-0.048 632 677 916 771 838 155 796 099 27 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.048 632 677 916 771 838 155 796 099 27(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
-0.048 632 677 916 771 838 155 796 099 27(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:

|-0.048 632 677 916 771 838 155 796 099 27| = 0.048 632 677 916 771 838 155 796 099 27


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: 0.048 632 677 916 771 838 155 796 099 27.

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) 0.048 632 677 916 771 838 155 796 099 27 × 2 = 0 + 0.097 265 355 833 543 676 311 592 198 54;
  • 2) 0.097 265 355 833 543 676 311 592 198 54 × 2 = 0 + 0.194 530 711 667 087 352 623 184 397 08;
  • 3) 0.194 530 711 667 087 352 623 184 397 08 × 2 = 0 + 0.389 061 423 334 174 705 246 368 794 16;
  • 4) 0.389 061 423 334 174 705 246 368 794 16 × 2 = 0 + 0.778 122 846 668 349 410 492 737 588 32;
  • 5) 0.778 122 846 668 349 410 492 737 588 32 × 2 = 1 + 0.556 245 693 336 698 820 985 475 176 64;
  • 6) 0.556 245 693 336 698 820 985 475 176 64 × 2 = 1 + 0.112 491 386 673 397 641 970 950 353 28;
  • 7) 0.112 491 386 673 397 641 970 950 353 28 × 2 = 0 + 0.224 982 773 346 795 283 941 900 706 56;
  • 8) 0.224 982 773 346 795 283 941 900 706 56 × 2 = 0 + 0.449 965 546 693 590 567 883 801 413 12;
  • 9) 0.449 965 546 693 590 567 883 801 413 12 × 2 = 0 + 0.899 931 093 387 181 135 767 602 826 24;
  • 10) 0.899 931 093 387 181 135 767 602 826 24 × 2 = 1 + 0.799 862 186 774 362 271 535 205 652 48;
  • 11) 0.799 862 186 774 362 271 535 205 652 48 × 2 = 1 + 0.599 724 373 548 724 543 070 411 304 96;
  • 12) 0.599 724 373 548 724 543 070 411 304 96 × 2 = 1 + 0.199 448 747 097 449 086 140 822 609 92;
  • 13) 0.199 448 747 097 449 086 140 822 609 92 × 2 = 0 + 0.398 897 494 194 898 172 281 645 219 84;
  • 14) 0.398 897 494 194 898 172 281 645 219 84 × 2 = 0 + 0.797 794 988 389 796 344 563 290 439 68;
  • 15) 0.797 794 988 389 796 344 563 290 439 68 × 2 = 1 + 0.595 589 976 779 592 689 126 580 879 36;
  • 16) 0.595 589 976 779 592 689 126 580 879 36 × 2 = 1 + 0.191 179 953 559 185 378 253 161 758 72;
  • 17) 0.191 179 953 559 185 378 253 161 758 72 × 2 = 0 + 0.382 359 907 118 370 756 506 323 517 44;
  • 18) 0.382 359 907 118 370 756 506 323 517 44 × 2 = 0 + 0.764 719 814 236 741 513 012 647 034 88;
  • 19) 0.764 719 814 236 741 513 012 647 034 88 × 2 = 1 + 0.529 439 628 473 483 026 025 294 069 76;
  • 20) 0.529 439 628 473 483 026 025 294 069 76 × 2 = 1 + 0.058 879 256 946 966 052 050 588 139 52;
  • 21) 0.058 879 256 946 966 052 050 588 139 52 × 2 = 0 + 0.117 758 513 893 932 104 101 176 279 04;
  • 22) 0.117 758 513 893 932 104 101 176 279 04 × 2 = 0 + 0.235 517 027 787 864 208 202 352 558 08;
  • 23) 0.235 517 027 787 864 208 202 352 558 08 × 2 = 0 + 0.471 034 055 575 728 416 404 705 116 16;
  • 24) 0.471 034 055 575 728 416 404 705 116 16 × 2 = 0 + 0.942 068 111 151 456 832 809 410 232 32;
  • 25) 0.942 068 111 151 456 832 809 410 232 32 × 2 = 1 + 0.884 136 222 302 913 665 618 820 464 64;
  • 26) 0.884 136 222 302 913 665 618 820 464 64 × 2 = 1 + 0.768 272 444 605 827 331 237 640 929 28;
  • 27) 0.768 272 444 605 827 331 237 640 929 28 × 2 = 1 + 0.536 544 889 211 654 662 475 281 858 56;
  • 28) 0.536 544 889 211 654 662 475 281 858 56 × 2 = 1 + 0.073 089 778 423 309 324 950 563 717 12;
