1.111 111 101 011 001 111 011 101 000 001 701 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.111 111 101 011 001 111 011 101 000 001 701(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
1.111 111 101 011 001 111 011 101 000 001 701(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: 0.111 111 101 011 001 111 011 101 000 001 701.

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) 0.111 111 101 011 001 111 011 101 000 001 701 × 2 = 0 + 0.222 222 202 022 002 222 022 202 000 003 402;
  • 2) 0.222 222 202 022 002 222 022 202 000 003 402 × 2 = 0 + 0.444 444 404 044 004 444 044 404 000 006 804;
  • 3) 0.444 444 404 044 004 444 044 404 000 006 804 × 2 = 0 + 0.888 888 808 088 008 888 088 808 000 013 608;
  • 4) 0.888 888 808 088 008 888 088 808 000 013 608 × 2 = 1 + 0.777 777 616 176 017 776 177 616 000 027 216;
  • 5) 0.777 777 616 176 017 776 177 616 000 027 216 × 2 = 1 + 0.555 555 232 352 035 552 355 232 000 054 432;
  • 6) 0.555 555 232 352 035 552 355 232 000 054 432 × 2 = 1 + 0.111 110 464 704 071 104 710 464 000 108 864;
  • 7) 0.111 110 464 704 071 104 710 464 000 108 864 × 2 = 0 + 0.222 220 929 408 142 209 420 928 000 217 728;
  • 8) 0.222 220 929 408 142 209 420 928 000 217 728 × 2 = 0 + 0.444 441 858 816 284 418 841 856 000 435 456;
  • 9) 0.444 441 858 816 284 418 841 856 000 435 456 × 2 = 0 + 0.888 883 717 632 568 837 683 712 000 870 912;
  • 10) 0.888 883 717 632 568 837 683 712 000 870 912 × 2 = 1 + 0.777 767 435 265 137 675 367 424 001 741 824;
  • 11) 0.777 767 435 265 137 675 367 424 001 741 824 × 2 = 1 + 0.555 534 870 530 275 350 734 848 003 483 648;
  • 12) 0.555 534 870 530 275 350 734 848 003 483 648 × 2 = 1 + 0.111 069 741 060 550 701 469 696 006 967 296;
  • 13) 0.111 069 741 060 550 701 469 696 006 967 296 × 2 = 0 + 0.222 139 482 121 101 402 939 392 013 934 592;
  • 14) 0.222 139 482 121 101 402 939 392 013 934 592 × 2 = 0 + 0.444 278 964 242 202 805 878 784 027 869 184;
  • 15) 0.444 278 964 242 202 805 878 784 027 869 184 × 2 = 0 + 0.888 557 928 484 405 611 757 568 055 738 368;
  • 16) 0.888 557 928 484 405 611 757 568 055 738 368 × 2 = 1 + 0.777 115 856 968 811 223 515 136 111 476 736;
  • 17) 0.777 115 856 968 811 223 515 136 111 476 736 × 2 = 1 + 0.554 231 713 937 622 447 030 272 222 953 472;
  • 18) 0.554 231 713 937 622 447 030 272 222 953 472 × 2 = 1 + 0.108 463 427 875 244 894 060 544 445 906 944;
  • 19) 0.108 463 427 875 244 894 060 544 445 906 944 × 2 = 0 + 0.216 926 855 750 489 788 121 088 891 813 888;
  • 20) 0.216 926 855 750 489 788 121 088 891 813 888 × 2 = 0 + 0.433 853 711 500 979 576 242 177 783 627 776;
  • 21) 0.433 853 711 500 979 576 242 177 783 627 776 × 2 = 0 + 0.867 707 423 001 959 152 484 355 567 255 552;
  • 22) 0.867 707 423 001 959 152 484 355 567 255 552 × 2 = 1 + 0.735 414 846 003 918 304 968 711 134 511 104;
  • 23) 0.735 414 846 003 918 304 968 711 134 511 104 × 2 = 1 + 0.470 829 692 007 836 609 937 422 269 022 208;
  • 24) 0.470 829 692 007 836 609 937 422 269 022 208 × 2 = 0 + 0.941 659 384 015 673 219 874 844 538 044 416;

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:


0.111 111 101 011 001 111 011 101 000 001 701(10) =

0.0001 1100 0111 0001 1100 0110(2)

5. Positive number before normalization:

1.111 111 101 011 001 111 011 101 000 001 701(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:


1.111 111 101 011 001 111 011 101 000 001 701(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 1.111 111 101 011 001 111 011 101 000 001 701 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