-0.016 738 891 601 562 496 530 147 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 147(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
-0.016 738 891 601 562 496 530 147(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:

|-0.016 738 891 601 562 496 530 147| = 0.016 738 891 601 562 496 530 147


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: 0.016 738 891 601 562 496 530 147.

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) 0.016 738 891 601 562 496 530 147 × 2 = 0 + 0.033 477 783 203 124 993 060 294;
  • 2) 0.033 477 783 203 124 993 060 294 × 2 = 0 + 0.066 955 566 406 249 986 120 588;
  • 3) 0.066 955 566 406 249 986 120 588 × 2 = 0 + 0.133 911 132 812 499 972 241 176;
  • 4) 0.133 911 132 812 499 972 241 176 × 2 = 0 + 0.267 822 265 624 999 944 482 352;
  • 5) 0.267 822 265 624 999 944 482 352 × 2 = 0 + 0.535 644 531 249 999 888 964 704;
  • 6) 0.535 644 531 249 999 888 964 704 × 2 = 1 + 0.071 289 062 499 999 777 929 408;
  • 7) 0.071 289 062 499 999 777 929 408 × 2 = 0 + 0.142 578 124 999 999 555 858 816;
  • 8) 0.142 578 124 999 999 555 858 816 × 2 = 0 + 0.285 156 249 999 999 111 717 632;
  • 9) 0.285 156 249 999 999 111 717 632 × 2 = 0 + 0.570 312 499 999 998 223 435 264;
  • 10) 0.570 312 499 999 998 223 435 264 × 2 = 1 + 0.140 624 999 999 996 446 870 528;
  • 11) 0.140 624 999 999 996 446 870 528 × 2 = 0 + 0.281 249 999 999 992 893 741 056;
  • 12) 0.281 249 999 999 992 893 741 056 × 2 = 0 + 0.562 499 999 999 985 787 482 112;
  • 13) 0.562 499 999 999 985 787 482 112 × 2 = 1 + 0.124 999 999 999 971 574 964 224;
  • 14) 0.124 999 999 999 971 574 964 224 × 2 = 0 + 0.249 999 999 999 943 149 928 448;
  • 15) 0.249 999 999 999 943 149 928 448 × 2 = 0 + 0.499 999 999 999 886 299 856 896;
  • 16) 0.499 999 999 999 886 299 856 896 × 2 = 0 + 0.999 999 999 999 772 599 713 792;
  • 17) 0.999 999 999 999 772 599 713 792 × 2 = 1 + 0.999 999 999 999 545 199 427 584;
  • 18) 0.999 999 999 999 545 199 427 584 × 2 = 1 + 0.999 999 999 999 090 398 855 168;
  • 19) 0.999 999 999 999 090 398 855 168 × 2 = 1 + 0.999 999 999 998 180 797 710 336;
  • 20) 0.999 999 999 998 180 797 710 336 × 2 = 1 + 0.999 999 999 996 361 595 420 672;
  • 21) 0.999 999 999 996 361 595 420 672 × 2 = 1 + 0.999 999 999 992 723 190 841 344;
  • 22) 0.999 999 999 992 723 190 841 344 × 2 = 1 + 0.999 999 999 985 446 381 682 688;
  • 23) 0.999 999 999 985 446 381 682 688 × 2 = 1 + 0.999 999 999 970 892 763 365 376;
  • 24) 0.999 999 999 970 892 763 365 376 × 2 = 1 + 0.999 999 999 941 785 526 730 752;
  • 25) 0.999 999 999 941 785 526 730 752 × 2 = 1 + 0.999 999 999 883 571 053 461 504;
  • 26) 0.999 999 999 883 571 053 461 504 × 2 = 1 + 0.999 999 999 767 142 106 923 008;
  • 27) 0.999 999 999 767 142 106 923 008 × 2 = 1 + 0.999 999 999 534 284 213 846 016;
  • 28) 0.999 999 999 534 284 213 846 016 × 2 = 1 + 0.999 999 999 068 568 427 692 032;
  • 29) 0.999 999 999 068 568 427 692 032 × 2 = 1 + 0.999 999 998 137 136 855 384 064;
