4.208 543 771 043 771 046 67 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 4.208 543 771 043 771 046 67(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
4.208 543 771 043 771 046 67(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: 4.
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;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

4(10) =

100(2)


3. Convert to binary (base 2) the fractional part: 0.208 543 771 043 771 046 67.

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) 0.208 543 771 043 771 046 67 × 2 = 0 + 0.417 087 542 087 542 093 34;
  • 2) 0.417 087 542 087 542 093 34 × 2 = 0 + 0.834 175 084 175 084 186 68;
  • 3) 0.834 175 084 175 084 186 68 × 2 = 1 + 0.668 350 168 350 168 373 36;
  • 4) 0.668 350 168 350 168 373 36 × 2 = 1 + 0.336 700 336 700 336 746 72;
  • 5) 0.336 700 336 700 336 746 72 × 2 = 0 + 0.673 400 673 400 673 493 44;
  • 6) 0.673 400 673 400 673 493 44 × 2 = 1 + 0.346 801 346 801 346 986 88;
  • 7) 0.346 801 346 801 346 986 88 × 2 = 0 + 0.693 602 693 602 693 973 76;
  • 8) 0.693 602 693 602 693 973 76 × 2 = 1 + 0.387 205 387 205 387 947 52;
  • 9) 0.387 205 387 205 387 947 52 × 2 = 0 + 0.774 410 774 410 775 895 04;
  • 10) 0.774 410 774 410 775 895 04 × 2 = 1 + 0.548 821 548 821 551 790 08;
  • 11) 0.548 821 548 821 551 790 08 × 2 = 1 + 0.097 643 097 643 103 580 16;
  • 12) 0.097 643 097 643 103 580 16 × 2 = 0 + 0.195 286 195 286 207 160 32;
  • 13) 0.195 286 195 286 207 160 32 × 2 = 0 + 0.390 572 390 572 414 320 64;
  • 14) 0.390 572 390 572 414 320 64 × 2 = 0 + 0.781 144 781 144 828 641 28;
  • 15) 0.781 144 781 144 828 641 28 × 2 = 1 + 0.562 289 562 289 657 282 56;
  • 16) 0.562 289 562 289 657 282 56 × 2 = 1 + 0.124 579 124 579 314 565 12;
  • 17) 0.124 579 124 579 314 565 12 × 2 = 0 + 0.249 158 249 158 629 130 24;
  • 18) 0.249 158 249 158 629 130 24 × 2 = 0 + 0.498 316 498 317 258 260 48;
  • 19) 0.498 316 498 317 258 260 48 × 2 = 0 + 0.996 632 996 634 516 520 96;
  • 20) 0.996 632 996 634 516 520 96 × 2 = 1 + 0.993 265 993 269 033 041 92;
  • 21) 0.993 265 993 269 033 041 92 × 2 = 1 + 0.986 531 986 538 066 083 84;
  • 22) 0.986 531 986 538 066 083 84 × 2 = 1 + 0.973 063 973 076 132 167 68;
  • 23) 0.973 063 973 076 132 167 68 × 2 = 1 + 0.946 127 946 152 264 335 36;
  • 24) 0.946 127 946 152 264 335 36 × 2 = 1 + 0.892 255 892 304 528 670 72;

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:


0.208 543 771 043 771 046 67(10) =

0.0011 0101 0110 0011 0001 1111(2)

5. Positive number before normalization:

4.208 543 771 043 771 046 67(10) =

100.0011 0101 0110 0011 0001 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


4.208 543 771 043 771 046 67(10) =

100.0011 0101 0110 0011 0001 1111(2) =

100.0011 0101 0110 0011 0001 1111(2) × 20 =

1.0000 1101 0101 1000 1100 0111 11(2) × 22


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): 2

Mantissa (not normalized):
1.0000 1101 0101 1000 1100 0111 11


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(8-1) - 1 =

(2 + 127)(10) =

129(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;
  • 129 ÷ 2 = 64 + 1;
  • 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) =

129(10) =

1000 0001(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 0110 1010 1100 0110 0011 111 =

000 0110 1010 1100 0110 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) =
1000 0001

Mantissa (23 bits) =
000 0110 1010 1100 0110 0011


Decimal number 4.208 543 771 043 771 046 67 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0001 - 000 0110 1010 1100 0110 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