1.745 459 324 169 999 826 281 696 182 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 182 1(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
1.745 459 324 169 999 826 281 696 182 1(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: 0.745 459 324 169 999 826 281 696 182 1.

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) 0.745 459 324 169 999 826 281 696 182 1 × 2 = 1 + 0.490 918 648 339 999 652 563 392 364 2;
  • 2) 0.490 918 648 339 999 652 563 392 364 2 × 2 = 0 + 0.981 837 296 679 999 305 126 784 728 4;
  • 3) 0.981 837 296 679 999 305 126 784 728 4 × 2 = 1 + 0.963 674 593 359 998 610 253 569 456 8;
  • 4) 0.963 674 593 359 998 610 253 569 456 8 × 2 = 1 + 0.927 349 186 719 997 220 507 138 913 6;
  • 5) 0.927 349 186 719 997 220 507 138 913 6 × 2 = 1 + 0.854 698 373 439 994 441 014 277 827 2;
  • 6) 0.854 698 373 439 994 441 014 277 827 2 × 2 = 1 + 0.709 396 746 879 988 882 028 555 654 4;
  • 7) 0.709 396 746 879 988 882 028 555 654 4 × 2 = 1 + 0.418 793 493 759 977 764 057 111 308 8;
  • 8) 0.418 793 493 759 977 764 057 111 308 8 × 2 = 0 + 0.837 586 987 519 955 528 114 222 617 6;
  • 9) 0.837 586 987 519 955 528 114 222 617 6 × 2 = 1 + 0.675 173 975 039 911 056 228 445 235 2;
  • 10) 0.675 173 975 039 911 056 228 445 235 2 × 2 = 1 + 0.350 347 950 079 822 112 456 890 470 4;
  • 11) 0.350 347 950 079 822 112 456 890 470 4 × 2 = 0 + 0.700 695 900 159 644 224 913 780 940 8;
  • 12) 0.700 695 900 159 644 224 913 780 940 8 × 2 = 1 + 0.401 391 800 319 288 449 827 561 881 6;
  • 13) 0.401 391 800 319 288 449 827 561 881 6 × 2 = 0 + 0.802 783 600 638 576 899 655 123 763 2;
  • 14) 0.802 783 600 638 576 899 655 123 763 2 × 2 = 1 + 0.605 567 201 277 153 799 310 247 526 4;
  • 15) 0.605 567 201 277 153 799 310 247 526 4 × 2 = 1 + 0.211 134 402 554 307 598 620 495 052 8;
  • 16) 0.211 134 402 554 307 598 620 495 052 8 × 2 = 0 + 0.422 268 805 108 615 197 240 990 105 6;
  • 17) 0.422 268 805 108 615 197 240 990 105 6 × 2 = 0 + 0.844 537 610 217 230 394 481 980 211 2;
  • 18) 0.844 537 610 217 230 394 481 980 211 2 × 2 = 1 + 0.689 075 220 434 460 788 963 960 422 4;
  • 19) 0.689 075 220 434 460 788 963 960 422 4 × 2 = 1 + 0.378 150 440 868 921 577 927 920 844 8;
  • 20) 0.378 150 440 868 921 577 927 920 844 8 × 2 = 0 + 0.756 300 881 737 843 155 855 841 689 6;
  • 21) 0.756 300 881 737 843 155 855 841 689 6 × 2 = 1 + 0.512 601 763 475 686 311 711 683 379 2;
  • 22) 0.512 601 763 475 686 311 711 683 379 2 × 2 = 1 + 0.025 203 526 951 372 623 423 366 758 4;
  • 23) 0.025 203 526 951 372 623 423 366 758 4 × 2 = 0 + 0.050 407 053 902 745 246 846 733 516 8;
  • 24) 0.050 407 053 902 745 246 846 733 516 8 × 2 = 0 + 0.100 814 107 805 490 493 693 467 033 6;
  • 25) 0.100 814 107 805 490 493 693 467 033 6 × 2 = 0 + 0.201 628 215 610 980 987 386 934 067 2;
  • 26) 0.201 628 215 610 980 987 386 934 067 2 × 2 = 0 + 0.403 256 431 221 961 974 773 868 134 4;
  • 27) 0.403 256 431 221 961 974 773 868 134 4 × 2 = 0 + 0.806 512 862 443 923 949 547 736 268 8;