  • 30) 0.999 999 998 137 136 855 384 064 × 2 = 1 + 0.999 999 996 274 273 710 768 128;
  • 31) 0.999 999 996 274 273 710 768 128 × 2 = 1 + 0.999 999 992 548 547 421 536 256;
  • 32) 0.999 999 992 548 547 421 536 256 × 2 = 1 + 0.999 999 985 097 094 843 072 512;
  • 33) 0.999 999 985 097 094 843 072 512 × 2 = 1 + 0.999 999 970 194 189 686 145 024;
  • 34) 0.999 999 970 194 189 686 145 024 × 2 = 1 + 0.999 999 940 388 379 372 290 048;
  • 35) 0.999 999 940 388 379 372 290 048 × 2 = 1 + 0.999 999 880 776 758 744 580 096;
  • 36) 0.999 999 880 776 758 744 580 096 × 2 = 1 + 0.999 999 761 553 517 489 160 192;
  • 37) 0.999 999 761 553 517 489 160 192 × 2 = 1 + 0.999 999 523 107 034 978 320 384;
  • 38) 0.999 999 523 107 034 978 320 384 × 2 = 1 + 0.999 999 046 214 069 956 640 768;
  • 39) 0.999 999 046 214 069 956 640 768 × 2 = 1 + 0.999 998 092 428 139 913 281 536;
  • 40) 0.999 998 092 428 139 913 281 536 × 2 = 1 + 0.999 996 184 856 279 826 563 072;
  • 41) 0.999 996 184 856 279 826 563 072 × 2 = 1 + 0.999 992 369 712 559 653 126 144;
  • 42) 0.999 992 369 712 559 653 126 144 × 2 = 1 + 0.999 984 739 425 119 306 252 288;
  • 43) 0.999 984 739 425 119 306 252 288 × 2 = 1 + 0.999 969 478 850 238 612 504 576;
  • 44) 0.999 969 478 850 238 612 504 576 × 2 = 1 + 0.999 938 957 700 477 225 009 152;
  • 45) 0.999 938 957 700 477 225 009 152 × 2 = 1 + 0.999 877 915 400 954 450 018 304;
  • 46) 0.999 877 915 400 954 450 018 304 × 2 = 1 + 0.999 755 830 801 908 900 036 608;
  • 47) 0.999 755 830 801 908 900 036 608 × 2 = 1 + 0.999 511 661 603 817 800 073 216;
  • 48) 0.999 511 661 603 817 800 073 216 × 2 = 1 + 0.999 023 323 207 635 600 146 432;
  • 49) 0.999 023 323 207 635 600 146 432 × 2 = 1 + 0.998 046 646 415 271 200 292 864;
  • 50) 0.998 046 646 415 271 200 292 864 × 2 = 1 + 0.996 093 292 830 542 400 585 728;
  • 51) 0.996 093 292 830 542 400 585 728 × 2 = 1 + 0.992 186 585 661 084 801 171 456;
  • 52) 0.992 186 585 661 084 801 171 456 × 2 = 1 + 0.984 373 171 322 169 602 342 912;
  • 53) 0.984 373 171 322 169 602 342 912 × 2 = 1 + 0.968 746 342 644 339 204 685 824;
  • 54) 0.968 746 342 644 339 204 685 824 × 2 = 1 + 0.937 492 685 288 678 409 371 648;
  • 55) 0.937 492 685 288 678 409 371 648 × 2 = 1 + 0.874 985 370 577 356 818 743 296;
  • 56) 0.874 985 370 577 356 818 743 296 × 2 = 1 + 0.749 970 741 154 713 637 486 592;
  • 57) 0.749 970 741 154 713 637 486 592 × 2 = 1 + 0.499 941 482 309 427 274 973 184;
  • 58) 0.499 941 482 309 427 274 973 184 × 2 = 0 + 0.999 882 964 618 854 549 946 368;

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:


0.016 738 891 601 562 496 530 147(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 147(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:


0.016 738 891 601 562 496 530 147(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 -0.016 738 891 601 562 496 530 147 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