1.745 459 324 169 999 826 281 696 186 918 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 918 5(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
1.745 459 324 169 999 826 281 696 186 918 5(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.745 459 324 169 999 826 281 696 186 918 5.

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) 0.745 459 324 169 999 826 281 696 186 918 5 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 837;
  • 2) 0.490 918 648 339 999 652 563 392 373 837 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 674;
  • 3) 0.981 837 296 679 999 305 126 784 747 674 × 2 = 1 + 0.963 674 593 359 998 610 253 569 495 348;
  • 4) 0.963 674 593 359 998 610 253 569 495 348 × 2 = 1 + 0.927 349 186 719 997 220 507 138 990 696;
  • 5) 0.927 349 186 719 997 220 507 138 990 696 × 2 = 1 + 0.854 698 373 439 994 441 014 277 981 392;
  • 6) 0.854 698 373 439 994 441 014 277 981 392 × 2 = 1 + 0.709 396 746 879 988 882 028 555 962 784;
  • 7) 0.709 396 746 879 988 882 028 555 962 784 × 2 = 1 + 0.418 793 493 759 977 764 057 111 925 568;
  • 8) 0.418 793 493 759 977 764 057 111 925 568 × 2 = 0 + 0.837 586 987 519 955 528 114 223 851 136;
  • 9) 0.837 586 987 519 955 528 114 223 851 136 × 2 = 1 + 0.675 173 975 039 911 056 228 447 702 272;
  • 10) 0.675 173 975 039 911 056 228 447 702 272 × 2 = 1 + 0.350 347 950 079 822 112 456 895 404 544;
  • 11) 0.350 347 950 079 822 112 456 895 404 544 × 2 = 0 + 0.700 695 900 159 644 224 913 790 809 088;
  • 12) 0.700 695 900 159 644 224 913 790 809 088 × 2 = 1 + 0.401 391 800 319 288 449 827 581 618 176;
  • 13) 0.401 391 800 319 288 449 827 581 618 176 × 2 = 0 + 0.802 783 600 638 576 899 655 163 236 352;
  • 14) 0.802 783 600 638 576 899 655 163 236 352 × 2 = 1 + 0.605 567 201 277 153 799 310 326 472 704;
  • 15) 0.605 567 201 277 153 799 310 326 472 704 × 2 = 1 + 0.211 134 402 554 307 598 620 652 945 408;
  • 16) 0.211 134 402 554 307 598 620 652 945 408 × 2 = 0 + 0.422 268 805 108 615 197 241 305 890 816;
  • 17) 0.422 268 805 108 615 197 241 305 890 816 × 2 = 0 + 0.844 537 610 217 230 394 482 611 781 632;
  • 18) 0.844 537 610 217 230 394 482 611 781 632 × 2 = 1 + 0.689 075 220 434 460 788 965 223 563 264;
  • 19) 0.689 075 220 434 460 788 965 223 563 264 × 2 = 1 + 0.378 150 440 868 921 577 930 447 126 528;
  • 20) 0.378 150 440 868 921 577 930 447 126 528 × 2 = 0 + 0.756 300 881 737 843 155 860 894 253 056;
  • 21) 0.756 300 881 737 843 155 860 894 253 056 × 2 = 1 + 0.512 601 763 475 686 311 721 788 506 112;
  • 22) 0.512 601 763 475 686 311 721 788 506 112 × 2 = 1 + 0.025 203 526 951 372 623 443 577 012 224;
  • 23) 0.025 203 526 951 372 623 443 577 012 224 × 2 = 0 + 0.050 407 053 902 745 246 887 154 024 448;
  • 24) 0.050 407 053 902 745 246 887 154 024 448 × 2 = 0 + 0.100 814 107 805 490 493 774 308 048 896;
  • 25) 0.100 814 107 805 490 493 774 308 048 896 × 2 = 0 + 0.201 628 215 610 980 987 548 616 097 792;
  • 26) 0.201 628 215 610 980 987 548 616 097 792 × 2 = 0 + 0.403 256 431 221 961 975 097 232 195 584;
  • 27) 0.403 256 431 221 961 975 097 232 195 584 × 2 = 0 + 0.806 512 862 443 923 950 194 464 391 168;
  • 28) 0.806 512 862 443 923 950 194 464 391 168 × 2 = 1 + 0.613 025 724 887 847 900 388 928 782 336;
  • 29) 0.613 025 724 887 847 900 388 928 782 336 × 2 = 1 + 0.226 051 449 775 695 800 777 857 564 672;
