1.793 662 034 335 765 850 782 373 861 32 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.793 662 034 335 765 850 782 373 861 32(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
1.793 662 034 335 765 850 782 373 861 32(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: 0.793 662 034 335 765 850 782 373 861 32.

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) 0.793 662 034 335 765 850 782 373 861 32 × 2 = 1 + 0.587 324 068 671 531 701 564 747 722 64;
  • 2) 0.587 324 068 671 531 701 564 747 722 64 × 2 = 1 + 0.174 648 137 343 063 403 129 495 445 28;
  • 3) 0.174 648 137 343 063 403 129 495 445 28 × 2 = 0 + 0.349 296 274 686 126 806 258 990 890 56;
  • 4) 0.349 296 274 686 126 806 258 990 890 56 × 2 = 0 + 0.698 592 549 372 253 612 517 981 781 12;
  • 5) 0.698 592 549 372 253 612 517 981 781 12 × 2 = 1 + 0.397 185 098 744 507 225 035 963 562 24;
  • 6) 0.397 185 098 744 507 225 035 963 562 24 × 2 = 0 + 0.794 370 197 489 014 450 071 927 124 48;
  • 7) 0.794 370 197 489 014 450 071 927 124 48 × 2 = 1 + 0.588 740 394 978 028 900 143 854 248 96;
  • 8) 0.588 740 394 978 028 900 143 854 248 96 × 2 = 1 + 0.177 480 789 956 057 800 287 708 497 92;
  • 9) 0.177 480 789 956 057 800 287 708 497 92 × 2 = 0 + 0.354 961 579 912 115 600 575 416 995 84;
  • 10) 0.354 961 579 912 115 600 575 416 995 84 × 2 = 0 + 0.709 923 159 824 231 201 150 833 991 68;
  • 11) 0.709 923 159 824 231 201 150 833 991 68 × 2 = 1 + 0.419 846 319 648 462 402 301 667 983 36;
  • 12) 0.419 846 319 648 462 402 301 667 983 36 × 2 = 0 + 0.839 692 639 296 924 804 603 335 966 72;
  • 13) 0.839 692 639 296 924 804 603 335 966 72 × 2 = 1 + 0.679 385 278 593 849 609 206 671 933 44;
  • 14) 0.679 385 278 593 849 609 206 671 933 44 × 2 = 1 + 0.358 770 557 187 699 218 413 343 866 88;
  • 15) 0.358 770 557 187 699 218 413 343 866 88 × 2 = 0 + 0.717 541 114 375 398 436 826 687 733 76;
  • 16) 0.717 541 114 375 398 436 826 687 733 76 × 2 = 1 + 0.435 082 228 750 796 873 653 375 467 52;
  • 17) 0.435 082 228 750 796 873 653 375 467 52 × 2 = 0 + 0.870 164 457 501 593 747 306 750 935 04;
  • 18) 0.870 164 457 501 593 747 306 750 935 04 × 2 = 1 + 0.740 328 915 003 187 494 613 501 870 08;
  • 19) 0.740 328 915 003 187 494 613 501 870 08 × 2 = 1 + 0.480 657 830 006 374 989 227 003 740 16;
  • 20) 0.480 657 830 006 374 989 227 003 740 16 × 2 = 0 + 0.961 315 660 012 749 978 454 007 480 32;
  • 21) 0.961 315 660 012 749 978 454 007 480 32 × 2 = 1 + 0.922 631 320 025 499 956 908 014 960 64;
  • 22) 0.922 631 320 025 499 956 908 014 960 64 × 2 = 1 + 0.845 262 640 050 999 913 816 029 921 28;
  • 23) 0.845 262 640 050 999 913 816 029 921 28 × 2 = 1 + 0.690 525 280 101 999 827 632 059 842 56;
  • 24) 0.690 525 280 101 999 827 632 059 842 56 × 2 = 1 + 0.381 050 560 203 999 655 264 119 685 12;

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:


0.793 662 034 335 765 850 782 373 861 32(10) =

0.1100 1011 0010 1101 0110 1111(2)

5. Positive number before normalization:

1.793 662 034 335 765 850 782 373 861 32(10) =

1.1100 1011 0010 1101 0110 1111(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:


1.793 662 034 335 765 850 782 373 861 32(10) =

1.1100 1011 0010 1101 0110 1111(2) =

1.1100 1011 0010 1101 0110 1111(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 1011 0010 1101 0110 1111


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 0101 1001 0110 1011 0111 1 =

110 0101 1001 0110 1011 0111


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 0101 1001 0110 1011 0111


Decimal number 1.793 662 034 335 765 850 782 373 861 32 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 110 0101 1001 0110 1011 0111

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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