-0.000 000 000 474 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 474(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
-0.000 000 000 474(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. Start with the positive version of the number:

|-0.000 000 000 474| = 0.000 000 000 474


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 474.

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) 0.000 000 000 474 × 2 = 0 + 0.000 000 000 948;
  • 2) 0.000 000 000 948 × 2 = 0 + 0.000 000 001 896;
  • 3) 0.000 000 001 896 × 2 = 0 + 0.000 000 003 792;
  • 4) 0.000 000 003 792 × 2 = 0 + 0.000 000 007 584;
  • 5) 0.000 000 007 584 × 2 = 0 + 0.000 000 015 168;
  • 6) 0.000 000 015 168 × 2 = 0 + 0.000 000 030 336;
  • 7) 0.000 000 030 336 × 2 = 0 + 0.000 000 060 672;
  • 8) 0.000 000 060 672 × 2 = 0 + 0.000 000 121 344;
  • 9) 0.000 000 121 344 × 2 = 0 + 0.000 000 242 688;
  • 10) 0.000 000 242 688 × 2 = 0 + 0.000 000 485 376;
  • 11) 0.000 000 485 376 × 2 = 0 + 0.000 000 970 752;
  • 12) 0.000 000 970 752 × 2 = 0 + 0.000 001 941 504;
  • 13) 0.000 001 941 504 × 2 = 0 + 0.000 003 883 008;
  • 14) 0.000 003 883 008 × 2 = 0 + 0.000 007 766 016;
  • 15) 0.000 007 766 016 × 2 = 0 + 0.000 015 532 032;
  • 16) 0.000 015 532 032 × 2 = 0 + 0.000 031 064 064;
  • 17) 0.000 031 064 064 × 2 = 0 + 0.000 062 128 128;
  • 18) 0.000 062 128 128 × 2 = 0 + 0.000 124 256 256;
  • 19) 0.000 124 256 256 × 2 = 0 + 0.000 248 512 512;
  • 20) 0.000 248 512 512 × 2 = 0 + 0.000 497 025 024;
  • 21) 0.000 497 025 024 × 2 = 0 + 0.000 994 050 048;
  • 22) 0.000 994 050 048 × 2 = 0 + 0.001 988 100 096;
  • 23) 0.001 988 100 096 × 2 = 0 + 0.003 976 200 192;
  • 24) 0.003 976 200 192 × 2 = 0 + 0.007 952 400 384;
  • 25) 0.007 952 400 384 × 2 = 0 + 0.015 904 800 768;
  • 26) 0.015 904 800 768 × 2 = 0 + 0.031 809 601 536;
  • 27) 0.031 809 601 536 × 2 = 0 + 0.063 619 203 072;
  • 28) 0.063 619 203 072 × 2 = 0 + 0.127 238 406 144;
  • 29) 0.127 238 406 144 × 2 = 0 + 0.254 476 812 288;
  • 30) 0.254 476 812 288 × 2 = 0 + 0.508 953 624 576;
  • 31) 0.508 953 624 576 × 2 = 1 + 0.017 907 249 152;
  • 32) 0.017 907 249 152 × 2 = 0 + 0.035 814 498 304;
  • 33) 0.035 814 498 304 × 2 = 0 + 0.071 628 996 608;
  • 34) 0.071 628 996 608 × 2 = 0 + 0.143 257 993 216;
  • 35) 0.143 257 993 216 × 2 = 0 + 0.286 515 986 432;
  • 36) 0.286 515 986 432 × 2 = 0 + 0.573 031 972 864;
  • 37) 0.573 031 972 864 × 2 = 1 + 0.146 063 945 728;
  • 38) 0.146 063 945 728 × 2 = 0 + 0.292 127 891 456;
  • 39) 0.292 127 891 456 × 2 = 0 + 0.584 255 782 912;
  • 40) 0.584 255 782 912 × 2 = 1 + 0.168 511 565 824;
  • 41) 0.168 511 565 824 × 2 = 0 + 0.337 023 131 648;
  • 42) 0.337 023 131 648 × 2 = 0 + 0.674 046 263 296;
  • 43) 0.674 046 263 296 × 2 = 1 + 0.348 092 526 592;
  • 44) 0.348 092 526 592 × 2 = 0 + 0.696 185 053 184;
  • 45) 0.696 185 053 184 × 2 = 1 + 0.392 370 106 368;
  • 46) 0.392 370 106 368 × 2 = 0 + 0.784 740 212 736;
  • 47) 0.784 740 212 736 × 2 = 1 + 0.569 480 425 472;
  • 48) 0.569 480 425 472 × 2 = 1 + 0.138 960 850 944;
  • 49) 0.138 960 850 944 × 2 = 0 + 0.277 921 701 888;
  • 50) 0.277 921 701 888 × 2 = 0 + 0.555 843 403 776;
  • 51) 0.555 843 403 776 × 2 = 1 + 0.111 686 807 552;
  • 52) 0.111 686 807 552 × 2 = 0 + 0.223 373 615 104;
  • 53) 0.223 373 615 104 × 2 = 0 + 0.446 747 230 208;
  • 54) 0.446 747 230 208 × 2 = 0 + 0.893 494 460 416;

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).


5. 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:


0.000 000 000 474(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 1001 0010 1011 0010 00(2)

6. Positive number before normalization:

0.000 000 000 474(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 1001 0010 1011 0010 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 474(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 1001 0010 1011 0010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 0000 1001 0010 1011 0010 00(2) × 20 =

1.0000 0100 1001 0101 1001 000(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.0000 0100 1001 0101 1001 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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. 000 0010 0100 1010 1100 1000 =

000 0010 0100 1010 1100 1000


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
000 0010 0100 1010 1100 1000


Decimal number -0.000 000 000 474 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 000 0010 0100 1010 1100 1000

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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