-0.000 000 000 741 16 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 16(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
-0.000 000 000 741 16(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:

|-0.000 000 000 741 16| = 0.000 000 000 741 16


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: 0.000 000 000 741 16.

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) 0.000 000 000 741 16 × 2 = 0 + 0.000 000 001 482 32;
  • 2) 0.000 000 001 482 32 × 2 = 0 + 0.000 000 002 964 64;
  • 3) 0.000 000 002 964 64 × 2 = 0 + 0.000 000 005 929 28;
  • 4) 0.000 000 005 929 28 × 2 = 0 + 0.000 000 011 858 56;
  • 5) 0.000 000 011 858 56 × 2 = 0 + 0.000 000 023 717 12;
  • 6) 0.000 000 023 717 12 × 2 = 0 + 0.000 000 047 434 24;
  • 7) 0.000 000 047 434 24 × 2 = 0 + 0.000 000 094 868 48;
  • 8) 0.000 000 094 868 48 × 2 = 0 + 0.000 000 189 736 96;
  • 9) 0.000 000 189 736 96 × 2 = 0 + 0.000 000 379 473 92;
  • 10) 0.000 000 379 473 92 × 2 = 0 + 0.000 000 758 947 84;
  • 11) 0.000 000 758 947 84 × 2 = 0 + 0.000 001 517 895 68;
  • 12) 0.000 001 517 895 68 × 2 = 0 + 0.000 003 035 791 36;
  • 13) 0.000 003 035 791 36 × 2 = 0 + 0.000 006 071 582 72;
  • 14) 0.000 006 071 582 72 × 2 = 0 + 0.000 012 143 165 44;
  • 15) 0.000 012 143 165 44 × 2 = 0 + 0.000 024 286 330 88;
  • 16) 0.000 024 286 330 88 × 2 = 0 + 0.000 048 572 661 76;
  • 17) 0.000 048 572 661 76 × 2 = 0 + 0.000 097 145 323 52;
  • 18) 0.000 097 145 323 52 × 2 = 0 + 0.000 194 290 647 04;
  • 19) 0.000 194 290 647 04 × 2 = 0 + 0.000 388 581 294 08;
  • 20) 0.000 388 581 294 08 × 2 = 0 + 0.000 777 162 588 16;
  • 21) 0.000 777 162 588 16 × 2 = 0 + 0.001 554 325 176 32;
  • 22) 0.001 554 325 176 32 × 2 = 0 + 0.003 108 650 352 64;
  • 23) 0.003 108 650 352 64 × 2 = 0 + 0.006 217 300 705 28;
  • 24) 0.006 217 300 705 28 × 2 = 0 + 0.012 434 601 410 56;
  • 25) 0.012 434 601 410 56 × 2 = 0 + 0.024 869 202 821 12;
  • 26) 0.024 869 202 821 12 × 2 = 0 + 0.049 738 405 642 24;
  • 27) 0.049 738 405 642 24 × 2 = 0 + 0.099 476 811 284 48;
  • 28) 0.099 476 811 284 48 × 2 = 0 + 0.198 953 622 568 96;
  • 29) 0.198 953 622 568 96 × 2 = 0 + 0.397 907 245 137 92;
  • 30) 0.397 907 245 137 92 × 2 = 0 + 0.795 814 490 275 84;
  • 31) 0.795 814 490 275 84 × 2 = 1 + 0.591 628 980 551 68;
  • 32) 0.591 628 980 551 68 × 2 = 1 + 0.183 257 961 103 36;
  • 33) 0.183 257 961 103 36 × 2 = 0 + 0.366 515 922 206 72;
  • 34) 0.366 515 922 206 72 × 2 = 0 + 0.733 031 844 413 44;
  • 35) 0.733 031 844 413 44 × 2 = 1 + 0.466 063 688 826 88;
  • 36) 0.466 063 688 826 88 × 2 = 0 + 0.932 127 377 653 76;
  • 37) 0.932 127 377 653 76 × 2 = 1 + 0.864 254 755 307 52;
  • 38) 0.864 254 755 307 52 × 2 = 1 + 0.728 509 510 615 04;
  • 39) 0.728 509 510 615 04 × 2 = 1 + 0.457 019 021 230 08;
  • 40) 0.457 019 021 230 08 × 2 = 0 + 0.914 038 042 460 16;
  • 41) 0.914 038 042 460 16 × 2 = 1 + 0.828 076 084 920 32;
  • 42) 0.828 076 084 920 32 × 2 = 1 + 0.656 152 169 840 64;
  • 43) 0.656 152 169 840 64 × 2 = 1 + 0.312 304 339 681 28;
  • 44) 0.312 304 339 681 28 × 2 = 0 + 0.624 608 679 362 56;
  • 45) 0.624 608 679 362 56 × 2 = 1 + 0.249 217 358 725 12;
  • 46) 0.249 217 358 725 12 × 2 = 0 + 0.498 434 717 450 24;
  • 47) 0.498 434 717 450 24 × 2 = 0 + 0.996 869 434 900 48;
  • 48) 0.996 869 434 900 48 × 2 = 1 + 0.993 738 869 800 96;
  • 49) 0.993 738 869 800 96 × 2 = 1 + 0.987 477 739 601 92;
  • 50) 0.987 477 739 601 92 × 2 = 1 + 0.974 955 479 203 84;
  • 51) 0.974 955 479 203 84 × 2 = 1 + 0.949 910 958 407 68;
  • 52) 0.949 910 958 407 68 × 2 = 1 + 0.899 821 916 815 36;
  • 53) 0.899 821 916 815 36 × 2 = 1 + 0.799 643 833 630 72;
  • 54) 0.799 643 833 630 72 × 2 = 1 + 0.599 287 667 261 44;

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:


0.000 000 000 741 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 1001 1111 11(2)

6. Positive number before normalization:

0.000 000 000 741 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 1001 1111 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:


0.000 000 000 741 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 1001 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 1110 1001 1111 11(2) × 20 =

1.1001 0111 0111 0100 1111 111(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 0111 0111 0100 1111 111


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 1011 1011 1010 0111 1111 =

100 1011 1011 1010 0111 1111


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 1011 1011 1010 0111 1111


Decimal number -0.000 000 000 741 16 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1011 1010 0111 1111

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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