0.000 000 017 669 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 017 669(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
0.000 000 017 669(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: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 017 669.

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) 0.000 000 017 669 × 2 = 0 + 0.000 000 035 338;
  • 2) 0.000 000 035 338 × 2 = 0 + 0.000 000 070 676;
  • 3) 0.000 000 070 676 × 2 = 0 + 0.000 000 141 352;
  • 4) 0.000 000 141 352 × 2 = 0 + 0.000 000 282 704;
  • 5) 0.000 000 282 704 × 2 = 0 + 0.000 000 565 408;
  • 6) 0.000 000 565 408 × 2 = 0 + 0.000 001 130 816;
  • 7) 0.000 001 130 816 × 2 = 0 + 0.000 002 261 632;
  • 8) 0.000 002 261 632 × 2 = 0 + 0.000 004 523 264;
  • 9) 0.000 004 523 264 × 2 = 0 + 0.000 009 046 528;
  • 10) 0.000 009 046 528 × 2 = 0 + 0.000 018 093 056;
  • 11) 0.000 018 093 056 × 2 = 0 + 0.000 036 186 112;
  • 12) 0.000 036 186 112 × 2 = 0 + 0.000 072 372 224;
  • 13) 0.000 072 372 224 × 2 = 0 + 0.000 144 744 448;
  • 14) 0.000 144 744 448 × 2 = 0 + 0.000 289 488 896;
  • 15) 0.000 289 488 896 × 2 = 0 + 0.000 578 977 792;
  • 16) 0.000 578 977 792 × 2 = 0 + 0.001 157 955 584;
  • 17) 0.001 157 955 584 × 2 = 0 + 0.002 315 911 168;
  • 18) 0.002 315 911 168 × 2 = 0 + 0.004 631 822 336;
  • 19) 0.004 631 822 336 × 2 = 0 + 0.009 263 644 672;
  • 20) 0.009 263 644 672 × 2 = 0 + 0.018 527 289 344;
  • 21) 0.018 527 289 344 × 2 = 0 + 0.037 054 578 688;
  • 22) 0.037 054 578 688 × 2 = 0 + 0.074 109 157 376;
  • 23) 0.074 109 157 376 × 2 = 0 + 0.148 218 314 752;
  • 24) 0.148 218 314 752 × 2 = 0 + 0.296 436 629 504;
  • 25) 0.296 436 629 504 × 2 = 0 + 0.592 873 259 008;
  • 26) 0.592 873 259 008 × 2 = 1 + 0.185 746 518 016;
  • 27) 0.185 746 518 016 × 2 = 0 + 0.371 493 036 032;
  • 28) 0.371 493 036 032 × 2 = 0 + 0.742 986 072 064;
  • 29) 0.742 986 072 064 × 2 = 1 + 0.485 972 144 128;
  • 30) 0.485 972 144 128 × 2 = 0 + 0.971 944 288 256;
  • 31) 0.971 944 288 256 × 2 = 1 + 0.943 888 576 512;
  • 32) 0.943 888 576 512 × 2 = 1 + 0.887 777 153 024;
  • 33) 0.887 777 153 024 × 2 = 1 + 0.775 554 306 048;
  • 34) 0.775 554 306 048 × 2 = 1 + 0.551 108 612 096;
  • 35) 0.551 108 612 096 × 2 = 1 + 0.102 217 224 192;
  • 36) 0.102 217 224 192 × 2 = 0 + 0.204 434 448 384;
  • 37) 0.204 434 448 384 × 2 = 0 + 0.408 868 896 768;
  • 38) 0.408 868 896 768 × 2 = 0 + 0.817 737 793 536;
  • 39) 0.817 737 793 536 × 2 = 1 + 0.635 475 587 072;
  • 40) 0.635 475 587 072 × 2 = 1 + 0.270 951 174 144;
  • 41) 0.270 951 174 144 × 2 = 0 + 0.541 902 348 288;
  • 42) 0.541 902 348 288 × 2 = 1 + 0.083 804 696 576;
  • 43) 0.083 804 696 576 × 2 = 0 + 0.167 609 393 152;
  • 44) 0.167 609 393 152 × 2 = 0 + 0.335 218 786 304;
  • 45) 0.335 218 786 304 × 2 = 0 + 0.670 437 572 608;
  • 46) 0.670 437 572 608 × 2 = 1 + 0.340 875 145 216;
  • 47) 0.340 875 145 216 × 2 = 0 + 0.681 750 290 432;
  • 48) 0.681 750 290 432 × 2 = 1 + 0.363 500 580 864;
  • 49) 0.363 500 580 864 × 2 = 0 + 0.727 001 161 728;

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:


0.000 000 017 669(10) =

0.0000 0000 0000 0000 0000 0000 0100 1011 1110 0011 0100 0101 0(2)

5. Positive number before normalization:

0.000 000 017 669(10) =

0.0000 0000 0000 0000 0000 0000 0100 1011 1110 0011 0100 0101 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 017 669(10) =

0.0000 0000 0000 0000 0000 0000 0100 1011 1110 0011 0100 0101 0(2) =

0.0000 0000 0000 0000 0000 0000 0100 1011 1110 0011 0100 0101 0(2) × 20 =

1.0010 1111 1000 1101 0001 010(2) × 2-26


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): -26

Mantissa (not normalized):
1.0010 1111 1000 1101 0001 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(8-1) - 1 =

(-26 + 127)(10) =

101(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;
  • 101 ÷ 2 = 50 + 1;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

101(10) =

0110 0101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 001 0111 1100 0110 1000 1010 =

001 0111 1100 0110 1000 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) =
0110 0101

Mantissa (23 bits) =
001 0111 1100 0110 1000 1010


Decimal number 0.000 000 017 669 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0101 - 001 0111 1100 0110 1000 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