  • 28) 0.806 512 862 443 923 949 547 736 268 8 × 2 = 1 + 0.613 025 724 887 847 899 095 472 537 6;
  • 29) 0.613 025 724 887 847 899 095 472 537 6 × 2 = 1 + 0.226 051 449 775 695 798 190 945 075 2;
  • 30) 0.226 051 449 775 695 798 190 945 075 2 × 2 = 0 + 0.452 102 899 551 391 596 381 890 150 4;
  • 31) 0.452 102 899 551 391 596 381 890 150 4 × 2 = 0 + 0.904 205 799 102 783 192 763 780 300 8;
  • 32) 0.904 205 799 102 783 192 763 780 300 8 × 2 = 1 + 0.808 411 598 205 566 385 527 560 601 6;
  • 33) 0.808 411 598 205 566 385 527 560 601 6 × 2 = 1 + 0.616 823 196 411 132 771 055 121 203 2;
  • 34) 0.616 823 196 411 132 771 055 121 203 2 × 2 = 1 + 0.233 646 392 822 265 542 110 242 406 4;
  • 35) 0.233 646 392 822 265 542 110 242 406 4 × 2 = 0 + 0.467 292 785 644 531 084 220 484 812 8;
  • 36) 0.467 292 785 644 531 084 220 484 812 8 × 2 = 0 + 0.934 585 571 289 062 168 440 969 625 6;
  • 37) 0.934 585 571 289 062 168 440 969 625 6 × 2 = 1 + 0.869 171 142 578 124 336 881 939 251 2;
  • 38) 0.869 171 142 578 124 336 881 939 251 2 × 2 = 1 + 0.738 342 285 156 248 673 763 878 502 4;
  • 39) 0.738 342 285 156 248 673 763 878 502 4 × 2 = 1 + 0.476 684 570 312 497 347 527 757 004 8;
  • 40) 0.476 684 570 312 497 347 527 757 004 8 × 2 = 0 + 0.953 369 140 624 994 695 055 514 009 6;
  • 41) 0.953 369 140 624 994 695 055 514 009 6 × 2 = 1 + 0.906 738 281 249 989 390 111 028 019 2;
  • 42) 0.906 738 281 249 989 390 111 028 019 2 × 2 = 1 + 0.813 476 562 499 978 780 222 056 038 4;
  • 43) 0.813 476 562 499 978 780 222 056 038 4 × 2 = 1 + 0.626 953 124 999 957 560 444 112 076 8;
  • 44) 0.626 953 124 999 957 560 444 112 076 8 × 2 = 1 + 0.253 906 249 999 915 120 888 224 153 6;
  • 45) 0.253 906 249 999 915 120 888 224 153 6 × 2 = 0 + 0.507 812 499 999 830 241 776 448 307 2;
  • 46) 0.507 812 499 999 830 241 776 448 307 2 × 2 = 1 + 0.015 624 999 999 660 483 552 896 614 4;
  • 47) 0.015 624 999 999 660 483 552 896 614 4 × 2 = 0 + 0.031 249 999 999 320 967 105 793 228 8;
  • 48) 0.031 249 999 999 320 967 105 793 228 8 × 2 = 0 + 0.062 499 999 998 641 934 211 586 457 6;
  • 49) 0.062 499 999 998 641 934 211 586 457 6 × 2 = 0 + 0.124 999 999 997 283 868 423 172 915 2;
  • 50) 0.124 999 999 997 283 868 423 172 915 2 × 2 = 0 + 0.249 999 999 994 567 736 846 345 830 4;
  • 51) 0.249 999 999 994 567 736 846 345 830 4 × 2 = 0 + 0.499 999 999 989 135 473 692 691 660 8;
  • 52) 0.499 999 999 989 135 473 692 691 660 8 × 2 = 0 + 0.999 999 999 978 270 947 385 383 321 6;
  • 53) 0.999 999 999 978 270 947 385 383 321 6 × 2 = 1 + 0.999 999 999 956 541 894 770 766 643 2;

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:


0.745 459 324 169 999 826 281 696 182 1(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 182 1(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:


1.745 459 324 169 999 826 281 696 182 1(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 1.745 459 324 169 999 826 281 696 182 1 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