-0.000 000 000 627 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 627(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
-0.000 000 000 627(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:

|-0.000 000 000 627| = 0.000 000 000 627


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: 0.000 000 000 627.

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) 0.000 000 000 627 × 2 = 0 + 0.000 000 001 254;
  • 2) 0.000 000 001 254 × 2 = 0 + 0.000 000 002 508;
  • 3) 0.000 000 002 508 × 2 = 0 + 0.000 000 005 016;
  • 4) 0.000 000 005 016 × 2 = 0 + 0.000 000 010 032;
  • 5) 0.000 000 010 032 × 2 = 0 + 0.000 000 020 064;
  • 6) 0.000 000 020 064 × 2 = 0 + 0.000 000 040 128;
  • 7) 0.000 000 040 128 × 2 = 0 + 0.000 000 080 256;
  • 8) 0.000 000 080 256 × 2 = 0 + 0.000 000 160 512;
  • 9) 0.000 000 160 512 × 2 = 0 + 0.000 000 321 024;
  • 10) 0.000 000 321 024 × 2 = 0 + 0.000 000 642 048;
  • 11) 0.000 000 642 048 × 2 = 0 + 0.000 001 284 096;
  • 12) 0.000 001 284 096 × 2 = 0 + 0.000 002 568 192;
  • 13) 0.000 002 568 192 × 2 = 0 + 0.000 005 136 384;
  • 14) 0.000 005 136 384 × 2 = 0 + 0.000 010 272 768;
  • 15) 0.000 010 272 768 × 2 = 0 + 0.000 020 545 536;
  • 16) 0.000 020 545 536 × 2 = 0 + 0.000 041 091 072;
  • 17) 0.000 041 091 072 × 2 = 0 + 0.000 082 182 144;
  • 18) 0.000 082 182 144 × 2 = 0 + 0.000 164 364 288;
  • 19) 0.000 164 364 288 × 2 = 0 + 0.000 328 728 576;
  • 20) 0.000 328 728 576 × 2 = 0 + 0.000 657 457 152;
  • 21) 0.000 657 457 152 × 2 = 0 + 0.001 314 914 304;
  • 22) 0.001 314 914 304 × 2 = 0 + 0.002 629 828 608;
  • 23) 0.002 629 828 608 × 2 = 0 + 0.005 259 657 216;
  • 24) 0.005 259 657 216 × 2 = 0 + 0.010 519 314 432;
  • 25) 0.010 519 314 432 × 2 = 0 + 0.021 038 628 864;
  • 26) 0.021 038 628 864 × 2 = 0 + 0.042 077 257 728;
  • 27) 0.042 077 257 728 × 2 = 0 + 0.084 154 515 456;
  • 28) 0.084 154 515 456 × 2 = 0 + 0.168 309 030 912;
  • 29) 0.168 309 030 912 × 2 = 0 + 0.336 618 061 824;
  • 30) 0.336 618 061 824 × 2 = 0 + 0.673 236 123 648;
  • 31) 0.673 236 123 648 × 2 = 1 + 0.346 472 247 296;
  • 32) 0.346 472 247 296 × 2 = 0 + 0.692 944 494 592;
  • 33) 0.692 944 494 592 × 2 = 1 + 0.385 888 989 184;
  • 34) 0.385 888 989 184 × 2 = 0 + 0.771 777 978 368;
  • 35) 0.771 777 978 368 × 2 = 1 + 0.543 555 956 736;
  • 36) 0.543 555 956 736 × 2 = 1 + 0.087 111 913 472;
  • 37) 0.087 111 913 472 × 2 = 0 + 0.174 223 826 944;
  • 38) 0.174 223 826 944 × 2 = 0 + 0.348 447 653 888;
  • 39) 0.348 447 653 888 × 2 = 0 + 0.696 895 307 776;
  • 40) 0.696 895 307 776 × 2 = 1 + 0.393 790 615 552;
  • 41) 0.393 790 615 552 × 2 = 0 + 0.787 581 231 104;
  • 42) 0.787 581 231 104 × 2 = 1 + 0.575 162 462 208;
  • 43) 0.575 162 462 208 × 2 = 1 + 0.150 324 924 416;
  • 44) 0.150 324 924 416 × 2 = 0 + 0.300 649 848 832;
  • 45) 0.300 649 848 832 × 2 = 0 + 0.601 299 697 664;
  • 46) 0.601 299 697 664 × 2 = 1 + 0.202 599 395 328;
  • 47) 0.202 599 395 328 × 2 = 0 + 0.405 198 790 656;
  • 48) 0.405 198 790 656 × 2 = 0 + 0.810 397 581 312;
  • 49) 0.810 397 581 312 × 2 = 1 + 0.620 795 162 624;
  • 50) 0.620 795 162 624 × 2 = 1 + 0.241 590 325 248;
  • 51) 0.241 590 325 248 × 2 = 0 + 0.483 180 650 496;
  • 52) 0.483 180 650 496 × 2 = 0 + 0.966 361 300 992;
  • 53) 0.966 361 300 992 × 2 = 1 + 0.932 722 601 984;
  • 54) 0.932 722 601 984 × 2 = 1 + 0.865 445 203 968;

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:


0.000 000 000 627(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0001 0110 0100 1100 11(2)

6. Positive number before normalization:

0.000 000 000 627(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0001 0110 0100 1100 11(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:


0.000 000 000 627(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0001 0110 0100 1100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 0001 0110 0100 1100 11(2) × 20 =

1.0101 1000 1011 0010 0110 011(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):
1.0101 1000 1011 0010 0110 011


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. 010 1100 0101 1001 0011 0011 =

010 1100 0101 1001 0011 0011


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) =
010 1100 0101 1001 0011 0011


Decimal number -0.000 000 000 627 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 010 1100 0101 1001 0011 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