-0.000 000 000 736 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 736 2(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
-0.000 000 000 736 2(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:

|-0.000 000 000 736 2| = 0.000 000 000 736 2


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: 0.000 000 000 736 2.

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) 0.000 000 000 736 2 × 2 = 0 + 0.000 000 001 472 4;
  • 2) 0.000 000 001 472 4 × 2 = 0 + 0.000 000 002 944 8;
  • 3) 0.000 000 002 944 8 × 2 = 0 + 0.000 000 005 889 6;
  • 4) 0.000 000 005 889 6 × 2 = 0 + 0.000 000 011 779 2;
  • 5) 0.000 000 011 779 2 × 2 = 0 + 0.000 000 023 558 4;
  • 6) 0.000 000 023 558 4 × 2 = 0 + 0.000 000 047 116 8;
  • 7) 0.000 000 047 116 8 × 2 = 0 + 0.000 000 094 233 6;
  • 8) 0.000 000 094 233 6 × 2 = 0 + 0.000 000 188 467 2;
  • 9) 0.000 000 188 467 2 × 2 = 0 + 0.000 000 376 934 4;
  • 10) 0.000 000 376 934 4 × 2 = 0 + 0.000 000 753 868 8;
  • 11) 0.000 000 753 868 8 × 2 = 0 + 0.000 001 507 737 6;
  • 12) 0.000 001 507 737 6 × 2 = 0 + 0.000 003 015 475 2;
  • 13) 0.000 003 015 475 2 × 2 = 0 + 0.000 006 030 950 4;
  • 14) 0.000 006 030 950 4 × 2 = 0 + 0.000 012 061 900 8;
  • 15) 0.000 012 061 900 8 × 2 = 0 + 0.000 024 123 801 6;
  • 16) 0.000 024 123 801 6 × 2 = 0 + 0.000 048 247 603 2;
  • 17) 0.000 048 247 603 2 × 2 = 0 + 0.000 096 495 206 4;
  • 18) 0.000 096 495 206 4 × 2 = 0 + 0.000 192 990 412 8;
  • 19) 0.000 192 990 412 8 × 2 = 0 + 0.000 385 980 825 6;
  • 20) 0.000 385 980 825 6 × 2 = 0 + 0.000 771 961 651 2;
  • 21) 0.000 771 961 651 2 × 2 = 0 + 0.001 543 923 302 4;
  • 22) 0.001 543 923 302 4 × 2 = 0 + 0.003 087 846 604 8;
  • 23) 0.003 087 846 604 8 × 2 = 0 + 0.006 175 693 209 6;
  • 24) 0.006 175 693 209 6 × 2 = 0 + 0.012 351 386 419 2;
  • 25) 0.012 351 386 419 2 × 2 = 0 + 0.024 702 772 838 4;
  • 26) 0.024 702 772 838 4 × 2 = 0 + 0.049 405 545 676 8;
  • 27) 0.049 405 545 676 8 × 2 = 0 + 0.098 811 091 353 6;
  • 28) 0.098 811 091 353 6 × 2 = 0 + 0.197 622 182 707 2;
  • 29) 0.197 622 182 707 2 × 2 = 0 + 0.395 244 365 414 4;
  • 30) 0.395 244 365 414 4 × 2 = 0 + 0.790 488 730 828 8;
  • 31) 0.790 488 730 828 8 × 2 = 1 + 0.580 977 461 657 6;
  • 32) 0.580 977 461 657 6 × 2 = 1 + 0.161 954 923 315 2;
  • 33) 0.161 954 923 315 2 × 2 = 0 + 0.323 909 846 630 4;
  • 34) 0.323 909 846 630 4 × 2 = 0 + 0.647 819 693 260 8;
  • 35) 0.647 819 693 260 8 × 2 = 1 + 0.295 639 386 521 6;
  • 36) 0.295 639 386 521 6 × 2 = 0 + 0.591 278 773 043 2;
  • 37) 0.591 278 773 043 2 × 2 = 1 + 0.182 557 546 086 4;
  • 38) 0.182 557 546 086 4 × 2 = 0 + 0.365 115 092 172 8;
  • 39) 0.365 115 092 172 8 × 2 = 0 + 0.730 230 184 345 6;
  • 40) 0.730 230 184 345 6 × 2 = 1 + 0.460 460 368 691 2;
  • 41) 0.460 460 368 691 2 × 2 = 0 + 0.920 920 737 382 4;
  • 42) 0.920 920 737 382 4 × 2 = 1 + 0.841 841 474 764 8;
  • 43) 0.841 841 474 764 8 × 2 = 1 + 0.683 682 949 529 6;
  • 44) 0.683 682 949 529 6 × 2 = 1 + 0.367 365 899 059 2;
  • 45) 0.367 365 899 059 2 × 2 = 0 + 0.734 731 798 118 4;
  • 46) 0.734 731 798 118 4 × 2 = 1 + 0.469 463 596 236 8;
  • 47) 0.469 463 596 236 8 × 2 = 0 + 0.938 927 192 473 6;
  • 48) 0.938 927 192 473 6 × 2 = 1 + 0.877 854 384 947 2;
  • 49) 0.877 854 384 947 2 × 2 = 1 + 0.755 708 769 894 4;
  • 50) 0.755 708 769 894 4 × 2 = 1 + 0.511 417 539 788 8;
  • 51) 0.511 417 539 788 8 × 2 = 1 + 0.022 835 079 577 6;
  • 52) 0.022 835 079 577 6 × 2 = 0 + 0.045 670 159 155 2;
  • 53) 0.045 670 159 155 2 × 2 = 0 + 0.091 340 318 310 4;
  • 54) 0.091 340 318 310 4 × 2 = 0 + 0.182 680 636 620 8;

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:


0.000 000 000 736 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1001 0111 0101 1110 00(2)

6. Positive number before normalization:

0.000 000 000 736 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1001 0111 0101 1110 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:


0.000 000 000 736 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1001 0111 0101 1110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1001 0111 0101 1110 00(2) × 20 =

1.1001 0100 1011 1010 1111 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):
1.1001 0100 1011 1010 1111 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. 100 1010 0101 1101 0111 1000 =

100 1010 0101 1101 0111 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) =
100 1010 0101 1101 0111 1000


Decimal number -0.000 000 000 736 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1010 0101 1101 0111 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