-0.000 000 000 759 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 759 6(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
-0.000 000 000 759 6(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:

|-0.000 000 000 759 6| = 0.000 000 000 759 6


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: 0.000 000 000 759 6.

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) 0.000 000 000 759 6 × 2 = 0 + 0.000 000 001 519 2;
  • 2) 0.000 000 001 519 2 × 2 = 0 + 0.000 000 003 038 4;
  • 3) 0.000 000 003 038 4 × 2 = 0 + 0.000 000 006 076 8;
  • 4) 0.000 000 006 076 8 × 2 = 0 + 0.000 000 012 153 6;
  • 5) 0.000 000 012 153 6 × 2 = 0 + 0.000 000 024 307 2;
  • 6) 0.000 000 024 307 2 × 2 = 0 + 0.000 000 048 614 4;
  • 7) 0.000 000 048 614 4 × 2 = 0 + 0.000 000 097 228 8;
  • 8) 0.000 000 097 228 8 × 2 = 0 + 0.000 000 194 457 6;
  • 9) 0.000 000 194 457 6 × 2 = 0 + 0.000 000 388 915 2;
  • 10) 0.000 000 388 915 2 × 2 = 0 + 0.000 000 777 830 4;
  • 11) 0.000 000 777 830 4 × 2 = 0 + 0.000 001 555 660 8;
  • 12) 0.000 001 555 660 8 × 2 = 0 + 0.000 003 111 321 6;
  • 13) 0.000 003 111 321 6 × 2 = 0 + 0.000 006 222 643 2;
  • 14) 0.000 006 222 643 2 × 2 = 0 + 0.000 012 445 286 4;
  • 15) 0.000 012 445 286 4 × 2 = 0 + 0.000 024 890 572 8;
  • 16) 0.000 024 890 572 8 × 2 = 0 + 0.000 049 781 145 6;
  • 17) 0.000 049 781 145 6 × 2 = 0 + 0.000 099 562 291 2;
  • 18) 0.000 099 562 291 2 × 2 = 0 + 0.000 199 124 582 4;
  • 19) 0.000 199 124 582 4 × 2 = 0 + 0.000 398 249 164 8;
  • 20) 0.000 398 249 164 8 × 2 = 0 + 0.000 796 498 329 6;
  • 21) 0.000 796 498 329 6 × 2 = 0 + 0.001 592 996 659 2;
  • 22) 0.001 592 996 659 2 × 2 = 0 + 0.003 185 993 318 4;
  • 23) 0.003 185 993 318 4 × 2 = 0 + 0.006 371 986 636 8;
  • 24) 0.006 371 986 636 8 × 2 = 0 + 0.012 743 973 273 6;
  • 25) 0.012 743 973 273 6 × 2 = 0 + 0.025 487 946 547 2;
  • 26) 0.025 487 946 547 2 × 2 = 0 + 0.050 975 893 094 4;
  • 27) 0.050 975 893 094 4 × 2 = 0 + 0.101 951 786 188 8;
  • 28) 0.101 951 786 188 8 × 2 = 0 + 0.203 903 572 377 6;
  • 29) 0.203 903 572 377 6 × 2 = 0 + 0.407 807 144 755 2;
  • 30) 0.407 807 144 755 2 × 2 = 0 + 0.815 614 289 510 4;
  • 31) 0.815 614 289 510 4 × 2 = 1 + 0.631 228 579 020 8;
  • 32) 0.631 228 579 020 8 × 2 = 1 + 0.262 457 158 041 6;
  • 33) 0.262 457 158 041 6 × 2 = 0 + 0.524 914 316 083 2;
  • 34) 0.524 914 316 083 2 × 2 = 1 + 0.049 828 632 166 4;
  • 35) 0.049 828 632 166 4 × 2 = 0 + 0.099 657 264 332 8;
  • 36) 0.099 657 264 332 8 × 2 = 0 + 0.199 314 528 665 6;
  • 37) 0.199 314 528 665 6 × 2 = 0 + 0.398 629 057 331 2;
  • 38) 0.398 629 057 331 2 × 2 = 0 + 0.797 258 114 662 4;
  • 39) 0.797 258 114 662 4 × 2 = 1 + 0.594 516 229 324 8;
  • 40) 0.594 516 229 324 8 × 2 = 1 + 0.189 032 458 649 6;
  • 41) 0.189 032 458 649 6 × 2 = 0 + 0.378 064 917 299 2;
  • 42) 0.378 064 917 299 2 × 2 = 0 + 0.756 129 834 598 4;
  • 43) 0.756 129 834 598 4 × 2 = 1 + 0.512 259 669 196 8;
  • 44) 0.512 259 669 196 8 × 2 = 1 + 0.024 519 338 393 6;
  • 45) 0.024 519 338 393 6 × 2 = 0 + 0.049 038 676 787 2;
  • 46) 0.049 038 676 787 2 × 2 = 0 + 0.098 077 353 574 4;
  • 47) 0.098 077 353 574 4 × 2 = 0 + 0.196 154 707 148 8;
  • 48) 0.196 154 707 148 8 × 2 = 0 + 0.392 309 414 297 6;
  • 49) 0.392 309 414 297 6 × 2 = 0 + 0.784 618 828 595 2;
  • 50) 0.784 618 828 595 2 × 2 = 1 + 0.569 237 657 190 4;
  • 51) 0.569 237 657 190 4 × 2 = 1 + 0.138 475 314 380 8;
  • 52) 0.138 475 314 380 8 × 2 = 0 + 0.276 950 628 761 6;
  • 53) 0.276 950 628 761 6 × 2 = 0 + 0.553 901 257 523 2;
  • 54) 0.553 901 257 523 2 × 2 = 1 + 0.107 802 515 046 4;

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:


0.000 000 000 759 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0011 0011 0000 0110 01(2)

6. Positive number before normalization:

0.000 000 000 759 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0011 0011 0000 0110 01(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:


0.000 000 000 759 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0011 0011 0000 0110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0100 0011 0011 0000 0110 01(2) × 20 =

1.1010 0001 1001 1000 0011 001(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):
1.1010 0001 1001 1000 0011 001


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. 101 0000 1100 1100 0001 1001 =

101 0000 1100 1100 0001 1001


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) =
101 0000 1100 1100 0001 1001


Decimal number -0.000 000 000 759 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 101 0000 1100 1100 0001 1001

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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