-0.000 000 000 426 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 426(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
-0.000 000 000 426(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:

|-0.000 000 000 426| = 0.000 000 000 426


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: 0.000 000 000 426.

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) 0.000 000 000 426 × 2 = 0 + 0.000 000 000 852;
  • 2) 0.000 000 000 852 × 2 = 0 + 0.000 000 001 704;
  • 3) 0.000 000 001 704 × 2 = 0 + 0.000 000 003 408;
  • 4) 0.000 000 003 408 × 2 = 0 + 0.000 000 006 816;
  • 5) 0.000 000 006 816 × 2 = 0 + 0.000 000 013 632;
  • 6) 0.000 000 013 632 × 2 = 0 + 0.000 000 027 264;
  • 7) 0.000 000 027 264 × 2 = 0 + 0.000 000 054 528;
  • 8) 0.000 000 054 528 × 2 = 0 + 0.000 000 109 056;
  • 9) 0.000 000 109 056 × 2 = 0 + 0.000 000 218 112;
  • 10) 0.000 000 218 112 × 2 = 0 + 0.000 000 436 224;
  • 11) 0.000 000 436 224 × 2 = 0 + 0.000 000 872 448;
  • 12) 0.000 000 872 448 × 2 = 0 + 0.000 001 744 896;
  • 13) 0.000 001 744 896 × 2 = 0 + 0.000 003 489 792;
  • 14) 0.000 003 489 792 × 2 = 0 + 0.000 006 979 584;
  • 15) 0.000 006 979 584 × 2 = 0 + 0.000 013 959 168;
  • 16) 0.000 013 959 168 × 2 = 0 + 0.000 027 918 336;
  • 17) 0.000 027 918 336 × 2 = 0 + 0.000 055 836 672;
  • 18) 0.000 055 836 672 × 2 = 0 + 0.000 111 673 344;
  • 19) 0.000 111 673 344 × 2 = 0 + 0.000 223 346 688;
  • 20) 0.000 223 346 688 × 2 = 0 + 0.000 446 693 376;
  • 21) 0.000 446 693 376 × 2 = 0 + 0.000 893 386 752;
  • 22) 0.000 893 386 752 × 2 = 0 + 0.001 786 773 504;
  • 23) 0.001 786 773 504 × 2 = 0 + 0.003 573 547 008;
  • 24) 0.003 573 547 008 × 2 = 0 + 0.007 147 094 016;
  • 25) 0.007 147 094 016 × 2 = 0 + 0.014 294 188 032;
  • 26) 0.014 294 188 032 × 2 = 0 + 0.028 588 376 064;
  • 27) 0.028 588 376 064 × 2 = 0 + 0.057 176 752 128;
  • 28) 0.057 176 752 128 × 2 = 0 + 0.114 353 504 256;
  • 29) 0.114 353 504 256 × 2 = 0 + 0.228 707 008 512;
  • 30) 0.228 707 008 512 × 2 = 0 + 0.457 414 017 024;
  • 31) 0.457 414 017 024 × 2 = 0 + 0.914 828 034 048;
  • 32) 0.914 828 034 048 × 2 = 1 + 0.829 656 068 096;
  • 33) 0.829 656 068 096 × 2 = 1 + 0.659 312 136 192;
  • 34) 0.659 312 136 192 × 2 = 1 + 0.318 624 272 384;
  • 35) 0.318 624 272 384 × 2 = 0 + 0.637 248 544 768;
  • 36) 0.637 248 544 768 × 2 = 1 + 0.274 497 089 536;
  • 37) 0.274 497 089 536 × 2 = 0 + 0.548 994 179 072;
  • 38) 0.548 994 179 072 × 2 = 1 + 0.097 988 358 144;
  • 39) 0.097 988 358 144 × 2 = 0 + 0.195 976 716 288;
  • 40) 0.195 976 716 288 × 2 = 0 + 0.391 953 432 576;
  • 41) 0.391 953 432 576 × 2 = 0 + 0.783 906 865 152;
  • 42) 0.783 906 865 152 × 2 = 1 + 0.567 813 730 304;
  • 43) 0.567 813 730 304 × 2 = 1 + 0.135 627 460 608;
  • 44) 0.135 627 460 608 × 2 = 0 + 0.271 254 921 216;
  • 45) 0.271 254 921 216 × 2 = 0 + 0.542 509 842 432;
  • 46) 0.542 509 842 432 × 2 = 1 + 0.085 019 684 864;
  • 47) 0.085 019 684 864 × 2 = 0 + 0.170 039 369 728;
  • 48) 0.170 039 369 728 × 2 = 0 + 0.340 078 739 456;
  • 49) 0.340 078 739 456 × 2 = 0 + 0.680 157 478 912;
  • 50) 0.680 157 478 912 × 2 = 1 + 0.360 314 957 824;
  • 51) 0.360 314 957 824 × 2 = 0 + 0.720 629 915 648;
  • 52) 0.720 629 915 648 × 2 = 1 + 0.441 259 831 296;
  • 53) 0.441 259 831 296 × 2 = 0 + 0.882 519 662 592;
  • 54) 0.882 519 662 592 × 2 = 1 + 0.765 039 325 184;
  • 55) 0.765 039 325 184 × 2 = 1 + 0.530 078 650 368;

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:


0.000 000 000 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1101 0100 0110 0100 0101 011(2)

6. Positive number before normalization:

0.000 000 000 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1101 0100 0110 0100 0101 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 426(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1101 0100 0110 0100 0101 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1101 0100 0110 0100 0101 011(2) × 20 =

1.1101 0100 0110 0100 0101 011(2) × 2-32


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): -32

Mantissa (not normalized):
1.1101 0100 0110 0100 0101 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

95(10) =

0101 1111(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. 110 1010 0011 0010 0010 1011 =

110 1010 0011 0010 0010 1011


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) =
0101 1111

Mantissa (23 bits) =
110 1010 0011 0010 0010 1011


Decimal number -0.000 000 000 426 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 110 1010 0011 0010 0010 1011

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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