-0.016 738 891 601 562 496 530 552 78 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 78(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
-0.016 738 891 601 562 496 530 552 78(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:

|-0.016 738 891 601 562 496 530 552 78| = 0.016 738 891 601 562 496 530 552 78


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: 0.016 738 891 601 562 496 530 552 78.

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) 0.016 738 891 601 562 496 530 552 78 × 2 = 0 + 0.033 477 783 203 124 993 061 105 56;
  • 2) 0.033 477 783 203 124 993 061 105 56 × 2 = 0 + 0.066 955 566 406 249 986 122 211 12;
  • 3) 0.066 955 566 406 249 986 122 211 12 × 2 = 0 + 0.133 911 132 812 499 972 244 422 24;
  • 4) 0.133 911 132 812 499 972 244 422 24 × 2 = 0 + 0.267 822 265 624 999 944 488 844 48;
  • 5) 0.267 822 265 624 999 944 488 844 48 × 2 = 0 + 0.535 644 531 249 999 888 977 688 96;
  • 6) 0.535 644 531 249 999 888 977 688 96 × 2 = 1 + 0.071 289 062 499 999 777 955 377 92;
  • 7) 0.071 289 062 499 999 777 955 377 92 × 2 = 0 + 0.142 578 124 999 999 555 910 755 84;
  • 8) 0.142 578 124 999 999 555 910 755 84 × 2 = 0 + 0.285 156 249 999 999 111 821 511 68;
  • 9) 0.285 156 249 999 999 111 821 511 68 × 2 = 0 + 0.570 312 499 999 998 223 643 023 36;
  • 10) 0.570 312 499 999 998 223 643 023 36 × 2 = 1 + 0.140 624 999 999 996 447 286 046 72;
  • 11) 0.140 624 999 999 996 447 286 046 72 × 2 = 0 + 0.281 249 999 999 992 894 572 093 44;
  • 12) 0.281 249 999 999 992 894 572 093 44 × 2 = 0 + 0.562 499 999 999 985 789 144 186 88;
  • 13) 0.562 499 999 999 985 789 144 186 88 × 2 = 1 + 0.124 999 999 999 971 578 288 373 76;
  • 14) 0.124 999 999 999 971 578 288 373 76 × 2 = 0 + 0.249 999 999 999 943 156 576 747 52;
  • 15) 0.249 999 999 999 943 156 576 747 52 × 2 = 0 + 0.499 999 999 999 886 313 153 495 04;
  • 16) 0.499 999 999 999 886 313 153 495 04 × 2 = 0 + 0.999 999 999 999 772 626 306 990 08;
  • 17) 0.999 999 999 999 772 626 306 990 08 × 2 = 1 + 0.999 999 999 999 545 252 613 980 16;
  • 18) 0.999 999 999 999 545 252 613 980 16 × 2 = 1 + 0.999 999 999 999 090 505 227 960 32;
  • 19) 0.999 999 999 999 090 505 227 960 32 × 2 = 1 + 0.999 999 999 998 181 010 455 920 64;
  • 20) 0.999 999 999 998 181 010 455 920 64 × 2 = 1 + 0.999 999 999 996 362 020 911 841 28;
  • 21) 0.999 999 999 996 362 020 911 841 28 × 2 = 1 + 0.999 999 999 992 724 041 823 682 56;
  • 22) 0.999 999 999 992 724 041 823 682 56 × 2 = 1 + 0.999 999 999 985 448 083 647 365 12;
  • 23) 0.999 999 999 985 448 083 647 365 12 × 2 = 1 + 0.999 999 999 970 896 167 294 730 24;
  • 24) 0.999 999 999 970 896 167 294 730 24 × 2 = 1 + 0.999 999 999 941 792 334 589 460 48;
  • 25) 0.999 999 999 941 792 334 589 460 48 × 2 = 1 + 0.999 999 999 883 584 669 178 920 96;
  • 26) 0.999 999 999 883 584 669 178 920 96 × 2 = 1 + 0.999 999 999 767 169 338 357 841 92;
  • 27) 0.999 999 999 767 169 338 357 841 92 × 2 = 1 + 0.999 999 999 534 338 676 715 683 84;
  • 28) 0.999 999 999 534 338 676 715 683 84 × 2 = 1 + 0.999 999 999 068 677 353 431 367 68;
  • 29) 0.999 999 999 068 677 353 431 367 68 × 2 = 1 + 0.999 999 998 137 354 706 862 735 36;
