-0.048 632 677 916 771 838 155 796 097 57 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 796 097 57(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
-0.048 632 677 916 771 838 155 796 097 57(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:

|-0.048 632 677 916 771 838 155 796 097 57| = 0.048 632 677 916 771 838 155 796 097 57


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: 0.048 632 677 916 771 838 155 796 097 57.

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) 0.048 632 677 916 771 838 155 796 097 57 × 2 = 0 + 0.097 265 355 833 543 676 311 592 195 14;
  • 2) 0.097 265 355 833 543 676 311 592 195 14 × 2 = 0 + 0.194 530 711 667 087 352 623 184 390 28;
  • 3) 0.194 530 711 667 087 352 623 184 390 28 × 2 = 0 + 0.389 061 423 334 174 705 246 368 780 56;
  • 4) 0.389 061 423 334 174 705 246 368 780 56 × 2 = 0 + 0.778 122 846 668 349 410 492 737 561 12;
  • 5) 0.778 122 846 668 349 410 492 737 561 12 × 2 = 1 + 0.556 245 693 336 698 820 985 475 122 24;
  • 6) 0.556 245 693 336 698 820 985 475 122 24 × 2 = 1 + 0.112 491 386 673 397 641 970 950 244 48;
  • 7) 0.112 491 386 673 397 641 970 950 244 48 × 2 = 0 + 0.224 982 773 346 795 283 941 900 488 96;
  • 8) 0.224 982 773 346 795 283 941 900 488 96 × 2 = 0 + 0.449 965 546 693 590 567 883 800 977 92;
  • 9) 0.449 965 546 693 590 567 883 800 977 92 × 2 = 0 + 0.899 931 093 387 181 135 767 601 955 84;
  • 10) 0.899 931 093 387 181 135 767 601 955 84 × 2 = 1 + 0.799 862 186 774 362 271 535 203 911 68;
  • 11) 0.799 862 186 774 362 271 535 203 911 68 × 2 = 1 + 0.599 724 373 548 724 543 070 407 823 36;
  • 12) 0.599 724 373 548 724 543 070 407 823 36 × 2 = 1 + 0.199 448 747 097 449 086 140 815 646 72;
  • 13) 0.199 448 747 097 449 086 140 815 646 72 × 2 = 0 + 0.398 897 494 194 898 172 281 631 293 44;
  • 14) 0.398 897 494 194 898 172 281 631 293 44 × 2 = 0 + 0.797 794 988 389 796 344 563 262 586 88;
  • 15) 0.797 794 988 389 796 344 563 262 586 88 × 2 = 1 + 0.595 589 976 779 592 689 126 525 173 76;
  • 16) 0.595 589 976 779 592 689 126 525 173 76 × 2 = 1 + 0.191 179 953 559 185 378 253 050 347 52;
  • 17) 0.191 179 953 559 185 378 253 050 347 52 × 2 = 0 + 0.382 359 907 118 370 756 506 100 695 04;
  • 18) 0.382 359 907 118 370 756 506 100 695 04 × 2 = 0 + 0.764 719 814 236 741 513 012 201 390 08;
  • 19) 0.764 719 814 236 741 513 012 201 390 08 × 2 = 1 + 0.529 439 628 473 483 026 024 402 780 16;
  • 20) 0.529 439 628 473 483 026 024 402 780 16 × 2 = 1 + 0.058 879 256 946 966 052 048 805 560 32;
  • 21) 0.058 879 256 946 966 052 048 805 560 32 × 2 = 0 + 0.117 758 513 893 932 104 097 611 120 64;
  • 22) 0.117 758 513 893 932 104 097 611 120 64 × 2 = 0 + 0.235 517 027 787 864 208 195 222 241 28;
  • 23) 0.235 517 027 787 864 208 195 222 241 28 × 2 = 0 + 0.471 034 055 575 728 416 390 444 482 56;
  • 24) 0.471 034 055 575 728 416 390 444 482 56 × 2 = 0 + 0.942 068 111 151 456 832 780 888 965 12;
  • 25) 0.942 068 111 151 456 832 780 888 965 12 × 2 = 1 + 0.884 136 222 302 913 665 561 777 930 24;
  • 26) 0.884 136 222 302 913 665 561 777 930 24 × 2 = 1 + 0.768 272 444 605 827 331 123 555 860 48;
  • 27) 0.768 272 444 605 827 331 123 555 860 48 × 2 = 1 + 0.536 544 889 211 654 662 247 111 720 96;
  • 28) 0.536 544 889 211 654 662 247 111 720 96 × 2 = 1 + 0.073 089 778 423 309 324 494 223 441 92;
  • 29) 0.073 089 778 423 309 324 494 223 441 92 × 2 = 0 + 0.146 179 556 846 618 648 988 446 883 84;
  • 30) 0.146 179 556 846 618 648 988 446 883 84 × 2 = 0 + 0.292 359 113 693 237 297 976 893 767 68;
  • 31) 0.292 359 113 693 237 297 976 893 767 68 × 2 = 0 + 0.584 718 227 386 474 595 953 787 535 36;
  • 32) 0.584 718 227 386 474 595 953 787 535 36 × 2 = 1 + 0.169 436 454 772 949 191 907 575 070 72;