  • 30) 0.226 051 449 775 695 800 777 857 564 672 × 2 = 0 + 0.452 102 899 551 391 601 555 715 129 344;
  • 31) 0.452 102 899 551 391 601 555 715 129 344 × 2 = 0 + 0.904 205 799 102 783 203 111 430 258 688;
  • 32) 0.904 205 799 102 783 203 111 430 258 688 × 2 = 1 + 0.808 411 598 205 566 406 222 860 517 376;
  • 33) 0.808 411 598 205 566 406 222 860 517 376 × 2 = 1 + 0.616 823 196 411 132 812 445 721 034 752;
  • 34) 0.616 823 196 411 132 812 445 721 034 752 × 2 = 1 + 0.233 646 392 822 265 624 891 442 069 504;
  • 35) 0.233 646 392 822 265 624 891 442 069 504 × 2 = 0 + 0.467 292 785 644 531 249 782 884 139 008;
  • 36) 0.467 292 785 644 531 249 782 884 139 008 × 2 = 0 + 0.934 585 571 289 062 499 565 768 278 016;
  • 37) 0.934 585 571 289 062 499 565 768 278 016 × 2 = 1 + 0.869 171 142 578 124 999 131 536 556 032;
  • 38) 0.869 171 142 578 124 999 131 536 556 032 × 2 = 1 + 0.738 342 285 156 249 998 263 073 112 064;
  • 39) 0.738 342 285 156 249 998 263 073 112 064 × 2 = 1 + 0.476 684 570 312 499 996 526 146 224 128;
  • 40) 0.476 684 570 312 499 996 526 146 224 128 × 2 = 0 + 0.953 369 140 624 999 993 052 292 448 256;
  • 41) 0.953 369 140 624 999 993 052 292 448 256 × 2 = 1 + 0.906 738 281 249 999 986 104 584 896 512;
  • 42) 0.906 738 281 249 999 986 104 584 896 512 × 2 = 1 + 0.813 476 562 499 999 972 209 169 793 024;
  • 43) 0.813 476 562 499 999 972 209 169 793 024 × 2 = 1 + 0.626 953 124 999 999 944 418 339 586 048;
  • 44) 0.626 953 124 999 999 944 418 339 586 048 × 2 = 1 + 0.253 906 249 999 999 888 836 679 172 096;
  • 45) 0.253 906 249 999 999 888 836 679 172 096 × 2 = 0 + 0.507 812 499 999 999 777 673 358 344 192;
  • 46) 0.507 812 499 999 999 777 673 358 344 192 × 2 = 1 + 0.015 624 999 999 999 555 346 716 688 384;
  • 47) 0.015 624 999 999 999 555 346 716 688 384 × 2 = 0 + 0.031 249 999 999 999 110 693 433 376 768;
  • 48) 0.031 249 999 999 999 110 693 433 376 768 × 2 = 0 + 0.062 499 999 999 998 221 386 866 753 536;
  • 49) 0.062 499 999 999 998 221 386 866 753 536 × 2 = 0 + 0.124 999 999 999 996 442 773 733 507 072;
  • 50) 0.124 999 999 999 996 442 773 733 507 072 × 2 = 0 + 0.249 999 999 999 992 885 547 467 014 144;
  • 51) 0.249 999 999 999 992 885 547 467 014 144 × 2 = 0 + 0.499 999 999 999 985 771 094 934 028 288;
  • 52) 0.499 999 999 999 985 771 094 934 028 288 × 2 = 0 + 0.999 999 999 999 971 542 189 868 056 576;
  • 53) 0.999 999 999 999 971 542 189 868 056 576 × 2 = 1 + 0.999 999 999 999 943 084 379 736 113 152;

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).


4. 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:


0.745 459 324 169 999 826 281 696 186 918 5(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 918 5(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.745 459 324 169 999 826 281 696 186 918 5(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(10)


9. 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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1023(10) =

011 1111 1111(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1111

Mantissa (52 bits) =
1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


Decimal number 1.745 459 324 169 999 826 281 696 186 918 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000

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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