1.414 213 562 373 095 048 801 685 01 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.414 213 562 373 095 048 801 685 01(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
1.414 213 562 373 095 048 801 685 01(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: 0.414 213 562 373 095 048 801 685 01.

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) 0.414 213 562 373 095 048 801 685 01 × 2 = 0 + 0.828 427 124 746 190 097 603 370 02;
  • 2) 0.828 427 124 746 190 097 603 370 02 × 2 = 1 + 0.656 854 249 492 380 195 206 740 04;
  • 3) 0.656 854 249 492 380 195 206 740 04 × 2 = 1 + 0.313 708 498 984 760 390 413 480 08;
  • 4) 0.313 708 498 984 760 390 413 480 08 × 2 = 0 + 0.627 416 997 969 520 780 826 960 16;
  • 5) 0.627 416 997 969 520 780 826 960 16 × 2 = 1 + 0.254 833 995 939 041 561 653 920 32;
  • 6) 0.254 833 995 939 041 561 653 920 32 × 2 = 0 + 0.509 667 991 878 083 123 307 840 64;
  • 7) 0.509 667 991 878 083 123 307 840 64 × 2 = 1 + 0.019 335 983 756 166 246 615 681 28;
  • 8) 0.019 335 983 756 166 246 615 681 28 × 2 = 0 + 0.038 671 967 512 332 493 231 362 56;
  • 9) 0.038 671 967 512 332 493 231 362 56 × 2 = 0 + 0.077 343 935 024 664 986 462 725 12;
  • 10) 0.077 343 935 024 664 986 462 725 12 × 2 = 0 + 0.154 687 870 049 329 972 925 450 24;
  • 11) 0.154 687 870 049 329 972 925 450 24 × 2 = 0 + 0.309 375 740 098 659 945 850 900 48;
  • 12) 0.309 375 740 098 659 945 850 900 48 × 2 = 0 + 0.618 751 480 197 319 891 701 800 96;
  • 13) 0.618 751 480 197 319 891 701 800 96 × 2 = 1 + 0.237 502 960 394 639 783 403 601 92;
  • 14) 0.237 502 960 394 639 783 403 601 92 × 2 = 0 + 0.475 005 920 789 279 566 807 203 84;
  • 15) 0.475 005 920 789 279 566 807 203 84 × 2 = 0 + 0.950 011 841 578 559 133 614 407 68;
  • 16) 0.950 011 841 578 559 133 614 407 68 × 2 = 1 + 0.900 023 683 157 118 267 228 815 36;
  • 17) 0.900 023 683 157 118 267 228 815 36 × 2 = 1 + 0.800 047 366 314 236 534 457 630 72;
  • 18) 0.800 047 366 314 236 534 457 630 72 × 2 = 1 + 0.600 094 732 628 473 068 915 261 44;
  • 19) 0.600 094 732 628 473 068 915 261 44 × 2 = 1 + 0.200 189 465 256 946 137 830 522 88;
  • 20) 0.200 189 465 256 946 137 830 522 88 × 2 = 0 + 0.400 378 930 513 892 275 661 045 76;
  • 21) 0.400 378 930 513 892 275 661 045 76 × 2 = 0 + 0.800 757 861 027 784 551 322 091 52;
  • 22) 0.800 757 861 027 784 551 322 091 52 × 2 = 1 + 0.601 515 722 055 569 102 644 183 04;
  • 23) 0.601 515 722 055 569 102 644 183 04 × 2 = 1 + 0.203 031 444 111 138 205 288 366 08;
  • 24) 0.203 031 444 111 138 205 288 366 08 × 2 = 0 + 0.406 062 888 222 276 410 576 732 16;

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:


0.414 213 562 373 095 048 801 685 01(10) =

0.0110 1010 0000 1001 1110 0110(2)

5. Positive number before normalization:

1.414 213 562 373 095 048 801 685 01(10) =

1.0110 1010 0000 1001 1110 0110(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:


1.414 213 562 373 095 048 801 685 01(10) =

1.0110 1010 0000 1001 1110 0110(2) =

1.0110 1010 0000 1001 1110 0110(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):
1.0110 1010 0000 1001 1110 0110


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. 011 0101 0000 0100 1111 0011 0 =

011 0101 0000 0100 1111 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) =
0111 1111

Mantissa (23 bits) =
011 0101 0000 0100 1111 0011


Decimal number 1.414 213 562 373 095 048 801 685 01 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 011 0101 0000 0100 1111 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