1.732 050 807 568 877 293 527 446 341 506 725 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.732 050 807 568 877 293 527 446 341 506 725(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
1.732 050 807 568 877 293 527 446 341 506 725(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: 0.732 050 807 568 877 293 527 446 341 506 725.

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) 0.732 050 807 568 877 293 527 446 341 506 725 × 2 = 1 + 0.464 101 615 137 754 587 054 892 683 013 45;
  • 2) 0.464 101 615 137 754 587 054 892 683 013 45 × 2 = 0 + 0.928 203 230 275 509 174 109 785 366 026 9;
  • 3) 0.928 203 230 275 509 174 109 785 366 026 9 × 2 = 1 + 0.856 406 460 551 018 348 219 570 732 053 8;
  • 4) 0.856 406 460 551 018 348 219 570 732 053 8 × 2 = 1 + 0.712 812 921 102 036 696 439 141 464 107 6;
  • 5) 0.712 812 921 102 036 696 439 141 464 107 6 × 2 = 1 + 0.425 625 842 204 073 392 878 282 928 215 2;
  • 6) 0.425 625 842 204 073 392 878 282 928 215 2 × 2 = 0 + 0.851 251 684 408 146 785 756 565 856 430 4;
  • 7) 0.851 251 684 408 146 785 756 565 856 430 4 × 2 = 1 + 0.702 503 368 816 293 571 513 131 712 860 8;
  • 8) 0.702 503 368 816 293 571 513 131 712 860 8 × 2 = 1 + 0.405 006 737 632 587 143 026 263 425 721 6;
  • 9) 0.405 006 737 632 587 143 026 263 425 721 6 × 2 = 0 + 0.810 013 475 265 174 286 052 526 851 443 2;
  • 10) 0.810 013 475 265 174 286 052 526 851 443 2 × 2 = 1 + 0.620 026 950 530 348 572 105 053 702 886 4;
  • 11) 0.620 026 950 530 348 572 105 053 702 886 4 × 2 = 1 + 0.240 053 901 060 697 144 210 107 405 772 8;
  • 12) 0.240 053 901 060 697 144 210 107 405 772 8 × 2 = 0 + 0.480 107 802 121 394 288 420 214 811 545 6;
  • 13) 0.480 107 802 121 394 288 420 214 811 545 6 × 2 = 0 + 0.960 215 604 242 788 576 840 429 623 091 2;
  • 14) 0.960 215 604 242 788 576 840 429 623 091 2 × 2 = 1 + 0.920 431 208 485 577 153 680 859 246 182 4;
  • 15) 0.920 431 208 485 577 153 680 859 246 182 4 × 2 = 1 + 0.840 862 416 971 154 307 361 718 492 364 8;
  • 16) 0.840 862 416 971 154 307 361 718 492 364 8 × 2 = 1 + 0.681 724 833 942 308 614 723 436 984 729 6;
  • 17) 0.681 724 833 942 308 614 723 436 984 729 6 × 2 = 1 + 0.363 449 667 884 617 229 446 873 969 459 2;
  • 18) 0.363 449 667 884 617 229 446 873 969 459 2 × 2 = 0 + 0.726 899 335 769 234 458 893 747 938 918 4;
  • 19) 0.726 899 335 769 234 458 893 747 938 918 4 × 2 = 1 + 0.453 798 671 538 468 917 787 495 877 836 8;
  • 20) 0.453 798 671 538 468 917 787 495 877 836 8 × 2 = 0 + 0.907 597 343 076 937 835 574 991 755 673 6;
  • 21) 0.907 597 343 076 937 835 574 991 755 673 6 × 2 = 1 + 0.815 194 686 153 875 671 149 983 511 347 2;
  • 22) 0.815 194 686 153 875 671 149 983 511 347 2 × 2 = 1 + 0.630 389 372 307 751 342 299 967 022 694 4;
  • 23) 0.630 389 372 307 751 342 299 967 022 694 4 × 2 = 1 + 0.260 778 744 615 502 684 599 934 045 388 8;
  • 24) 0.260 778 744 615 502 684 599 934 045 388 8 × 2 = 0 + 0.521 557 489 231 005 369 199 868 090 777 6;

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:


0.732 050 807 568 877 293 527 446 341 506 725(10) =

0.1011 1011 0110 0111 1010 1110(2)

5. Positive number before normalization:

1.732 050 807 568 877 293 527 446 341 506 725(10) =

1.1011 1011 0110 0111 1010 1110(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:


1.732 050 807 568 877 293 527 446 341 506 725(10) =

1.1011 1011 0110 0111 1010 1110(2) =

1.1011 1011 0110 0111 1010 1110(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):
1.1011 1011 0110 0111 1010 1110


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. 101 1101 1011 0011 1101 0111 0 =

101 1101 1011 0011 1101 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) =
101 1101 1011 0011 1101 0111


Decimal number 1.732 050 807 568 877 293 527 446 341 506 725 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 101 1101 1011 0011 1101 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