) 0.071 289 062 499 999 777 949 888 × 2 = 0 + 0.142 578 124 999 999 555 899 776;
  • 8) 0.142 578 124 999 999 555 899 776 × 2 = 0 + 0.285 156 249 999 999 111 799 552;
  • 9) 0.285 156 249 999 999 111 799 552 × 2 = 0 + 0.570 312 499 999 998 223 599 104;
  • 10) 0.570 312 499 999 998 223 599 104 × 2 = 1 + 0.140 624 999 999 996 447 198 208;
  • 11) 0.140 624 999 999 996 447 198 208 × 2 = 0 + 0.281 249 999 999 992 894 396 416;
  • 12) 0.281 249 999 999 992 894 396 416 × 2 = 0 + 0.562 499 999 999 985 788 792 832;
  • 13) 0.562 499 999 999 985 788 792 832 × 2 = 1 + 0.124 999 999 999 971 577 585 664;
  • 14) 0.124 999 999 999 971 577 585 664 × 2 = 0 + 0.249 999 999 999 943 155 171 328;
  • 15) 0.249 999 999 999 943 155 171 328 × 2 = 0 + 0.499 999 999 999 886 310 342 656;
  • 16) 0.499 999 999 999 886 310 342 656 × 2 = 0 + 0.999 999 999 999 772 620 685 312;
  • 17) 0.999 999 999 999 772 620 685 312 × 2 = 1 + 0.999 999 999 999 545 241 370 624;
  • 18) 0.999 999 999 999 545 241 370 624 × 2 = 1 + 0.999 999 999 999 090 482 741 248;
  • 19) 0.999 999 999 999 090 482 741 248 × 2 = 1 + 0.999 999 999 998 180 965 482 496;
  • 20) 0.999 999 999 998 180 965 482 496 × 2 = 1 + 0.999 999 999 996 361 930 964 992;
  • 21) 0.999 999 999 996 361 930 964 992 × 2 = 1 + 0.999 999 999 992 723 861 929 984;
  • 22) 0.999 999 999 992 723 861 929 984 × 2 = 1 + 0.999 999 999 985 447 723 859 968;
  • 23) 0.999 999 999 985 447 723 859 968 × 2 = 1 + 0.999 999 999 970 895 447 719 936;
  • 24) 0.999 999 999 970 895 447 719 936 × 2 = 1 + 0.999 999 999 941 790 895 439 872;
  • 25) 0.999 999 999 941 790 895 439 872 × 2 = 1 + 0.999 999 999 883 581 790 879 744;
  • 26) 0.999 999 999 883 581 790 879 744 × 2 = 1 + 0.999 999 999 767 163 581 759 488;
  • 27) 0.999 999 999 767 163 581 759 488 × 2 = 1 + 0.999 999 999 534 327 163 518 976;
  • 28) 0.999 999 999 534 327 163 518 976 × 2 = 1 + 0.999 999 999 068 654 327 037 952;
  • 29) 0.999 999 999 068 654 327 037 952 × 2 = 1 + 0.999 999 998 137 308 654 075 904;
  • 30) 0.999 999 998 137 308 654 075 904 × 2 = 1 + 0.999 999 996 274 617 308 151 808;
  • 31) 0.999 999 996 274 617 308 151 808 × 2 = 1 + 0.999 999 992 549 234 616 303 616;
  • 32) 0.999 999 992 549 234 616 303 616 × 2 = 1 + 0.999 999 985 098 469 232 607 232;
  • 33) 0.999 999 985 098 469 232 607 232 × 2 = 1 + 0.999 999 970 196 938 465 214 464;
  • 34) 0.999 999 970 196 938 465 214 464 × 2 = 1 + 0.999 999 940 393 876 930 428 928;
  • 35) 0.999 999 940 393 876 930 428 928 × 2 = 1 + 0.999 999 880 787 753 860 857 856;
  • 36) 0.999 999 880 787 753 860 857 856 × 2 = 1 + 0.999 999 761 575 507 721 715 712;
  • 37) 0.999 999 761 575 507 721 715 712 × 2 = 1 + 0.999 999 523 151 015 443 431 424;
  • 38) 0.999 999 523 151 015 443 431 424 × 2 = 1 + 0.999 999 046 302 030 886 862 848;
  • 39) 0.999 999 046 302 030 886 862 848 × 2 = 1 + 0.999 998 092 604 061 773 725 696;
  • 40) 0.999 998 092 604 061 773 725 696 × 2 = 1 + 0.999 996 185 208 123 547 451 392;
  • 41) 0.999 996 185 208 123 547 451 392 × 2 = 1 + 0.999 992 370 416 247 094 902 784;
  • 42) 0.999 992 370 416 247 094 902 784 × 2 = 1 + 0.999 984 740 832 494 189 805 568;
  • 43) 0.999 984 740 832 494 189 805 568 × 2 = 1 + 0.999 969 481 664 988 379 611 136;
  • 44) 0.999 969 481 664 988 379 611 136 × 2 = 1 + 0.999 938 963 329 976 759 222 272;
  • 45) 0.999 938 963 329 976 759 222 272 × 2 = 1 + 0.999 877 926 659 953 518 444 544;
  • 46) 0.999 877 926 659 953 518 444 544 × 2 = 1 + 0.999 755 853 319 907 036 889 088;
  • 47) 0.999 755 853 319 907 036 889 088 × 2 = 1 + 0.999 511 706 639 814 073 778 176;
  • 48) 0.999 511 706 639 814 073 778 176 × 2 = 1 + 0.999 023 413 279 628 147 556 352;
  • 49) 0.999 023 413 279 628 147 556 352 × 2 = 1 + 0.998 046 826 559 256 295 112 704;
  • 50) 0.998 046 826 559 256 295 112 704 × 2 = 1 + 0.996 093 653 118 512 590 225 408;
  • 51) 0.996 093 653 118 512 590 225 408 × 2 = 1 + 0.992 187 306 237 025 180 450 816;
  • 52) 0.992 187 306 237 025 180 450 816 × 2 = 1 + 0.984 374 612 474 050 360 901 632;
  • 53) 0.984 374 612 474 050 360 901 632 × 2 = 1 + 0.968 749 224 948 100 721 803 264;
  • 54) 0.968 749 224 948 100 721 803 264 × 2 = 1 + 0.937 498 449 896 201 443 606 528;
  • 55) 0.937 498 449 896 201 443 606 528 × 2 = 1 + 0.874 996 899 792 402 887 213 056;
  • 56) 0.874 996 899 792 402 887 213 056 × 2 = 1 + 0.749 993 799 584 805 774 426 112;
  • 57) 0.749 993 799 584 805 774 426 112 × 2 = 1 + 0.499 987 599 169 611 548 852 224;
  • 58) 0.499 987 599 169 611 548 852 224 × 2 = 0 + 0.999 975 198 339 223 097 704 448;

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:


0.016 738 891 601 562 496 530 467(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 467(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:


0.016 738 891 601 562 496 530 467(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 -0.016 738 891 601 562 496 530 467 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