1.761 828 530 288 944 705 262 110 801 413 655 281 062 76 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.761 828 530 288 944 705 262 110 801 413 655 281 062 76(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
1.761 828 530 288 944 705 262 110 801 413 655 281 062 76(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: 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: 0.761 828 530 288 944 705 262 110 801 413 655 281 062 76.

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) 0.761 828 530 288 944 705 262 110 801 413 655 281 062 76 × 2 = 1 + 0.523 657 060 577 889 410 524 221 602 827 310 562 125 52;
  • 2) 0.523 657 060 577 889 410 524 221 602 827 310 562 125 52 × 2 = 1 + 0.047 314 121 155 778 821 048 443 205 654 621 124 251 04;
  • 3) 0.047 314 121 155 778 821 048 443 205 654 621 124 251 04 × 2 = 0 + 0.094 628 242 311 557 642 096 886 411 309 242 248 502 08;
  • 4) 0.094 628 242 311 557 642 096 886 411 309 242 248 502 08 × 2 = 0 + 0.189 256 484 623 115 284 193 772 822 618 484 497 004 16;
  • 5) 0.189 256 484 623 115 284 193 772 822 618 484 497 004 16 × 2 = 0 + 0.378 512 969 246 230 568 387 545 645 236 968 994 008 32;
  • 6) 0.378 512 969 246 230 568 387 545 645 236 968 994 008 32 × 2 = 0 + 0.757 025 938 492 461 136 775 091 290 473 937 988 016 64;
  • 7) 0.757 025 938 492 461 136 775 091 290 473 937 988 016 64 × 2 = 1 + 0.514 051 876 984 922 273 550 182 580 947 875 976 033 28;
  • 8) 0.514 051 876 984 922 273 550 182 580 947 875 976 033 28 × 2 = 1 + 0.028 103 753 969 844 547 100 365 161 895 751 952 066 56;
  • 9) 0.028 103 753 969 844 547 100 365 161 895 751 952 066 56 × 2 = 0 + 0.056 207 507 939 689 094 200 730 323 791 503 904 133 12;
  • 10) 0.056 207 507 939 689 094 200 730 323 791 503 904 133 12 × 2 = 0 + 0.112 415 015 879 378 188 401 460 647 583 007 808 266 24;
  • 11) 0.112 415 015 879 378 188 401 460 647 583 007 808 266 24 × 2 = 0 + 0.224 830 031 758 756 376 802 921 295 166 015 616 532 48;
  • 12) 0.224 830 031 758 756 376 802 921 295 166 015 616 532 48 × 2 = 0 + 0.449 660 063 517 512 753 605 842 590 332 031 233 064 96;
  • 13) 0.449 660 063 517 512 753 605 842 590 332 031 233 064 96 × 2 = 0 + 0.899 320 127 035 025 507 211 685 180 664 062 466 129 92;
  • 14) 0.899 320 127 035 025 507 211 685 180 664 062 466 129 92 × 2 = 1 + 0.798 640 254 070 051 014 423 370 361 328 124 932 259 84;
  • 15) 0.798 640 254 070 051 014 423 370 361 328 124 932 259 84 × 2 = 1 + 0.597 280 508 140 102 028 846 740 722 656 249 864 519 68;
  • 16) 0.597 280 508 140 102 028 846 740 722 656 249 864 519 68 × 2 = 1 + 0.194 561 016 280 204 057 693 481 445 312 499 729 039 36;
  • 17) 0.194 561 016 280 204 057 693 481 445 312 499 729 039 36 × 2 = 0 + 0.389 122 032 560 408 115 386 962 890 624 999 458 078 72;
  • 18) 0.389 122 032 560 408 115 386 962 890 624 999 458 078 72 × 2 = 0 + 0.778 244 065 120 816 230 773 925 781 249 998 916 157 44;
  • 19) 0.778 244 065 120 816 230 773 925 781 249 998 916 157 44 × 2 = 1 + 0.556 488 130 241 632 461 547 851 562 499 997 832 314 88;
  • 20) 0.556 488 130 241 632 461 547 851 562 499 997 832 314 88 × 2 = 1 + 0.112 976 260 483 264 923 095 703 124 999 995 664 629 76;
  • 21) 0.112 976 260 483 264 923 095 703 124 999 995 664 629 76 × 2 = 0 + 0.225 952 520 966 529 846 191 406 249 999 991 329 259 52;
  • 22) 0.225 952 520 966 529 846 191 406 249 999 991 329 259 52 × 2 = 0 + 0.451 905 041 933 059 692 382 812 499 999 982 658 519 04;
  • 23) 0.451 905 041 933 059 692 382 812 499 999 982 658 519 04 × 2 = 0 + 0.903 810 083 866 119 384 765 624 999 999 965 317 038 08;
  • 24) 0.903 810 083 866 119 384 765 624 999 999 965 317 038 08 × 2 = 1 + 0.807 620 167 732 238 769 531 249 999 999 930 634 076 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:


0.761 828 530 288 944 705 262 110 801 413 655 281 062 76(10) =

0.1100 0011 0000 0111 0011 0001(2)

5. Positive number before normalization:

1.761 828 530 288 944 705 262 110 801 413 655 281 062 76(10) =

1.1100 0011 0000 0111 0011 0001(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:


1.761 828 530 288 944 705 262 110 801 413 655 281 062 76(10) =

1.1100 0011 0000 0111 0011 0001(2) =

1.1100 0011 0000 0111 0011 0001(2) × 20


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): 0

Mantissa (not normalized):
1.1100 0011 0000 0111 0011 0001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(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;
  • 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) =

127(10) =

0111 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 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. 110 0001 1000 0011 1001 1000 1 =

110 0001 1000 0011 1001 1000


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) =
0111 1111

Mantissa (23 bits) =
110 0001 1000 0011 1001 1000


Decimal number 1.761 828 530 288 944 705 262 110 801 413 655 281 062 76 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 110 0001 1000 0011 1001 1000

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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