1.793 662 034 335 765 850 782 373 863 02 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.793 662 034 335 765 850 782 373 863 02(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
1.793 662 034 335 765 850 782 373 863 02(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: 0.793 662 034 335 765 850 782 373 863 02.

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) 0.793 662 034 335 765 850 782 373 863 02 × 2 = 1 + 0.587 324 068 671 531 701 564 747 726 04;
  • 2) 0.587 324 068 671 531 701 564 747 726 04 × 2 = 1 + 0.174 648 137 343 063 403 129 495 452 08;
  • 3) 0.174 648 137 343 063 403 129 495 452 08 × 2 = 0 + 0.349 296 274 686 126 806 258 990 904 16;
  • 4) 0.349 296 274 686 126 806 258 990 904 16 × 2 = 0 + 0.698 592 549 372 253 612 517 981 808 32;
  • 5) 0.698 592 549 372 253 612 517 981 808 32 × 2 = 1 + 0.397 185 098 744 507 225 035 963 616 64;
  • 6) 0.397 185 098 744 507 225 035 963 616 64 × 2 = 0 + 0.794 370 197 489 014 450 071 927 233 28;
  • 7) 0.794 370 197 489 014 450 071 927 233 28 × 2 = 1 + 0.588 740 394 978 028 900 143 854 466 56;
  • 8) 0.588 740 394 978 028 900 143 854 466 56 × 2 = 1 + 0.177 480 789 956 057 800 287 708 933 12;
  • 9) 0.177 480 789 956 057 800 287 708 933 12 × 2 = 0 + 0.354 961 579 912 115 600 575 417 866 24;
  • 10) 0.354 961 579 912 115 600 575 417 866 24 × 2 = 0 + 0.709 923 159 824 231 201 150 835 732 48;
  • 11) 0.709 923 159 824 231 201 150 835 732 48 × 2 = 1 + 0.419 846 319 648 462 402 301 671 464 96;
  • 12) 0.419 846 319 648 462 402 301 671 464 96 × 2 = 0 + 0.839 692 639 296 924 804 603 342 929 92;
  • 13) 0.839 692 639 296 924 804 603 342 929 92 × 2 = 1 + 0.679 385 278 593 849 609 206 685 859 84;
  • 14) 0.679 385 278 593 849 609 206 685 859 84 × 2 = 1 + 0.358 770 557 187 699 218 413 371 719 68;
  • 15) 0.358 770 557 187 699 218 413 371 719 68 × 2 = 0 + 0.717 541 114 375 398 436 826 743 439 36;
  • 16) 0.717 541 114 375 398 436 826 743 439 36 × 2 = 1 + 0.435 082 228 750 796 873 653 486 878 72;
  • 17) 0.435 082 228 750 796 873 653 486 878 72 × 2 = 0 + 0.870 164 457 501 593 747 306 973 757 44;
  • 18) 0.870 164 457 501 593 747 306 973 757 44 × 2 = 1 + 0.740 328 915 003 187 494 613 947 514 88;
  • 19) 0.740 328 915 003 187 494 613 947 514 88 × 2 = 1 + 0.480 657 830 006 374 989 227 895 029 76;
  • 20) 0.480 657 830 006 374 989 227 895 029 76 × 2 = 0 + 0.961 315 660 012 749 978 455 790 059 52;
  • 21) 0.961 315 660 012 749 978 455 790 059 52 × 2 = 1 + 0.922 631 320 025 499 956 911 580 119 04;
  • 22) 0.922 631 320 025 499 956 911 580 119 04 × 2 = 1 + 0.845 262 640 050 999 913 823 160 238 08;
  • 23) 0.845 262 640 050 999 913 823 160 238 08 × 2 = 1 + 0.690 525 280 101 999 827 646 320 476 16;
  • 24) 0.690 525 280 101 999 827 646 320 476 16 × 2 = 1 + 0.381 050 560 203 999 655 292 640 952 32;

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:


0.793 662 034 335 765 850 782 373 863 02(10) =

0.1100 1011 0010 1101 0110 1111(2)

5. Positive number before normalization:

1.793 662 034 335 765 850 782 373 863 02(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:


1.793 662 034 335 765 850 782 373 863 02(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 1.793 662 034 335 765 850 782 373 863 02 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