1.111 111 101 011 001 111 011 101 000 001 100 112 05 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.111 111 101 011 001 111 011 101 000 001 100 112 05(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
1.111 111 101 011 001 111 011 101 000 001 100 112 05(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: 0.111 111 101 011 001 111 011 101 000 001 100 112 05.

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) 0.111 111 101 011 001 111 011 101 000 001 100 112 05 × 2 = 0 + 0.222 222 202 022 002 222 022 202 000 002 200 224 1;
  • 2) 0.222 222 202 022 002 222 022 202 000 002 200 224 1 × 2 = 0 + 0.444 444 404 044 004 444 044 404 000 004 400 448 2;
  • 3) 0.444 444 404 044 004 444 044 404 000 004 400 448 2 × 2 = 0 + 0.888 888 808 088 008 888 088 808 000 008 800 896 4;
  • 4) 0.888 888 808 088 008 888 088 808 000 008 800 896 4 × 2 = 1 + 0.777 777 616 176 017 776 177 616 000 017 601 792 8;
  • 5) 0.777 777 616 176 017 776 177 616 000 017 601 792 8 × 2 = 1 + 0.555 555 232 352 035 552 355 232 000 035 203 585 6;
  • 6) 0.555 555 232 352 035 552 355 232 000 035 203 585 6 × 2 = 1 + 0.111 110 464 704 071 104 710 464 000 070 407 171 2;
  • 7) 0.111 110 464 704 071 104 710 464 000 070 407 171 2 × 2 = 0 + 0.222 220 929 408 142 209 420 928 000 140 814 342 4;
  • 8) 0.222 220 929 408 142 209 420 928 000 140 814 342 4 × 2 = 0 + 0.444 441 858 816 284 418 841 856 000 281 628 684 8;
  • 9) 0.444 441 858 816 284 418 841 856 000 281 628 684 8 × 2 = 0 + 0.888 883 717 632 568 837 683 712 000 563 257 369 6;
  • 10) 0.888 883 717 632 568 837 683 712 000 563 257 369 6 × 2 = 1 + 0.777 767 435 265 137 675 367 424 001 126 514 739 2;
  • 11) 0.777 767 435 265 137 675 367 424 001 126 514 739 2 × 2 = 1 + 0.555 534 870 530 275 350 734 848 002 253 029 478 4;
  • 12) 0.555 534 870 530 275 350 734 848 002 253 029 478 4 × 2 = 1 + 0.111 069 741 060 550 701 469 696 004 506 058 956 8;
  • 13) 0.111 069 741 060 550 701 469 696 004 506 058 956 8 × 2 = 0 + 0.222 139 482 121 101 402 939 392 009 012 117 913 6;
  • 14) 0.222 139 482 121 101 402 939 392 009 012 117 913 6 × 2 = 0 + 0.444 278 964 242 202 805 878 784 018 024 235 827 2;
  • 15) 0.444 278 964 242 202 805 878 784 018 024 235 827 2 × 2 = 0 + 0.888 557 928 484 405 611 757 568 036 048 471 654 4;
  • 16) 0.888 557 928 484 405 611 757 568 036 048 471 654 4 × 2 = 1 + 0.777 115 856 968 811 223 515 136 072 096 943 308 8;
  • 17) 0.777 115 856 968 811 223 515 136 072 096 943 308 8 × 2 = 1 + 0.554 231 713 937 622 447 030 272 144 193 886 617 6;
  • 18) 0.554 231 713 937 622 447 030 272 144 193 886 617 6 × 2 = 1 + 0.108 463 427 875 244 894 060 544 288 387 773 235 2;
  • 19) 0.108 463 427 875 244 894 060 544 288 387 773 235 2 × 2 = 0 + 0.216 926 855 750 489 788 121 088 576 775 546 470 4;
  • 20) 0.216 926 855 750 489 788 121 088 576 775 546 470 4 × 2 = 0 + 0.433 853 711 500 979 576 242 177 153 551 092 940 8;
  • 21) 0.433 853 711 500 979 576 242 177 153 551 092 940 8 × 2 = 0 + 0.867 707 423 001 959 152 484 354 307 102 185 881 6;
  • 22) 0.867 707 423 001 959 152 484 354 307 102 185 881 6 × 2 = 1 + 0.735 414 846 003 918 304 968 708 614 204 371 763 2;
  • 23) 0.735 414 846 003 918 304 968 708 614 204 371 763 2 × 2 = 1 + 0.470 829 692 007 836 609 937 417 228 408 743 526 4;
  • 24) 0.470 829 692 007 836 609 937 417 228 408 743 526 4 × 2 = 0 + 0.941 659 384 015 673 219 874 834 456 817 487 052 8;

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:


0.111 111 101 011 001 111 011 101 000 001 100 112 05(10) =

0.0001 1100 0111 0001 1100 0110(2)

5. Positive number before normalization:

1.111 111 101 011 001 111 011 101 000 001 100 112 05(10) =

1.0001 1100 0111 0001 1100 0110(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:


1.111 111 101 011 001 111 011 101 000 001 100 112 05(10) =

1.0001 1100 0111 0001 1100 0110(2) =

1.0001 1100 0111 0001 1100 0110(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):
1.0001 1100 0111 0001 1100 0110


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. 000 1110 0011 1000 1110 0011 0 =

000 1110 0011 1000 1110 0011


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) =
000 1110 0011 1000 1110 0011


Decimal number 1.111 111 101 011 001 111 011 101 000 001 100 112 05 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 000 1110 0011 1000 1110 0011

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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