2.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
2.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26.

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) 0.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26 × 2 = 1 + 0.436 563 656 918 090 470 720 574 942 705 324 995 514 488 52;
  • 2) 0.436 563 656 918 090 470 720 574 942 705 324 995 514 488 52 × 2 = 0 + 0.873 127 313 836 180 941 441 149 885 410 649 991 028 977 04;
  • 3) 0.873 127 313 836 180 941 441 149 885 410 649 991 028 977 04 × 2 = 1 + 0.746 254 627 672 361 882 882 299 770 821 299 982 057 954 08;
  • 4) 0.746 254 627 672 361 882 882 299 770 821 299 982 057 954 08 × 2 = 1 + 0.492 509 255 344 723 765 764 599 541 642 599 964 115 908 16;
  • 5) 0.492 509 255 344 723 765 764 599 541 642 599 964 115 908 16 × 2 = 0 + 0.985 018 510 689 447 531 529 199 083 285 199 928 231 816 32;
  • 6) 0.985 018 510 689 447 531 529 199 083 285 199 928 231 816 32 × 2 = 1 + 0.970 037 021 378 895 063 058 398 166 570 399 856 463 632 64;
  • 7) 0.970 037 021 378 895 063 058 398 166 570 399 856 463 632 64 × 2 = 1 + 0.940 074 042 757 790 126 116 796 333 140 799 712 927 265 28;
  • 8) 0.940 074 042 757 790 126 116 796 333 140 799 712 927 265 28 × 2 = 1 + 0.880 148 085 515 580 252 233 592 666 281 599 425 854 530 56;
  • 9) 0.880 148 085 515 580 252 233 592 666 281 599 425 854 530 56 × 2 = 1 + 0.760 296 171 031 160 504 467 185 332 563 198 851 709 061 12;
  • 10) 0.760 296 171 031 160 504 467 185 332 563 198 851 709 061 12 × 2 = 1 + 0.520 592 342 062 321 008 934 370 665 126 397 703 418 122 24;
  • 11) 0.520 592 342 062 321 008 934 370 665 126 397 703 418 122 24 × 2 = 1 + 0.041 184 684 124 642 017 868 741 330 252 795 406 836 244 48;
  • 12) 0.041 184 684 124 642 017 868 741 330 252 795 406 836 244 48 × 2 = 0 + 0.082 369 368 249 284 035 737 482 660 505 590 813 672 488 96;
  • 13) 0.082 369 368 249 284 035 737 482 660 505 590 813 672 488 96 × 2 = 0 + 0.164 738 736 498 568 071 474 965 321 011 181 627 344 977 92;
  • 14) 0.164 738 736 498 568 071 474 965 321 011 181 627 344 977 92 × 2 = 0 + 0.329 477 472 997 136 142 949 930 642 022 363 254 689 955 84;
  • 15) 0.329 477 472 997 136 142 949 930 642 022 363 254 689 955 84 × 2 = 0 + 0.658 954 945 994 272 285 899 861 284 044 726 509 379 911 68;
  • 16) 0.658 954 945 994 272 285 899 861 284 044 726 509 379 911 68 × 2 = 1 + 0.317 909 891 988 544 571 799 722 568 089 453 018 759 823 36;
  • 17) 0.317 909 891 988 544 571 799 722 568 089 453 018 759 823 36 × 2 = 0 + 0.635 819 783 977 089 143 599 445 136 178 906 037 519 646 72;
  • 18) 0.635 819 783 977 089 143 599 445 136 178 906 037 519 646 72 × 2 = 1 + 0.271 639 567 954 178 287 198 890 272 357 812 075 039 293 44;
  • 19) 0.271 639 567 954 178 287 198 890 272 357 812 075 039 293 44 × 2 = 0 + 0.543 279 135 908 356 574 397 780 544 715 624 150 078 586 88;
  • 20) 0.543 279 135 908 356 574 397 780 544 715 624 150 078 586 88 × 2 = 1 + 0.086 558 271 816 713 148 795 561 089 431 248 300 157 173 76;
  • 21) 0.086 558 271 816 713 148 795 561 089 431 248 300 157 173 76 × 2 = 0 + 0.173 116 543 633 426 297 591 122 178 862 496 600 314 347 52;
  • 22) 0.173 116 543 633 426 297 591 122 178 862 496 600 314 347 52 × 2 = 0 + 0.346 233 087 266 852 595 182 244 357 724 993 200 628 695 04;
  • 23) 0.346 233 087 266 852 595 182 244 357 724 993 200 628 695 04 × 2 = 0 + 0.692 466 174 533 705 190 364 488 715 449 986 401 257 390 08;
  • 24) 0.692 466 174 533 705 190 364 488 715 449 986 401 257 390 08 × 2 = 1 + 0.384 932 349 067 410 380 728 977 430 899 972 802 514 780 16;

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:


0.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26(10) =

0.1011 0111 1110 0001 0101 0001(2)

5. Positive number before normalization:

2.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26(10) =

10.1011 0111 1110 0001 0101 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26(10) =

10.1011 0111 1110 0001 0101 0001(2) =

10.1011 0111 1110 0001 0101 0001(2) × 20 =

1.0101 1011 1111 0000 1010 1000 1(2) × 21


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.0101 1011 1111 0000 1010 1000 1


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

128(10) =

1000 0000(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 23 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. 010 1101 1111 1000 0101 0100 01 =

010 1101 1111 1000 0101 0100


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
1000 0000

Mantissa (23 bits) =
010 1101 1111 1000 0101 0100


Decimal number 2.718 281 828 459 045 235 360 287 471 352 662 497 757 244 26 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 010 1101 1111 1000 0101 0100

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111