  • 33) 0.169 436 454 772 949 191 907 575 070 72 × 2 = 0 + 0.338 872 909 545 898 383 815 150 141 44;
  • 34) 0.338 872 909 545 898 383 815 150 141 44 × 2 = 0 + 0.677 745 819 091 796 767 630 300 282 88;
  • 35) 0.677 745 819 091 796 767 630 300 282 88 × 2 = 1 + 0.355 491 638 183 593 535 260 600 565 76;
  • 36) 0.355 491 638 183 593 535 260 600 565 76 × 2 = 0 + 0.710 983 276 367 187 070 521 201 131 52;
  • 37) 0.710 983 276 367 187 070 521 201 131 52 × 2 = 1 + 0.421 966 552 734 374 141 042 402 263 04;
  • 38) 0.421 966 552 734 374 141 042 402 263 04 × 2 = 0 + 0.843 933 105 468 748 282 084 804 526 08;
  • 39) 0.843 933 105 468 748 282 084 804 526 08 × 2 = 1 + 0.687 866 210 937 496 564 169 609 052 16;
  • 40) 0.687 866 210 937 496 564 169 609 052 16 × 2 = 1 + 0.375 732 421 874 993 128 339 218 104 32;
  • 41) 0.375 732 421 874 993 128 339 218 104 32 × 2 = 0 + 0.751 464 843 749 986 256 678 436 208 64;
  • 42) 0.751 464 843 749 986 256 678 436 208 64 × 2 = 1 + 0.502 929 687 499 972 513 356 872 417 28;
  • 43) 0.502 929 687 499 972 513 356 872 417 28 × 2 = 1 + 0.005 859 374 999 945 026 713 744 834 56;
  • 44) 0.005 859 374 999 945 026 713 744 834 56 × 2 = 0 + 0.011 718 749 999 890 053 427 489 669 12;
  • 45) 0.011 718 749 999 890 053 427 489 669 12 × 2 = 0 + 0.023 437 499 999 780 106 854 979 338 24;
  • 46) 0.023 437 499 999 780 106 854 979 338 24 × 2 = 0 + 0.046 874 999 999 560 213 709 958 676 48;
  • 47) 0.046 874 999 999 560 213 709 958 676 48 × 2 = 0 + 0.093 749 999 999 120 427 419 917 352 96;
  • 48) 0.093 749 999 999 120 427 419 917 352 96 × 2 = 0 + 0.187 499 999 998 240 854 839 834 705 92;
  • 49) 0.187 499 999 998 240 854 839 834 705 92 × 2 = 0 + 0.374 999 999 996 481 709 679 669 411 84;
  • 50) 0.374 999 999 996 481 709 679 669 411 84 × 2 = 0 + 0.749 999 999 992 963 419 359 338 823 68;
  • 51) 0.749 999 999 992 963 419 359 338 823 68 × 2 = 1 + 0.499 999 999 985 926 838 718 677 647 36;
  • 52) 0.499 999 999 985 926 838 718 677 647 36 × 2 = 0 + 0.999 999 999 971 853 677 437 355 294 72;
  • 53) 0.999 999 999 971 853 677 437 355 294 72 × 2 = 1 + 0.999 999 999 943 707 354 874 710 589 44;
  • 54) 0.999 999 999 943 707 354 874 710 589 44 × 2 = 1 + 0.999 999 999 887 414 709 749 421 178 88;
  • 55) 0.999 999 999 887 414 709 749 421 178 88 × 2 = 1 + 0.999 999 999 774 829 419 498 842 357 76;
  • 56) 0.999 999 999 774 829 419 498 842 357 76 × 2 = 1 + 0.999 999 999 549 658 838 997 684 715 52;
  • 57) 0.999 999 999 549 658 838 997 684 715 52 × 2 = 1 + 0.999 999 999 099 317 677 995 369 431 04;

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:


0.048 632 677 916 771 838 155 796 097 57(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(2)

6. Positive number before normalization:

0.048 632 677 916 771 838 155 796 097 57(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.048 632 677 916 771 838 155 796 097 57(10) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(2) =

0.0000 1100 0111 0011 0011 0000 1111 0001 0010 1011 0110 0000 0010 1111 1(2) × 20 =

1.1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 1111(2) × 2-5


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): -5

Mantissa (not normalized):
1.1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 1111 =

1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 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 1010

Mantissa (52 bits) =
1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 1111


Decimal number -0.048 632 677 916 771 838 155 796 097 57 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1000 1110 0110 0110 0001 1110 0010 0101 0110 1100 0000 0101 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