  • 29) 0.073 089 778 423 309 324 950 563 717 12 × 2 = 0 + 0.146 179 556 846 618 649 901 127 434 24;
  • 30) 0.146 179 556 846 618 649 901 127 434 24 × 2 = 0 + 0.292 359 113 693 237 299 802 254 868 48;
  • 31) 0.292 359 113 693 237 299 802 254 868 48 × 2 = 0 + 0.584 718 227 386 474 599 604 509 736 96;
  • 32) 0.584 718 227 386 474 599 604 509 736 96 × 2 = 1 + 0.169 436 454 772 949 199 209 019 473 92;
  • 33) 0.169 436 454 772 949 199 209 019 473 92 × 2 = 0 + 0.338 872 909 545 898 398 418 038 947 84;
  • 34) 0.338 872 909 545 898 398 418 038 947 84 × 2 = 0 + 0.677 745 819 091 796 796 836 077 895 68;
  • 35) 0.677 745 819 091 796 796 836 077 895 68 × 2 = 1 + 0.355 491 638 183 593 593 672 155 791 36;
  • 36) 0.355 491 638 183 593 593 672 155 791 36 × 2 = 0 + 0.710 983 276 367 187 187 344 311 582 72;
  • 37) 0.710 983 276 367 187 187 344 311 582 72 × 2 = 1 + 0.421 966 552 734 374 374 688 623 165 44;
  • 38) 0.421 966 552 734 374 374 688 623 165 44 × 2 = 0 + 0.843 933 105 468 748 749 377 246 330 88;
  • 39) 0.843 933 105 468 748 749 377 246 330 88 × 2 = 1 + 0.687 866 210 937 497 498 754 492 661 76;
  • 40) 0.687 866 210 937 497 498 754 492 661 76 × 2 = 1 + 0.375 732 421 874 994 997 508 985 323 52;
  • 41) 0.375 732 421 874 994 997 508 985 323 52 × 2 = 0 + 0.751 464 843 749 989 995 017 970 647 04;
  • 42) 0.751 464 843 749 989 995 017 970 647 04 × 2 = 1 + 0.502 929 687 499 979 990 035 941 294 08;
  • 43) 0.502 929 687 499 979 990 035 941 294 08 × 2 = 1 + 0.005 859 374 999 959 980 071 882 588 16;
  • 44) 0.005 859 374 999 959 980 071 882 588 16 × 2 = 0 + 0.011 718 749 999 919 960 143 765 176 32;
  • 45) 0.011 718 749 999 919 960 143 765 176 32 × 2 = 0 + 0.023 437 499 999 839 920 287 530 352 64;
  • 46) 0.023 437 499 999 839 920 287 530 352 64 × 2 = 0 + 0.046 874 999 999 679 840 575 060 705 28;
  • 47) 0.046 874 999 999 679 840 575 060 705 28 × 2 = 0 + 0.093 749 999 999 359 681 150 121 410 56;
  • 48) 0.093 749 999 999 359 681 150 121 410 56 × 2 = 0 + 0.187 499 999 998 719 362 300 242 821 12;
  • 49) 0.187 499 999 998 719 362 300 242 821 12 × 2 = 0 + 0.374 999 999 997 438 724 600 485 642 24;
  • 50) 0.374 999 999 997 438 724 600 485 642 24 × 2 = 0 + 0.749 999 999 994 877 449 200 971 284 48;
  • 51) 0.749 999 999 994 877 449 200 971 284 48 × 2 = 1 + 0.499 999 999 989 754 898 401 942 568 96;
  • 52) 0.499 999 999 989 754 898 401 942 568 96 × 2 = 0 + 0.999 999 999 979 509 796 803 885 137 92;
  • 53) 0.999 999 999 979 509 796 803 885 137 92 × 2 = 1 + 0.999 999 999 959 019 593 607 770 275 84;
  • 54) 0.999 999 999 959 019 593 607 770 275 84 × 2 = 1 + 0.999 999 999 918 039 187 215 540 551 68;
  • 55) 0.999 999 999 918 039 187 215 540 551 68 × 2 = 1 + 0.999 999 999 836 078 374 431 081 103 36;
  • 56) 0.999 999 999 836 078 374 431 081 103 36 × 2 = 1 + 0.999 999 999 672 156 748 862 162 206 72;
  • 57) 0.999 999 999 672 156 748 862 162 206 72 × 2 = 1 + 0.999 999 999 344 313 497 724 324 413 44;

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:


0.048 632 677 916 771 838 155 796 099 27(10) =

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

6. Positive number before normalization:

0.048 632 677 916 771 838 155 796 099 27(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:


0.048 632 677 916 771 838 155 796 099 27(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 -0.048 632 677 916 771 838 155 796 099 27 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