  • 30) 0.999 999 998 137 354 706 862 735 36 × 2 = 1 + 0.999 999 996 274 709 413 725 470 72;
  • 31) 0.999 999 996 274 709 413 725 470 72 × 2 = 1 + 0.999 999 992 549 418 827 450 941 44;
  • 32) 0.999 999 992 549 418 827 450 941 44 × 2 = 1 + 0.999 999 985 098 837 654 901 882 88;
  • 33) 0.999 999 985 098 837 654 901 882 88 × 2 = 1 + 0.999 999 970 197 675 309 803 765 76;
  • 34) 0.999 999 970 197 675 309 803 765 76 × 2 = 1 + 0.999 999 940 395 350 619 607 531 52;
  • 35) 0.999 999 940 395 350 619 607 531 52 × 2 = 1 + 0.999 999 880 790 701 239 215 063 04;
  • 36) 0.999 999 880 790 701 239 215 063 04 × 2 = 1 + 0.999 999 761 581 402 478 430 126 08;
  • 37) 0.999 999 761 581 402 478 430 126 08 × 2 = 1 + 0.999 999 523 162 804 956 860 252 16;
  • 38) 0.999 999 523 162 804 956 860 252 16 × 2 = 1 + 0.999 999 046 325 609 913 720 504 32;
  • 39) 0.999 999 046 325 609 913 720 504 32 × 2 = 1 + 0.999 998 092 651 219 827 441 008 64;
  • 40) 0.999 998 092 651 219 827 441 008 64 × 2 = 1 + 0.999 996 185 302 439 654 882 017 28;
  • 41) 0.999 996 185 302 439 654 882 017 28 × 2 = 1 + 0.999 992 370 604 879 309 764 034 56;
  • 42) 0.999 992 370 604 879 309 764 034 56 × 2 = 1 + 0.999 984 741 209 758 619 528 069 12;
  • 43) 0.999 984 741 209 758 619 528 069 12 × 2 = 1 + 0.999 969 482 419 517 239 056 138 24;
  • 44) 0.999 969 482 419 517 239 056 138 24 × 2 = 1 + 0.999 938 964 839 034 478 112 276 48;
  • 45) 0.999 938 964 839 034 478 112 276 48 × 2 = 1 + 0.999 877 929 678 068 956 224 552 96;
  • 46) 0.999 877 929 678 068 956 224 552 96 × 2 = 1 + 0.999 755 859 356 137 912 449 105 92;
  • 47) 0.999 755 859 356 137 912 449 105 92 × 2 = 1 + 0.999 511 718 712 275 824 898 211 84;
  • 48) 0.999 511 718 712 275 824 898 211 84 × 2 = 1 + 0.999 023 437 424 551 649 796 423 68;
  • 49) 0.999 023 437 424 551 649 796 423 68 × 2 = 1 + 0.998 046 874 849 103 299 592 847 36;
  • 50) 0.998 046 874 849 103 299 592 847 36 × 2 = 1 + 0.996 093 749 698 206 599 185 694 72;
  • 51) 0.996 093 749 698 206 599 185 694 72 × 2 = 1 + 0.992 187 499 396 413 198 371 389 44;
  • 52) 0.992 187 499 396 413 198 371 389 44 × 2 = 1 + 0.984 374 998 792 826 396 742 778 88;
  • 53) 0.984 374 998 792 826 396 742 778 88 × 2 = 1 + 0.968 749 997 585 652 793 485 557 76;
  • 54) 0.968 749 997 585 652 793 485 557 76 × 2 = 1 + 0.937 499 995 171 305 586 971 115 52;
  • 55) 0.937 499 995 171 305 586 971 115 52 × 2 = 1 + 0.874 999 990 342 611 173 942 231 04;
  • 56) 0.874 999 990 342 611 173 942 231 04 × 2 = 1 + 0.749 999 980 685 222 347 884 462 08;
  • 57) 0.749 999 980 685 222 347 884 462 08 × 2 = 1 + 0.499 999 961 370 444 695 768 924 16;
  • 58) 0.499 999 961 370 444 695 768 924 16 × 2 = 0 + 0.999 999 922 740 889 391 537 848 32;

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:


0.016 738 891 601 562 496 530 552 78(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 78(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:


0.016 738 891 601 562 496 530 552 78(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 -0.016 738 891 601 562 496 530 552 78 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