3.141 592 653 589 793 115 997 963 473 21 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.141 592 653 589 793 115 997 963 473 21(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
3.141 592 653 589 793 115 997 963 473 21(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.141 592 653 589 793 115 997 963 473 21.

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) 0.141 592 653 589 793 115 997 963 473 21 × 2 = 0 + 0.283 185 307 179 586 231 995 926 946 42;
  • 2) 0.283 185 307 179 586 231 995 926 946 42 × 2 = 0 + 0.566 370 614 359 172 463 991 853 892 84;
  • 3) 0.566 370 614 359 172 463 991 853 892 84 × 2 = 1 + 0.132 741 228 718 344 927 983 707 785 68;
  • 4) 0.132 741 228 718 344 927 983 707 785 68 × 2 = 0 + 0.265 482 457 436 689 855 967 415 571 36;
  • 5) 0.265 482 457 436 689 855 967 415 571 36 × 2 = 0 + 0.530 964 914 873 379 711 934 831 142 72;
  • 6) 0.530 964 914 873 379 711 934 831 142 72 × 2 = 1 + 0.061 929 829 746 759 423 869 662 285 44;
  • 7) 0.061 929 829 746 759 423 869 662 285 44 × 2 = 0 + 0.123 859 659 493 518 847 739 324 570 88;
  • 8) 0.123 859 659 493 518 847 739 324 570 88 × 2 = 0 + 0.247 719 318 987 037 695 478 649 141 76;
  • 9) 0.247 719 318 987 037 695 478 649 141 76 × 2 = 0 + 0.495 438 637 974 075 390 957 298 283 52;
  • 10) 0.495 438 637 974 075 390 957 298 283 52 × 2 = 0 + 0.990 877 275 948 150 781 914 596 567 04;
  • 11) 0.990 877 275 948 150 781 914 596 567 04 × 2 = 1 + 0.981 754 551 896 301 563 829 193 134 08;
  • 12) 0.981 754 551 896 301 563 829 193 134 08 × 2 = 1 + 0.963 509 103 792 603 127 658 386 268 16;
  • 13) 0.963 509 103 792 603 127 658 386 268 16 × 2 = 1 + 0.927 018 207 585 206 255 316 772 536 32;
  • 14) 0.927 018 207 585 206 255 316 772 536 32 × 2 = 1 + 0.854 036 415 170 412 510 633 545 072 64;
  • 15) 0.854 036 415 170 412 510 633 545 072 64 × 2 = 1 + 0.708 072 830 340 825 021 267 090 145 28;
  • 16) 0.708 072 830 340 825 021 267 090 145 28 × 2 = 1 + 0.416 145 660 681 650 042 534 180 290 56;
  • 17) 0.416 145 660 681 650 042 534 180 290 56 × 2 = 0 + 0.832 291 321 363 300 085 068 360 581 12;
  • 18) 0.832 291 321 363 300 085 068 360 581 12 × 2 = 1 + 0.664 582 642 726 600 170 136 721 162 24;
  • 19) 0.664 582 642 726 600 170 136 721 162 24 × 2 = 1 + 0.329 165 285 453 200 340 273 442 324 48;
  • 20) 0.329 165 285 453 200 340 273 442 324 48 × 2 = 0 + 0.658 330 570 906 400 680 546 884 648 96;
  • 21) 0.658 330 570 906 400 680 546 884 648 96 × 2 = 1 + 0.316 661 141 812 801 361 093 769 297 92;
  • 22) 0.316 661 141 812 801 361 093 769 297 92 × 2 = 0 + 0.633 322 283 625 602 722 187 538 595 84;
  • 23) 0.633 322 283 625 602 722 187 538 595 84 × 2 = 1 + 0.266 644 567 251 205 444 375 077 191 68;
  • 24) 0.266 644 567 251 205 444 375 077 191 68 × 2 = 0 + 0.533 289 134 502 410 888 750 154 383 36;

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:


0.141 592 653 589 793 115 997 963 473 21(10) =

0.0010 0100 0011 1111 0110 1010(2)

5. Positive number before normalization:

3.141 592 653 589 793 115 997 963 473 21(10) =

11.0010 0100 0011 1111 0110 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.141 592 653 589 793 115 997 963 473 21(10) =

11.0010 0100 0011 1111 0110 1010(2) =

11.0010 0100 0011 1111 0110 1010(2) × 20 =

1.1001 0010 0001 1111 1011 0101 0(2) × 21


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): 1

Mantissa (not normalized):
1.1001 0010 0001 1111 1011 0101 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

128(10) =

1000 0000(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. 100 1001 0000 1111 1101 1010 10 =

100 1001 0000 1111 1101 1010


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) =
1000 0000

Mantissa (23 bits) =
100 1001 0000 1111 1101 1010


Decimal number 3.141 592 653 589 793 115 997 963 473 21 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 100 1001 0000 1111 1101 1010

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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